# User:Stockequation/sandbox

## Introduction to Chart to Scalar Theory

Chart to Scalar Theory says there is a correspondence or relation between discrete buy and sell orders in the stock market volume and stock market geometry on the boundary. The boundary represents the charts which we can see. If you peel it back, there are buy and sell orders which are contained in the volume. The idea of the scalar is to encode the properties of the stock market into a scalar which represents a constraint of a hypothetical stock chart constrained by boundary conditions.

The motivation behind Chart to Scalar is to convert stock charts and trading volume into a mathematical framework to better understand how the stock market works, and make predictions. This has applications such as measuring capital inflows, market impact, simulating how external events such as insider trading affect how a stock will trade in the expectation that the market is flat or rising, explaining how crashes occur, and pricing options.

The name Chart to Scalar arises from a variational problem that involves finding a hypothetical RHS (right hand side) chart or function ${\displaystyle f_{r}}$ that represents the path of a stock constrained by a scalar value ${\displaystyle d}$ that encodes information about the LHS (left hand side) chart, volume, geometry and other characteristics of the charts and external events, in addition to the constraints of the boundary conditions and endpoints.

${\displaystyle d=\int _{x_{1}}^{x_{2}}f_{r}(x)\,dx}$ where ${\displaystyle f_{r}}$ is a function of minimum length

There are many models that try to explain how the markets work from a wide variety of research areas, such as behavior economics, behavior finance, fundamental valuation, but there are fewer papers that envisage the stock market itself as a closed physical system. Dilip Abreu and Markus K. Brunnermeier(2003) discusses in the context of behavior finance how presence of rational arbitrageurs can create persistent bubbles. A paper by Bouchaud (1998) posits a non linear Langevin equation as a model for stock market fluctuations and crashes. Racorean (2014) proposes a model that encapsulates all of the trading activity of a group of stocks as a high dimensional polygon. C Zhang (2010) appropriates classical physics establish the Schrödinger equation for a stock price. Hsinan Hsu (2001) applies the kinematic and kinetic theories of physics to derive price behavior equations for the stock markets.

In Chart to Scalar, the stock market is simplified to 1-dimensional 'universe' with two particles; a buy and a sell order and the action of the system is restricted in such a way where along the x axis there is only allowed to be one y intercept (obviously because stocks cannot go 'back' in time). The variational equation outputs either a monotonically increasing/decreasing concave, linear, or convex RHS (right hand side) curve ${\displaystyle f_{r}}$ of how the stock should theoretically trade given the values in the equilibrium equation ${\displaystyle \Lambda _{L}=\Lambda _{R}}$. If ${\displaystyle \eta _{l}=1}$, the equilibrium equation can be interpreted to mean that the amount of money flowing into a stock on the LHS must equal the amount that flows out on the RHS. Or if ${\displaystyle \eta _{l}=-1}$, the amount of money that flows into the stock on the RHS must equal the amount that flowed out on the LHS.

Rather than looking at forces, we reduce the market to static scalars that contain information about the market. The equation of motion is the path of least action that encloses the area of the scalar. This is a Lagrangian approach instead of a Newtonian one. Second, we introduce LHS 'resistance' which is encoded into the scalar. This resistance is analogous to the resistance of gravity or friction in Newtonian mechanics or a energy barrier in quantum mechanics.

The first section covers the notation, the formulas that arise out of the buy/sell order annihilation process, and some examples. Then the model is extended to include a geometric component to provide an additional degree of freedom when needed. The second part will cover simulations and data extraction. In the third part of the paper an option ricing formula is derived via Chart to Scalar and the results are compared to Black Scholes. The inclusion of volume and trade size could explain phenomena observed in real live stocks that isn't accounted for in conventional option pricing models. This is important because traditional models (such as Black-Scholes and Binomial) have many significant limitations which Chart to Scalar may be able to resolve, such as the inability to account for historical market movements (Baggett & Thompson (2006))[1] and their frequent overpricing of options, with the overpricing increasing with the time to maturity. (Hull & White (1987)) [2] In our paper, we conclude that Chart to Scalar options grow at a slower rate ${\displaystyle t^{1/4}}$ than Black Scholes options ${\displaystyle t^{1/2}}$, and secondly, that rare events according to Black-Scholes are more common in Chart to Scalar.

### Key Points

1. The tendency of a stock to rise or fall is determined by it's volume of the RHS (right hand side) of the stock graph functional relative to the volume of LHS (left hand side) functional, in addition to other variables in the stock market equilibrium equation.

2. Buy and sell orders and their interaction are the propagators of price change.

3. In a restricted 1-d trading space, stocks trade along a geodesic, swathing out convex, linear, or concave curves as the solution of a variational equation constrained by the equilibrium equation.

4. Most equations are scale invariant meaning that adjusting time frames by a scaling factor will yield the same results.

The trading space is a quasi two-dimensional space that consists of two charts. We call it quasi-two dimensional because backward motion is prohibited and hence all curves cannot have more than one y intercept. One chart is on LHS ${\displaystyle f_{l}}$ (left-hand side) and the second on the RHS (right-hand side) ${\displaystyle f_{r}}$ and are adjoined at some point, denoted as ${\displaystyle (x_{1},p_{1})}$. The resistance and support forces denoted by ${\displaystyle \Lambda _{L}}$ occupies the LHS. A horizontal line drawn through ${\displaystyle (x_{2},p_{2})}$ will intersect the LHS curve at ${\displaystyle (x_{a},p_{2})}$ where ${\displaystyle x_{a}=f_{l}^{-1}(p_{2})}$.

The curve ${\displaystyle f_{l}}$ is either concave, convex or a line. Depending on ${\displaystyle \Lambda _{L}}$ and additional variables like buy or sell orders, a ${\displaystyle f_{r}}$ RHS curve is produced, which is either concave, convex or linear.

On the LHS, ${\displaystyle f_{l}}$ for either a convex , concave curve or line is signed either + 1 or -1 denoted by ${\displaystyle \eta _{l}}$ . A positive sign means that the price on the LHS is rising. This is a support curve. A resistance curve has a negative sign. The RHS always has the opposite sign as the LHS or ${\displaystyle -\eta _{l}=\eta _{r}}$. If the RHS is rising it will run up against the negatively signed LHS resistance curve. If the RHS is falling it will fall against the positively signed LHS support curve, as shown in the appended diagram. (the shaded region represents volume)

Negative values are not permitted

${\displaystyle 0\leq f_{r}(x)}$ for all ${\displaystyle x}$

${\displaystyle 0\leq f_{l}(x)}$ for all ${\displaystyle x}$

${\displaystyle x_{0}\leq x_{a}

${\displaystyle f_{r}(x_{2})=p_{2}=f_{l}(x_{a})}$

${\displaystyle f_{l}(x_{1})=p_{1}=f_{r}(x_{1})}$

For ${\displaystyle \eta _{l}=-1}$:

${\displaystyle 0

For ${\displaystyle \eta _{l}=1}$ the parameters are reversed:

${\displaystyle p_{1}>p_{2}>0}$

Imagine a diagonal line that connects the points ${\displaystyle (x_{a},p_{2}),(x_{1},p_{1})}$ for a LHS curve ${\displaystyle f_{l}}$ and ${\displaystyle (x_{1},p_{1}),(x_{2},p_{2})}$ for a RHS curve ${\displaystyle f_{r}}$. For either LHS or RHS, a concave curve cannot exceed this line and convex curve cannot fall below it.

With the new notation, ${\displaystyle |\ln(p_{2}/p_{1})|}$ can be re-written as ${\displaystyle \eta _{r}\ln(p_{2}/p_{1})}$. As a rule of thumb, ${\displaystyle \eta _{r}\ln(p_{2}/p_{1})}$ must always be positive, for any problem. ${\displaystyle p_{2},p_{1}}$ must be chosen in such a way that this holds.

### Annihilation Process

In Chart to Scalar theory, the interaction of buy orders ${\displaystyle b_{o}}$and sell orders ${\displaystyle s_{o}}$ is the propagator is price change. This is expressed by the simple system of equations:

${\displaystyle {\begin{cases}b_{o}+s_{o}=v_{r}/a_{v}\\b_{o}-s_{o}=v(p_{2})/a_{v}\end{cases}}}$

Where ${\displaystyle v(p_{2})}$ is some function inputs price ${\displaystyle p_{2}}$ and returns volume. And ${\displaystyle v_{r}}$ is RHS volume and ${\displaystyle a_{v}}$ is the average order size.

The challenge is finding ${\displaystyle v(p_{2})}$. In the subsequent sections, we'll try to derive a relationship between price and volume using simple geometric shapes such as triangles, lines, and arcs to represent the paths of stocks. An alternative derivation of ${\displaystyle f(p,v)}$ using an order book can be found here: User:Optionpricing

As a stock is trading in some time frame there exists a value called an energy level that is the product of two or three components: a slope/time component ${\displaystyle w_{l,r}}$, a volume ${\displaystyle v_{l,r}}$ component and geometric component ${\displaystyle \phi _{l,r}}$, which is added for more complex problems and will be discussed in further detail later. The LHS (left-hand side) values are denoted by subscript ${\displaystyle l}$ such as ${\displaystyle v_{l},w_{l}}$ and RHS by ${\displaystyle r}$ such as ${\displaystyle v_{r},w_{r}}$. The RHS product is ${\displaystyle w_{r}v_{r}}$; the LHS is ${\displaystyle w_{l}v_{l}}$. Setting them equal, we have the equilibrium equation: ${\displaystyle w_{l}v_{l}=w_{r}v_{r}}$.

In this trading space a buy and sell order of equal price and quantity will instantly annihilate producing no net energy (represented as a flat line on a chart). An excess of buy orders will cause a price rise appearing as a rising slope on a chart or ${\displaystyle dp/dt}$. However, the actual price of the stock need not matter. Instead, the ratio of buy orders to total orders does. Thus a stock trading at 10 cents will experience the same percentage change in price as one trading at $10 if the ratios of buy and sell orders are the same. We're looking for a function where ${\displaystyle f(x)d/dx=f(nx)d/dx}$ where ${\displaystyle n}$ is a scaling factor; the solution is ${\displaystyle \ln(x)}$. All else (volume, time) being equal, a stock rising from$90 to $100 should have the same properties as one rising from$9 to $10. We also need some way to define the x component of a slope in a manner that is time or scale invariant. Define the variable ${\displaystyle z}$ that relate the time duration of events on the LHS to the RHS. The actual units of time are not important. For the LHS, the default time is ${\displaystyle 1}$. This serves as our reference point. For example, if ${\displaystyle z=1}$ then it means the time or duration of the LHS of the chart is the same as the RHS. We define ${\displaystyle w_{r,l}}$ as a function that ranges between 0 and 1 and measures the percentage of volume 'buys' or 'sells' remaining after buy and sell order annihilation has occurred, and is dependent on the rate of change of a stock in some time frame or ${\displaystyle m}$. ${\displaystyle w_{r,l}}$ can be interpreted as some function of ${\displaystyle m}$ or ${\displaystyle f(m)}$. For the RHS, we need to define ${\displaystyle m}$ both in terms of the ratio buy & sell orders and the rate of change. In terms of rate of change of price over the 'base' or time duration (where ${\displaystyle \Delta _{p}=\eta _{r}\ln(p_{2}/p_{1})}$) and ${\displaystyle \Delta _{b}=z}$: ${\displaystyle m={\frac {\Delta _{p}}{\Delta _{b}}}}$ and in terms of buy and sell orders: ${\displaystyle m=\left({\frac {b_{o}}{s_{o}}}-1\right)}$ Intuitively this makes sense. If ${\displaystyle b_{o}=s_{o}}$ then ${\displaystyle m=0}$ or no change in price. If ${\displaystyle b_{o}}$ vastly exceeds ${\displaystyle s_{o}}$ then ${\displaystyle m}$ increases. Because the total number of buy and sell orders ${\displaystyle t_{o}}$ is ${\displaystyle b_{o}+s_{o}}$ then ${\displaystyle s_{o}=t_{o}-b_{o}}$. Thus ${\displaystyle m}$ can be rewritten as: ${\displaystyle m=\left({\frac {2b_{o}-t_{o}}{t_{o}-b_{o}}}\right)}$ We're looking for a function ${\displaystyle f(m)}$ that converges to 1 as the ratio of ${\displaystyle b_{o}/s_{o}}$ increases, and 0 if ${\displaystyle b_{o}=s_{o}}$. The example below suffices: ${\displaystyle f(m)={\frac {2b_{o}-t_{o}}{t_{o}}}}$ Define ${\displaystyle t_{o}=v_{r}/a_{v}}$ The total number of orders is equal to the volume divided by the average order size. ${\displaystyle b_{o}}$ and ${\displaystyle s_{o}}$ are recovered from the equation above: ${\displaystyle b_{o}={\frac {v_{r}}{2a_{v}}}\left(1+f(m)\right)}$ ${\displaystyle s_{o}={\frac {v_{r}}{2a_{v}}}\left(1-f(m)\right)}$ Consider there are 5 total orders: three buys and two sells in some infinitesimal trading space. The two buys and sells will cancel and one buy order will be left. Hence, ${\displaystyle f(m)=1/5}$. If ${\displaystyle b_{o}=t_{o}/2}$ then all the buys and sells have cancelled and there is none left over, ${\displaystyle f(m)=0}$. If ${\displaystyle b_{o}=t_{o}}$ then ${\displaystyle f(m)=1}$ because all of the volume is buy orders. We can convert charts into buy and sell orders and vice versa through the relation: ${\displaystyle \left({\frac {2b_{o}-t_{o}}{t_{o}-b_{o}}}\right)={\frac {\Delta _{p}}{\Delta _{b}}}}$ The solution: ${\displaystyle b_{o}=\Delta _{p}+\Delta _{b}}$ ${\displaystyle t_{o}=\Delta _{p}+2\Delta _{b}}$ Putting these values back into ${\displaystyle f(m)}$ and ${\displaystyle w_{r}}$ and ${\displaystyle w_{l}}$ are obtained. ${\displaystyle w_{r}={\frac {\eta _{r}\ln(p_{2}/p_{1})}{\eta _{r}\ln(p_{2}/p_{1})+2z}}}$ ${\displaystyle w_{l}}$ is our time/slope reference triangle: ${\displaystyle w_{l}={\frac {\eta _{r}\ln(p_{2}/p_{1})}{\eta _{r}\ln(p_{2}/p_{1})+2}}}$ The buy and sell order equations: ${\displaystyle b_{o}={\frac {v_{r}}{2a_{v}}}\left(1-\eta _{l}w_{r}\right)}$ ${\displaystyle s_{o}={\frac {v_{r}}{2a_{v}}}\left(1+\eta _{l}w_{r}\right)}$ ${\displaystyle \eta _{l}}$ is added to account for negative RHS slopes, so to preserve symmetry between the buys and sells. ### Example 1: Suppose a stock rises from$50 to $51 in a one day period with a volume of ${\displaystyle 10^{7}}$. How much volume would be required for the stock to fall back to$50 instantaneously? ${\displaystyle a_{v}}$ is irrelevant.

The time duration of events on the RHS is infinitely small relative to the LHS. Because ${\displaystyle \eta _{r}=-1}$ and ${\displaystyle \eta _{r}\ln(p_{2}/p_{1})>0}$, we have ${\displaystyle p_{2}=50,p_{1}=51}$

Setting up the equilibrium equation ${\displaystyle v_{l}w_{l}=v_{r}w_{r}}$

${\displaystyle {\frac {-10^{7}\times \ln(50/51)}{-\ln(50/51)+2}}={\frac {-v_{r}\times \ln(50/51)}{-\ln(50/51)+2\times 1/\infty }}}$

The solution ${\displaystyle v_{r}=98,042}$ shares need to be sold.

Part 2: How many buy and sell orders?

With ${\displaystyle \eta _{l}=1}$ (the LHS is rising), when we plug ${\displaystyle w_{r}=1}$ into the buy order equation we get ${\displaystyle b_{o}={\frac {98,042(1-w_{r})}{2a_{v}}}=0}$. This answer is interpreted to mean that there are zero buy orders as the stock falls from $51 to$50 in an infinitesimal time frame. ${\displaystyle s_{o}={\frac {98,042(1+w_{r})}{2a_{v}}}={\frac {98,042}{a_{v}}}}$. Thus we have 100% sell orders for the ${\displaystyle v_{r}}$ volume, for any ${\displaystyle a_{v}}$.

### Example 2:

A stock has risen in from $10 to$11 in 30 trading days with an arbitrary daily volume. How much volume is requires for the stock to fall back to $10 in one trading day? ${\displaystyle {\frac {30}{-\ln(10/11)+2}}={\frac {v_{r}}{-\ln(10/11)+2\times 1/30}}}$ The solution is 2.3 which means that it requires 2.3 times the daily volume for the stock to fall back to$10 in a one day trading period. This agrees with empirical observations of how stocks and indexes can erase months or weeks worth of gains in single day or a couple days during a crash. An increase of 130% of the daily volume or 7.8% of the total LHS volume is sufficient for the stock to forfeit a month worth of gains.

Part 2: How many buys and sell orders?

When we plug ${\displaystyle w_{r}=.59}$ into the buy order equation we get ${\displaystyle b_{o}={\frac {2.3(.41)}{2a_{v}}}}$ This answer is interpreted to mean that 20.5% of the RHS volume consists of buy orders and 79.5% sell orders, for any ${\displaystyle a_{v}}$. Also note how ${\displaystyle b_{o}+s_{o}=t_{o}={\frac {2.3(.41)}{2a_{v}}}+{\frac {2.3(1+.59)}{2a_{v}}}={\frac {2.3}{a_{v}}}}$

This price/volume dynamic could explain why flash crashes occur and how preventing them would be difficult because a relatively little amount of high density volume is required for a stock experience a rapid decline.

## Geometry

### Motivation

For certain problems the linearized versions of ${\displaystyle \Lambda _{L}}$ and ${\displaystyle \Lambda _{R}}$ doesn't give enough degrees of freedom, hence a geometric component ${\displaystyle \phi _{l,r}}$ is introduced. ${\displaystyle \phi _{l,r}}$ measures efficiency of a path ${\displaystyle f_{l,r}}$. ${\displaystyle \phi }$ is a value between 0 and 1 that measures how much a stock path defects from a strait line, with 1 being a line. A line is the most efficient path and has a ${\displaystyle \phi =1}$, but in some other circumstances such as modeling insider selling then a different path is stipulated that may be convex or concave shaped. To see why an extra variable is needed consider we have our buy order equation with ${\displaystyle \eta _{l}=1}$

Consider the equilibrium equation ${\displaystyle w_{l}v_{l}=w_{r}v_{r}}$

If we wish to simulate churning ${\displaystyle v_{r}=v_{r}+c}$ we have: ${\displaystyle w_{l}v_{l}=(v_{r}+c)w_{r}}$

(Churning volume is volume that is evenly split between buy and sell orders)

Obviously, the equilibrium equation doesn't hold, but introducing a new variable ${\displaystyle \phi _{r}}$ bypasses this.

We have: ${\displaystyle \phi _{r}={\frac {w_{l}v_{l}}{w_{r}(v_{r}+c)}}}$

As ${\displaystyle c}$ grows the path ${\displaystyle f_{r}}$ becomes less efficient as indicated by falling ${\displaystyle \phi _{r}}$. By defining ${\displaystyle \phi _{r}}$ to be a functional path, we can give the output as a solution of a variational equation as the path of least action as constrained by the equilibrium equation. This path may not be linear, but concave or convex depending on our criteria. The path ${\displaystyle f_{r}}$ is how the stock should hypothetically trade on the RHS with the introduction of constraints, such as extra buy or sell orders or churning volume.

The buy & sell order equations include ${\displaystyle \phi _{r}}$

${\displaystyle b_{o}={\frac {v_{r}+c}{2a_{v}}}\left(1-\phi _{r}w_{r}\right)}$

As ${\displaystyle c\to \infty }$, then ${\displaystyle \phi _{r}\to 0}$ and ${\displaystyle b_{o}\to (v_{r}+c)/(2a_{v})}$. As more churning volume is added, the number of buy orders ${\displaystyle b_{o}}$ approaches half the total volume ${\displaystyle v_{r}+c}$. Either an inefficient path or a small ${\displaystyle w_{r}}$ results in an even admixture of buy and sell orders.

### Defining ${\displaystyle \phi }$

Define ${\displaystyle \phi _{l,r}}$ as a function that ranges between 0 and 1 depending on how much the area under the curve ${\displaystyle f_{l}(x)}$ or ${\displaystyle f_{r}(x)}$ differs from the area under the right-triangle adjoining the points ${\displaystyle (x_{a},p_{2}),(x_{1},p_{1})}$ (LHS) or ${\displaystyle (x_{1},p_{1}),(x_{2},p_{2})}$ (RHS), and approaches zero for increasing deviation or 1 for a diagonal line.

Concave and convex curves and lines arise out of solutions to the variational equation when constrained by the equilibrium equation. The linear path is the most optimal and therefore has the highest 'energy density'

An expression of ${\displaystyle \phi _{r}}$ or ${\displaystyle \phi _{l}}$ for concave curves:

${\displaystyle \phi _{r\ concave}={\frac {2\int _{x_{1}}^{x_{2}}f_{r}(x)\,dx+(x_{1}-x_{2})(p_{2}(1+\eta _{l})+p_{1}(1-\eta _{l}))}{\eta _{l}(p_{1}-p_{2})(x_{2}-x_{1})}}}$

${\displaystyle \phi _{l\ concave}={\frac {2\int _{x_{a}}^{x_{1}}f_{l}(x)\,dx+(x_{a}-x_{1})(p_{2}(1+\eta _{l})+p_{1}(1-\eta _{l}))}{\eta _{l}(p_{1}-p_{2})(x_{1}-x_{a})}}}$

The convex curves:

${\displaystyle \phi _{r\ convex}=2-{\frac {2\int _{x_{1}}^{x_{2}}f_{rc}(x)\,dx+(x_{1}-x_{2})(p_{2}(1+\eta _{l})+p_{1}(1-\eta _{l}))}{\eta _{l}(p_{1}-p_{2})(x_{2}-x_{1})}}}$

${\displaystyle \phi _{l\ convex}=2-{\frac {2\int _{x_{a}}^{x_{1}}f_{lc}(x)\,dx+(x_{a}-x_{1})(p_{2}(1+\eta _{l})+p_{1}(1-\eta _{l}))}{\eta _{l}(p_{1}-p_{2})(x_{1}-x_{a})}}}$

where ${\displaystyle f_{rc},f_{lc}}$ are convex curves

If ${\displaystyle f_{l}(x),f_{r}(x)}$ are expressed as a column vector, the complementary convex curves are found by using a transformation matrix to reflect ${\displaystyle f_{l}(x),f_{r}(x)}$ about the line adjoining ${\displaystyle (x_{a},p_{2}),(x_{1},p_{1})}$ for the LHS and ${\displaystyle (x_{1},p_{1}),(x_{2},p_{2})}$ for the RHS. These new curves have the same ${\displaystyle \phi _{l,r}}$ as the original curves about the endpoints of the region.

For example, let's assume our RHS space is a square with the coordinates ${\displaystyle (0,0),(0,1),(1,0),(1,1)}$ and we have a concave curve ${\displaystyle f_{r}=x^{2}}$. We have: ${\displaystyle \phi _{r\ concave}=2/3}$ and for ${\displaystyle f_{rc}=x^{1/2}}$ we also have ${\displaystyle \phi _{r\ convex}=2/3}$

The following table determines is you use a convex or a concave curve

Convex LHS ${\displaystyle \eta _{l}=-1}$ ${\displaystyle \to }$ Convex RHS ${\displaystyle \eta _{r}=1}$

Convex LHS ${\displaystyle \eta _{l}=1}$ ${\displaystyle \to }$ Convex RHS ${\displaystyle \eta _{r}=-1}$

Concave LHS ${\displaystyle \eta _{l}=-1}$ ${\displaystyle \to }$ Concave RHS ${\displaystyle \eta _{r}=1}$

Concave LHS ${\displaystyle \eta _{l}=1}$ ${\displaystyle \to }$ Concave RHS ${\displaystyle \eta _{r}=-1}$

### Invariance of ${\displaystyle \phi _{l},\phi _{r}}$

Crucially, ${\displaystyle \phi _{l,r}}$ exhibits scale invariance. Imagine that a chart occupying a computer screen is re-sized in such a way that the curves ${\displaystyle f_{l}}$ or ${\displaystyle f_{r}}$ appear stretched, compressed, or shifted. Another person on a different computer terminal looks at the same stock with a different time frame. Both should arrive at the same value of ${\displaystyle \phi _{l,r}}$.

Begin with the transformation:

${\displaystyle f_{r}(x)\to f_{r}(a\times x)}$

${\displaystyle x_{1}\to x_{1}/a}$

${\displaystyle x_{2}\to x_{2}/a}$

The limits of integration are compressed and ${\displaystyle p_{1}}$ and ${\displaystyle p_{2}}$ are the same. Define ${\displaystyle u=ax}$. Via integration by substitution obtain:

${\displaystyle {\frac {\int _{x_{1}}^{x_{2}}f_{r}(u)\,du}{a}}}$

Plugging in the transformed values of ${\displaystyle x_{1}}$ and ${\displaystyle x_{2}}$ into 1.0 the 'a' variable cancels out and we're left with the original equation.

For example, consider a concave RHS chart given by the potion of the equation ${\displaystyle y=x^{2}+c}$where ${\displaystyle 0\leq x\leq 1}$. We know ${\displaystyle \eta _{l}=-1}$ because the RHS is rising. Also: ${\displaystyle f_{r}=x^{2}+c}$, ${\displaystyle x_{1}=0}$, ${\displaystyle x_{2}=1}$, ${\displaystyle p_{1}=c}$, ${\displaystyle p_{2}=1+c}$. Plugging everything into 1.0 gives ${\displaystyle \phi _{r}=2/3}$

Now consider the the person viewing the graph wants a longer time frame on the same screen. ${\displaystyle f_{r}}$ will be compressed on the x axis by some factor ${\displaystyle a}$. Substituting ${\displaystyle x=>ax}$ gives ${\displaystyle a^{2}x^{2}+c}$. The transformed values of ${\displaystyle x_{1},x_{2}}$ are now ${\displaystyle 0,1/a}$ because the time frame is compressed. As expected ${\displaystyle p_{1}}$ and ${\displaystyle p_{2}}$ are unchanged because all we're doing is making the graph compressed, not changing the starting and final price. Plugging these new values into 1.0 still gives ${\displaystyle \phi _{r}=2/3}$.

### Invariance of Volume

Let's assume ${\displaystyle f_{l}}$ is bounded between ${\displaystyle x_{0}}$ and ${\displaystyle x_{1}}$ where ${\displaystyle x_{1}>x_{0}}$. There also exists a LHS volume function ${\displaystyle v_{l}(x)}$ between the aforementioned boundaries. For simplicity, ${\displaystyle v_{l}(x)}$ will be linear. The total volume between ${\displaystyle x_{0}}$ and ${\displaystyle x_{1}}$ will be denoted as ${\displaystyle v_{0}}$. Hence, we can express ${\displaystyle v_{l}}$ as a proportional relationship ${\displaystyle v_{l}(x_{a})=v_{0}(x_{1}-x_{a})/(x_{1}-x_{0})}$.

Suppose 'Tom' on computer 'A' observes that stock XYZ has carved a convex chart shape by ${\displaystyle -3x^{2}+6x+10}$ bounded by ${\displaystyle (0,10),(1,10),(1,13)}$. ${\displaystyle x_{1}-x_{0}=1}$ 'Mark' observes this this shape as well but his chart is stretched/resized on his computer and he gets ${\displaystyle -x^{2}/3+2x+10}$ bounded by ${\displaystyle (0,10),(3,10),(3,13)}$. ${\displaystyle x_{1}-x_{0}=3}$. For example, assuming ${\displaystyle p_{2}=12}$, they will both arrive at:

${\displaystyle \phi _{l}(x_{a})v_{l}(x_{a})={\frac {2v_{0}{\sqrt {3}}}{9}}}$

Or for general volume ${\displaystyle v(x)}$

${\displaystyle v_{0}=(x_{1}-x_{0})^{-1}\int _{x_{0}}^{x_{1}}v(x)\,dx}$

${\displaystyle v_{l}(x_{a})=(x_{1}-x_{0})^{-1}\int _{x_{a}}^{x_{1}}v(x)\,dx=v_{0}-(x_{1}-x_{0})^{-1}\int _{x_{0}}^{x_{a}}v(x)\,dx}$

### Variational equation for ${\displaystyle f_{r}}$

With ${\displaystyle \phi }$, we have the complete equilibrium equation:

${\displaystyle \Lambda _{l}=\Lambda _{R}}$ or ${\displaystyle v_{l}w_{l}\phi _{l}=v_{r}w_{r}\phi _{r}}$

The geodesic is the path ${\displaystyle f_{r}}$ bounded between the endpoints ${\displaystyle (x_{1},p_{1})}$ and ${\displaystyle (x_{2},p_{2})}$ that has the shortest length and satisfies the equilibrium equation.

This is a Isoperimetric problem where L is minimized while constrained by ${\displaystyle \phi _{r}}$ and ${\displaystyle f_{r}}$ , bounded by ${\displaystyle x=x_{1}}$, ${\displaystyle y=p_{1}}$, ${\displaystyle y=p_{2}}$, and ${\displaystyle x=x_{2}}$:

Variational equation:

${\displaystyle L=|x_{2}-x_{1}|+|p_{2}-p_{1}|+\int _{x_{1}}^{x_{2}}{\sqrt {1+[f_{r}'(x)]^{2}}}\,dx}$

Constrained by the equilibrium equation and appropriate boundary conditions

${\displaystyle {\frac {\Lambda _{l}}{v_{r}w_{r}}}=\phi _{r}}$

This can be solved using the Euler Formula and Lagrange multipliers, although the boundary conditions makes the problem difficult and closed form solutions often don't exist.

### Examples computing ${\displaystyle \phi _{l}(x_{a})v_{l}(x_{a})}$

The formulas below assume LHS volume does not vary ${\displaystyle v(x)d/dx=0}$

For curved ${\displaystyle f_{l}}$, computing ${\displaystyle \phi _{l}(x_{a})v_{l}(x_{a})}$ can be difficult, but there are some functions such as quadratics and ramps that are easier. Consider ${\displaystyle f_{l}}$ with ${\displaystyle \eta _{l}=1}$ with the boundary ${\displaystyle (0,p_{3}),(0,p_{1}),(1,p_{1}),(1,p_{3})}$, ${\displaystyle \eta _{l}=1}$ and ${\displaystyle p_{1}>p_{2}>p_{3}}$.

For quadratics ${\displaystyle f_{l}}$ of the form:

${\displaystyle f_{l}(x)=(p_{3}-p_{1})x^{2}-2(p_{3}-p_{1})x+p_{3};\eta _{l}=1}$

Upon simplification we have: ${\displaystyle \phi _{l}(x_{a})v_{l}(x_{a})={\frac {2v_{0}}{3}}{\sqrt {p_{1}-p_{2} \over p_{1}-p_{3}}}}$

Or the concave curve: ${\displaystyle f_{l}(x)=(p_{1}-p_{3})x^{2}+p_{3}}$

After some labor: ${\displaystyle \phi _{l}(x_{a})v_{l}(x_{a})={\frac {2v_{0}}{3(p_{1}-p_{2})}}\left(p_{1}-3p_{2}+2p_{3}+2(p_{2}-p_{3}){\sqrt {p_{2}-p_{3} \over p_{1}-p_{3}}}\right)}$

Consider we have a LHS ramp function connected by the coordinates ${\displaystyle (0,p_{3}),(j,p_{1}),(1,p_{1})}$ where ${\displaystyle 0\leq j\leq 1}$

We have: ${\displaystyle \phi _{l}(x_{a})={\frac {p_{1}-p_{2}}{p_{1}-p_{3}-j(p_{2}-p_{3})}}}$ , ${\displaystyle v(x_{a})=(1-j)v_{0}+j{\frac {(p_{1}-p_{2})v_{l}}{p_{1}-p_{3}}}}$ and

${\displaystyle \phi _{l}(x_{a})v_{l}(x_{a})={\frac {jv_{0}(p_{1}-p_{2})}{p_{1}-p_{3}}}}$

## Flat Market Simulations

### Derivation

In this section, Chart to Scalar will be applied to examples where the expectation is that the market will be flat on the RHS an we wish to simulate how changing or adding variables will affect the hypothetical RHS chart. In a flat market the expected number number of buy & sells on the RHS is equal. Because ${\displaystyle b_{o}=v_{r}/(2a_{v})}$ we have a particularly simple way of writing the sell order equation as ${\displaystyle 0=s_{o}-\Lambda _{L}}$. To derive this, begin with the equilibrium and buy order equation:

${\displaystyle \Lambda _{l}=\Lambda _{R}}$ or ${\displaystyle v_{l}w_{l}\phi _{l}=v_{r}w_{r}\phi _{r}}$

${\displaystyle b_{o}={\frac {v_{r}}{2a_{v}}}\left(1-\eta _{l}\phi _{r}w_{r}\right)}$

Note: to simulate a stock sale in flat market we have ${\displaystyle \eta _{l}=1}$. The LHS resistor is positively sloped.

The information on the LHS ${\displaystyle \Lambda _{l}}$ is known ahead of time, but the RHS information is unknown aside from ${\displaystyle v_{r}}$ which is evenly split between buy and sell orders (because the market is anticipated to be flat). To make the buy order equation udeful for simulation purposes, we can replace tho unknown RHS variables with the known LHS ones by rearranging the equilibrium equation:

${\displaystyle w_{r}\phi _{r}={\frac {\Lambda _{l}}{v_{r}}}}$

Plugging this into the buy order equation:

${\displaystyle b_{o}={\frac {v_{r}-\eta _{l}\ \Lambda _{l}}{2a_{v}}}}$

As we showed in a previous section, ${\displaystyle \Lambda _{l}}$ is a function of ${\displaystyle p_{2}}$, giving us a relationship between buy orders and price. Written out completely, we have:

${\displaystyle b_{o}={\frac {1}{2a_{v}}}\left(v_{r}+{\frac {\phi _{l}(x_{a})v_{l}(x_{a})\ln(p_{2}/p_{1})}{\eta _{r}\ln(p_{2}/p_{1})+2}}\right)}$

And the complimentary sell order equation:

${\displaystyle s_{o}={\frac {1}{2a_{v}}}\left(v_{r}-{\frac {\phi _{l}(x_{a})v_{l}(x_{a})\ln(p_{2}/p_{1})}{\eta _{r}\ln(p_{2}/p_{1})+2}}\right)}$

Where ${\displaystyle 0\leq b_{o}\leq v_{r}/a_{v}}$ and ${\displaystyle b_{o}+s_{o}=v_{r}/a_{v}}$

In the flat market we know ${\displaystyle b_{o}=v_{r}/(2a_{v})}$ hence we have: ${\displaystyle 0=\Lambda _{L}}$ or ${\displaystyle p_{2}=p_{1}}$ as a solution. If we add extra sell orders ${\displaystyle s_{o}}$ but leave ${\displaystyle b_{o}}$ unchanged, then we have ${\displaystyle p_{2} upon solving ${\displaystyle 0=s_{o}-\Lambda _{L}}$. This makes sense because we would expect extra sell orders to result in a lower price.

### Simple Example

Consider a linear LHS ${\displaystyle f_{l}}$ for stock has fallen from $33 to$30 in a 1 month period with a volume of ${\displaystyle 10^{8}}$ and the RHS is a reflection such that ${\displaystyle v_{r}=v_{l}=10^{8},\phi _{l}=1=\phi _{r},w_{r}=w_{l}}$. The LRS/LHS triangles are orientated in such a way that the LHS ${\displaystyle (0,30)(0,33)(1,30)}$ and RHS ${\displaystyle (1,30)(2,33)(2,30)}$ vertexes.${\displaystyle x_{2}-x_{1}=1}$, ${\displaystyle p_{2}=33}$, ${\displaystyle p_{1}=30}$, ${\displaystyle \eta _{l}=-1}$ and ${\displaystyle x_{1}-x_{0}=1}$

It's trivial to show that the equilibrium ${\displaystyle \Lambda _{l}=\Lambda _{R}}$ equation holds. But if we increase volume on the RHS by ${\displaystyle v}$ and we still want the equilibrium equation to hold and we require that ${\displaystyle x_{2}-x_{1}=1}$, then we must modify ${\displaystyle \phi _{r}}$. Solving gives ${\displaystyle \phi _{r}=10^{8}/(v+10^{8})}$ To generate the ${\displaystyle f_{r}}$ curve use the ${\displaystyle \phi _{r\ concave}}$ or ${\displaystyle \phi _{r\ convex}}$ equation with the appropriate boundary variables.

We obtain: ${\displaystyle {\frac {3\times 10^{8}}{2(v+10^{8})}}+30=\int _{1}^{2}f_{r}(x)\,dx}$ With the endpoints: ${\displaystyle f_{r}(1)=30}$ and ${\displaystyle f_{r}(2)=33}$

The ${\displaystyle f_{r}(x)}$ is a monotonically increasing curve of minimum length that lies between the boundaries. It's easy to see that when ${\displaystyle v=0}$ the solution is a line. Increasing ${\displaystyle v}$ results in an convex curve that almost completely hugs the boundaries ${\displaystyle y=30}$ and ${\displaystyle x=2}$ while enclosing an area that approaches 30. The diagram below illustrates how increasing ${\displaystyle v}$ results in a concave path

If we let ${\displaystyle a_{v}=1}$ we can compute the proportion of buy orders to total volume with and without the addition of extra volume ${\displaystyle v}$

Originally, we have ${\displaystyle b_{o}={\frac {10^{8}}{2}}\left(1+{\frac {\ln(33/30)}{\ln(33/30)+2}}\right)}$ or about 52% of the total 10^8 volume are buy orders. If ${\displaystyle v=10^{8}}$ then we have ${\displaystyle 2\times 10^{8}}$ total orders and only 51% of the total volume are buy orders.

### Insider Selling

Next simulate how insider selling (or a fund selling a specified number of shares) changes the pathway of a stock. The additional sell orders result in a concave/convex curve with a lower final price. The ${\displaystyle f_{r}}$ itself has to satisfy both the buy order equation (to factor in the additional sell orders) and the equilibrium equation.

We will use most of the same information in the prior problem but change the coordinates to ${\displaystyle (0,30)(1,30)(1,33)}$ for the LHS and ${\displaystyle (1,33)(2,33)}$ for the RHS. When plotted the LHS serves as the support triangle and the RHS shows a equal number of buy/seller orders, denoted as a line between its two points. We'll begin by solving for ${\displaystyle p_{2}}$ or the final price Let ${\displaystyle p_{1}=33}$, ${\displaystyle v_{r}=10^{8}}$, ${\displaystyle p_{3}=30}$, ${\displaystyle v_{l}=10^{8}}$, ${\displaystyle a_{v}=1}$ (irrelevant for this problem) and ${\displaystyle \eta _{r}=-1}$(to indicate the stock is falling). We'll assume an insider is selling ${\displaystyle 10^{6}}$ shares.

Because the RHS is initially a line, we have an even admixture of buy & sells and then we add ${\displaystyle 10^{6}}$ extra sells. And since ${\displaystyle f_{l}}$ is linear we have ${\displaystyle \phi _{l}(x_{a})=1}$. After some manipulation, we find that ${\displaystyle v_{l}(x_{a})={\frac {10^{8}(33-p_{2})}{3}}}$ Using the buy order equation we have ${\displaystyle \Lambda _{l}=10^{6}}$ or in expanded form:

${\displaystyle 10^{6}=-{\frac {10^{8}(33-p_{2})\ln(p_{2}/33)}{3(-\ln(p_{2}/33)+2)}}}$

Solving gives ${\displaystyle p_{2}=31.6}$, or a four percent decline attributed to insider selling.

Solving the equilibrium equation we find ${\displaystyle \phi _{r}=.98}$. Because ${\displaystyle \phi _{r}}$ is so close to 1 we know the resulting shape is approximately a line with the endpoints ${\displaystyle f_{r}(1)=33}$ and ${\displaystyle f_{r}(2)=31.6}$.

### Raising Capital

An additional application is if a firm or individual shareholder wants to sell to a private investor to raise money. How should the shares be priced? We can use Chart to Scalar to simulate the instantaneous sale of some ${\displaystyle s_{o}}$ number of shares required to raise y dollars. If the present price is ${\displaystyle p_{1}}$, the number of shares sold is ${\displaystyle s_{o}=2y/(p_{1}+p_{2})}$ where ${\displaystyle p_{2}}$ is the final price after the shares are sold. Since the transaction is an instantaneous event there are no buy orders on the RHS and hence ${\displaystyle b_{o}=0}$ and obviously ${\displaystyle \eta _{r}=-1}$ since the stock is falling on the RHS from the selling.

We have: ${\displaystyle \Lambda _{l}={\frac {2y}{p_{1}+p_{2}}}}$

For example, consider a stock has risen from $160 to its present price of ${\displaystyle p_{1}=\240}$ in 60 trading days with an average daily volume of ${\displaystyle 2\times 10^{6}}$. The shape of the LHS is linear ${\displaystyle \phi _{l}(x_{a})=1}$. If a large shareholder wants to raise$1 billion ${\displaystyle y=10^{9}}$, calculate how many shares should be sold and at what price?

The volume function: ${\displaystyle v_{l}(x_{a})=(240-x){\frac {60\times 2\times 10^{6}}{240-160}}}$

Solving, we find ${\displaystyle x=p_{2}=\202}$ assuming the shares as sold instantly on the open market and 4.5 million shares sold to raise $1 billion. But some shares will be sold at around$240 and others at $202. Taking the midpoint$221 is a fair value for the private secondary offering.

In summery, we see that the buy order equation tells us the final price of the stock; the equilibrium equation gives us the shape.

### Bubble Theorem

Chart to Scalar can be used to show that charts that resemble 'bubbles' are more susceptible to collapse, in agreement with real life results. A bubble on a stock chart typically resembles a parabola. Using the results in section 3, we will compute ${\displaystyle \phi _{l}}$ for our bubble and ${\displaystyle \phi _{l}}$ for an 'anti-bubble'. Then we'll show with some calculus that fewer shares are needed for a stock with a 'bubble' chart to fall, and hence is more vulnerable to collapse than the 'anti-bubble'.

Lets assume that ${\displaystyle p_{1}=1,p_{3}=0,x_{1}=1,x_{0}=0}$ The simplest 'bubble' that passes through the endpoints is ${\displaystyle f_{l}(x)=x^{2}}$. Using methods of linear algebra, it's trivial to show the refection about ${\displaystyle y=x}$ or 'anti-bubble' is ${\displaystyle f_{l}(x)=x^{1/2}}$. Both these curves have the same ${\displaystyle \phi }$ equal to 2/3 when the integrals are evaluated between 0 and 1. Showing that the 'bubble' chart is more vulnerable to collapse is the same as showing that ${\displaystyle \phi _{l}(x_{a})v_{l}(x_{a})<\phi _{l}(x_{a2})v_{l}(x_{a2})}$ for some ${\displaystyle p_{2}}$ Furthermore, using the properties of inverse function we know that ${\displaystyle x_{a}=f_{l}^{-1}(p_{2}),x_{a2}=f_{l}(p_{2})}$. Let ${\displaystyle p_{2}=1-\epsilon }$ and we have ${\displaystyle x_{a}=(1-\epsilon )^{1/2}}$ and ${\displaystyle x_{a2}=(1-\epsilon )^{2}}$. After some labor we have the following inequality (where the right hand side is the anti-bubble):

${\displaystyle -{\frac {2}{3}}+{\frac {2(1-\epsilon )^{3/2}}{3}}+\epsilon <\epsilon ^{2}-{\frac {2\epsilon ^{3}}{3}}{}}$

Using the binomial theorem on the square root and dividing both sides by ${\displaystyle \epsilon ^{2}}$ we have

${\displaystyle {\frac {1}{4}}+{\frac {\epsilon }{24}}...<1-{\frac {2\epsilon }{3}}}$ Now it's obvious that as ${\displaystyle \epsilon }$ approaches zero the anti-bubble is bigger.

If we let ${\displaystyle f_{l}(y)=y^{n}}$ with the same boundary conditions as mentioned earlier, we can simulate the behavior ${\displaystyle \phi _{l}(x_{a})v_{l}(x_{a})}$ as ${\displaystyle f_{l}(y)}$ becomes increasingly concave - that is letting ${\displaystyle n}$ approach infinity. Such a curve will appear to hug the boundaries fur sufficiency large ${\displaystyle n}$.

Using methods given in section 3 and letting ${\displaystyle p_{2}=p_{1}-\epsilon }$, we compute:

${\displaystyle \phi _{l}(x_{a})v_{l}(x_{a})=\left\{{\frac {n}{n+1}}(1-\epsilon )^{(n+1)/n}+{\frac {1}{n+1}}-1+\epsilon \right\}{\frac {2v_{l}}{\epsilon }}}$

Using the binomial theorem and upon simplification we have an infinite series that begins: ${\displaystyle \epsilon v_{l}(n+1)/n^{2}}$

Now it's obvious that ${\displaystyle \phi _{l}(x_{a})v_{l}(x_{a})}$ goes to zero as n becomes increasingly large. What this means is as a stock chart appears increasingly parabolic, it becomes less stable and more prone to falling. This is because of the increasingly small cross-sectional area of ${\displaystyle f_{l}}$ between ${\displaystyle x_{a}}$ and ${\displaystyle 1}$.

## Pricing Options

### Introduction

The Chart to Scalar option pricing formula is a consequence of the broader Chart to Scalar theory through an equivalency relation between the statistical property of buy and sell orders and price.

Differences between conventional options pricing models and Chart to Scalar:

1. Volatility-like variable ${\displaystyle \sigma _{p}}$ in Chart to Scalar is a combination of variables such as time until expiration, volume/price differentials, and trade size ${\displaystyle a_{v}}$.

2. Normal distribution instead of log-normal.

3. Options prices grow at ${\displaystyle t^{1/4}}$ versus ${\displaystyle t^{1/2}}$ for Black Scholes. Short term options are more expensive and longer term cheaper in Chart to Scalar than Black Scholes. (this is explained in more detail in the final section)

### Derivation, Part 1: The Displacement Equivalence Relation

The idea is to establish a relationship between discrete 'steps' and price displacement.

Consider a hypothetical stock chart where a stock has moved some amount ${\displaystyle a}$(usually a small percentage) over some time duration ${\displaystyle t_{1}}$ (often measured in years). It can be visualized as a triangle, with the vertices being ${\displaystyle t_{1}}$and ${\displaystyle p_{1}}$ and where ${\displaystyle p_{1}}$is the present price of the stock and the displacement is ${\displaystyle ap_{1}}$.

Denote ${\displaystyle t_{2}}$as the time until expiration

Consider a discrete sum of up ${\displaystyle u}$and down ${\displaystyle d}$stock orders denoted by

${\displaystyle u+d=t_{2}/t_{1}}$

Each 'up' and 'down' represents a 'time unit'. Adding the 'ups' and 'downs' gives the sum of units.

The second part of the fundamental interaction is the difference between 'ups' and 'downs':

${\displaystyle u-d=f(p_{2}-p_{1})}$

This means that the difference between 'ups' and 'downs' gives a function in terms of a new displacement, ${\displaystyle p_{2}-p_{1}}$where ${\displaystyle p_{2}\geq p_{1}}$.

The pair of linear equations solves for ${\displaystyle u}$and ${\displaystyle d}$:

${\displaystyle u={\frac {1}{2}}\left({\frac {t_{2}}{t_{1}}}+f(p_{2}-p_{1})\right)}$

${\displaystyle d={\frac {1}{2}}\left({\frac {t_{2}}{t_{1}}}-f(p_{2}-p_{1})\right)}$

Consider the proportional relation between two displacements, the base one with ${\displaystyle t_{1}}$and our new one, ${\displaystyle p_{2}-p_{1}}$

${\displaystyle {\frac {ap_{1}}{t_{1}}}={\frac {p_{2}-p_{1}}{x}}}$

Solving for ${\displaystyle x}$gives the needed function in terms ${\displaystyle p_{2}-p_{1}}$, which is plugged into ${\displaystyle u}$:

${\displaystyle u={\frac {1}{2}}\left({\frac {t_{2}}{t_{1}}}+{\frac {t_{1}(p_{2}-p_{1})}{ap_{1}}}\right)}$

When ${\displaystyle p_{2}=p_{1}}$, the stock is unchanged, meaning that the number of 'up' units is equal to the 'down' ones.

What we've done is establish a relationship between displacement of price and 'up' and 'down' units.

'Up' and 'down' units, analogous to tossing a coin, also obey a normal distribution:

${\displaystyle \mu _{1}+\sigma _{1}={\frac {t_{2}}{2t_{1}}}+{\frac {1}{2}}{\sqrt {\frac {t_{2}}{t_{1}}}}}$

There is also ${\displaystyle \mu _{2},\sigma _{2}}$for the price.

${\displaystyle \mu _{2}=p_{1}e^{rt}}$

(this is because if ${\displaystyle p_{2}=p_{1}}$the stock is unchanged, hence ${\displaystyle \mu _{1}={\frac {t_{2}}{2t_{1}}}}$meaning that the number of 'up' units is the same as 'down', resulting in no displacement.

We have to find ${\displaystyle \sigma _{2}}$

Because of the equivalence between units and price displacement, the ${\displaystyle \sigma _{2}}$ can be solved by setting ${\displaystyle p_{2}=p_{1}+\sigma _{2}}$

From the equivalence:

${\displaystyle \mu _{1}+\sigma _{1}=u}$

We have:

${\displaystyle {\sqrt {\frac {t_{2}}{t_{1}}}}={\frac {t_{1}\sigma _{2}}{ap_{1}}}}$

Rearranging gives the classic result: ${\displaystyle \sigma _{2}=ap_{1}{\sqrt {t_{2}}}}$setting ${\displaystyle t_{1}=1}$(for a single year and ${\displaystyle t_{2}}$is the fraction of the year)

### Derivation, Part 2

Let ${\displaystyle f_{l}(x)}$ be linear so that ${\displaystyle \phi _{l}(x_{a})=1}$

The LHS resistor force is denoted by a triangle with the vertexes ${\displaystyle (x_{0},p_{3}),(x_{0},p_{1}),(x_{1},p_{1})}$.

The total volume ${\displaystyle v_{l}}$ between ${\displaystyle x_{0}}$ and ${\displaystyle x_{1}}$ is distributed evenly.

Thus the volume between segment ${\displaystyle x_{a}}$ and ${\displaystyle x_{1}}$ is ${\displaystyle v_{l}(x_{a})=v_{o}{\frac {x_{1}-x_{a}}{x_{1}-x_{0}}}}$

Plugging ${\displaystyle v_{l}(x_{a})\phi _{l}(x_{a})}$ into the buy order equation we have (first order Taylor approximation for ${\displaystyle w_{l}}$ is used):

${\displaystyle b_{o}=b(p_{2})={\frac {1}{2a_{v}}}\left(v_{r}+{\frac {v_{o}(p_{2}-p_{1})\ln(p_{2}/p_{1})}{(p_{3}-p_{1})(\ln(p_{2}/p_{1})+2)}}\right)\approx {\frac {1}{2a_{v}}}\left(v_{r}+{\frac {v_{o}(p_{2}-p_{1})^{2}}{2p_{1}(p_{3}-p_{1})}}\right)}$

Like above, let ${\displaystyle p_{2}=p_{1}+\sigma _{2}}$ Also ${\displaystyle \mu _{2}=p_{1}e^{rt}}$

${\displaystyle \mu _{1}={\frac {v_{r}}{2a_{v}}},\sigma _{1}={\frac {1}{2}}{\sqrt {\frac {v_{r}}{a_{v}}}}}$

${\displaystyle \sigma _{2}={\frac {\sqrt {2p_{1}(p_{3}-p_{1})}}{\sqrt {v_{o}}}}(v_{r}a_{v})^{1/4}}$

The normal distribution is a fundamental solution to the heat equation. This is an initial value problem on (0,∞) with homogeneous Dirichlet boundary conditions.

${\displaystyle {\begin{cases}u_{t}=ku_{xx}&(x,t)\in [0,\infty )\times (0,\infty )\\u(x,0)=g(x)&\\u(0,t)=0&\end{cases}}}$
${\displaystyle u(x,t)={\frac {1}{\sqrt {4\pi kt}}}\int _{0}^{\infty }\left[\exp \left(-{\frac {(x-y)^{2}}{4kt}}\right)-\exp \left(-{\frac {(x+y)^{2}}{4kt}}\right)\right]g(y)\,dy}$

${\displaystyle v_{r}}$ is a function of ${\displaystyle v_{o}}$ and time, hence ${\displaystyle v_{r}=f(t,v_{o})}$

So we have (for some constant b) ${\displaystyle \sigma _{2}=b(f(t,v_{o}))^{1/4}}$

And: ${\displaystyle {\frac {\sigma _{2}^{2}}{2t}}=k={\frac {b^{2}(f(t,v_{o}))^{1/2}}{2t}}}$

Because k is a function of t, integration is necessary to solve the PDE ${\displaystyle k={\frac {b^{2}}{2t}}\int _{0}^{t}{\frac {(f(t,v_{o}))^{1/2}}{t}}\,dt}$

Letting ${\displaystyle g(y)=e^{-rt}(y-p_{s})}$, ${\displaystyle x=p_{1}e^{rt}}$ and restricting the bounds of the integration from ${\displaystyle p_{s}}$ to ${\displaystyle \infty }$ and evaluate the integral to compute the call:

${\displaystyle C=p_{1}+{\sqrt {\frac {kt}{\pi }}}\left(e^{-{\frac {(p_{1}e^{rt}+p_{s})^{2}}{4kt}}-rt}\right)\left(e^{\frac {p_{1}p_{s}e^{rt}}{kt}}-1\right)-{\frac {(p_{1}-p_{s}e^{-rt})}{2}}\left(\operatorname {erf} \left({\frac {p_{s}-p_{1}e^{rt}}{2{\sqrt {kt}}}}\right)\right)-{\frac {(p_{1}+p_{s}e^{-rt})}{2}}\left(\operatorname {erf} \left({\frac {p_{s}+p_{1}e^{rt}}{2{\sqrt {kt}}}}\right)\right)}$

### Example

If ${\displaystyle v_{r}}$ and ${\displaystyle v_{o}}$ are of the form ${\displaystyle v_{r}=mt_{2},v_{o}=mt_{1}}$ with the time factors ${\displaystyle (t_{2},t_{1})}$ multiplied by a constant m, then:

${\displaystyle k={\sqrt {\frac {a_{v}}{mt_{2}}}}{\frac {2p_{1}(p_{3}-p_{1})}{t_{1}}}}$

Where ${\displaystyle t_{2}}$ is the time until expiration.

Suppose a hypothetical stock has fallen from $40 to$30 in a 6 month period (120 days) with a daily volume of ${\displaystyle 10^{6}}$ and and ${\displaystyle a_{v}=10^{3}}$. Calculate the $30 call with an expiration of 20 days. Assume no interest. ${\displaystyle p_{s}=p_{1}=30,p_{3}=40,r=0,t_{1}=120,t_{2}=20,m=10^{6}}$ We find ${\displaystyle k={\sqrt {2}}/40}$ Since ${\displaystyle p_{s}=p_{1}}$, the call option pricing formula simplifies dramatically and we have ${\displaystyle C\approx .47}$. ### Chart to Scalar Option Pricing Vs. Black Scholes Consider the example above where ${\displaystyle p_{s}=p_{1}}$ (at-the-money). The call price for Chart to Scalar can be approximated as: ${\displaystyle C\approx \left({\frac {a_{v}t_{2}}{m}}\right)^{1/4}{\sqrt {\frac {2p_{1}(p_{3}-p_{1})}{\pi t_{1}}}}}$ The approximation with Black Scholes for at-the-money calls is: ${\displaystyle C\approx {\frac {p_{1}\alpha }{\sqrt {2\pi }}}{\sqrt {\frac {t_{2}}{256}}}}$ In both instances, ${\displaystyle t_{2}}$ is the number of days until expiration. The notable difference between Chart to Scalar and conventional option pricing models is that sigma is proportional to ${\displaystyle t^{1/4}}$ instead of ${\displaystyle t^{1/2}}$.This generates fat tails due to the intrinsic property of the price/volume dynamic, without the need for added fat tail parameters or the market being incomplete. To test the difference, using the same variables as in the earlier example for the at-the-money call, the strike prices for a variety of times ${\displaystyle t_{2}}$ is plotted. The difficulty is converting price displacement ${\displaystyle p_{3}-p_{1}}$ into a volatility, but I think ${\displaystyle \alpha =.25}$ seems reasonable based on empirical evidence. This is more volatile than an index fund, which has volatility that fluctuates between .15 to .20. For the example above, Chart to Scalar (blue) v. Black Sholes (purple) for an at the money call. The vertical axis is the call price. The horizontal is the number of trading days until expiration. A noted limitation of Black Scholes is that it tends to underestimate the probability of uncommon events, even though these events are observed more frequently in real life. For the example given in this section, if we raise the strike to ${\displaystyle p_{s}=31.4}$, reduce the time until expiration to one day ${\displaystyle t_{2}=1}$, and keep all other variables unchanged, Black Scholes gives a probability of about 1/4000 of the option being exercised, or to put it another way one would have to wait a decade for the underlying to rise 4.7% in a single day. Chart to Scalar, on the other hand, gives a 1/150 probability, which is more realistic. ### High Frequency Trading (HFT) We hypothesise HFT has a stabilizing effect on the market. E. Renshaw (1995) showed that the market is more stable than it used to be. Chart to Scalar could provide a mathematical explanation for why this may be. Consider ${\displaystyle \sigma _{p}}$ which we derived earlier If the trade size ${\displaystyle a_{v}}$ is small relative to ${\displaystyle v_{r}}$, then the variability of moves is smaller. (${\displaystyle a_{v}=0}$ for example) . On the other hand, if we increase ${\displaystyle a_{v}}$, ${\displaystyle v_{r}}$,and ${\displaystyle v_{o}}$ by a scaling factor it cancels out and ${\displaystyle \sigma _{p}}$ is unchanged. So in the context of this model, HFT is either stabilizing or neutral. For example, without HFT let’s assume we have: ${\displaystyle v_{o}=10^{6},v_{r}=10^{6},a_{v}=10^{3}}$ Now introduce HFT: ${\displaystyle v_{o}=10^{7},v_{r}=10^{7},a_{v}=10^{3}}$ We’ve increased the volume ${\displaystyle v_{o}}$ of the stock on the LHS (left hand side) and the RHS ${\displaystyle v_{r}}$ but the trade size is unchanged because we’re assuming the high frequency trades are no larger than normal trades. Because: ${\displaystyle {\frac {\sqrt {10^{6}10^{3}}}{10^{6}}}>{\frac {\sqrt {10^{7}10^{3}}}{10^{7}}}}$ We see that the noise orders have a stabilizing effect. The limitations of this model is the assumption that HFT is the same as random trades and ignoring possible feedback effects. ## Single Solution Solutions to problems in Chart to Scalar where ${\displaystyle \phi _{r}<1}$ are not unique, meaning there are two solutions: a concave and a convex one for the RHS (right-hand-side). The introduction of integrals ${\displaystyle a_{l},a_{r}}$ on the LHS and RHS equilibrium equations, respectively, allow for a single solution. This is because solving the equilibrium equation results in two unique scalar values ${\displaystyle d_{1},d_{2}}$. Due to the variational principle, the value of d which confers with a longer ${\displaystyle f_{r}}$ can be discarded, leaving a path of least distance ${\displaystyle f_{r}}$ that adheres to the constraints. Define ${\displaystyle a_{l},a_{r}}$ as the average price of ${\displaystyle f_{l},f_{r}}$: ${\displaystyle a_{l}=(x_{1}-x_{a})^{-1}\int _{x_{a}}^{x_{1}}f_{l}(x)\,dx}$ ${\displaystyle a_{r}={\frac {d}{x_{2}-x_{1}}}}$ The equilibrium equation with ${\displaystyle a_{l},a_{r}}$: ${\displaystyle \Lambda _{l}=\Lambda _{R}}$ or ${\displaystyle a_{l}v_{l}w_{l}\phi _{l}=a_{r}v_{r}w_{r}\phi _{r}}$ The buy order equation: ${\displaystyle b_{o}={\frac {1}{2a_{v}}}\left(v_{r}-{\frac {\eta _{l}\Lambda _{l}}{a_{r}}}\right)}$ Sell order equation: ${\displaystyle s_{o}=v_{r}/a_{v}-b_{o}}$ Like before, ${\displaystyle w_{r}={\frac {-\ln(p_{2}/p_{1})}{-\ln(p_{2}/p_{1})+2(x_{2}-x_{1})/(x_{1}-x_{a})}}}$ RHS Convex: ${\displaystyle \phi _{r}={\frac {2p_{1}-2d}{p_{1}-p_{2}}}}$ RHS Concave: ${\displaystyle \phi _{r}={\frac {2d-2p_{2}}{p_{1}-p_{2}}}}$ ### Example 1: Bursting of a Stock Market Bubble Consider a simulation of a simple stock market market to show how the concavity of the RHS of the bubble bursting must match the LHS of the bubble inflating The inflation of the bubble on the LHS is a concave curve given by ${\displaystyle f_{l}(x)=(p_{1}-p_{3})x^{2}+p_{3}}$ bounded between ${\displaystyle (0,p_{3}),(1,p_{1})}$. The RHS is bounded between ${\displaystyle (1,p_{1}),(2,p_{3})}$. For this simulation, the hypothetical stock rises from ${\displaystyle p_{3}}$ to ${\displaystyle p_{1}}$ and it falls back to ${\displaystyle p_{3}}$ as the bubble deflates. (${\displaystyle \eta _{l}=1}$) is chosen to indicate the LHS is rising. The LHS and RHS volume is equal and ${\displaystyle v_{l}(x)d/dx=0,v_{r}(x)d/dx=0}$. Thus the conditions are imposed: The time symmetry condition: ${\displaystyle x_{1}-x_{0}=x_{2}-x_{1}=1}$ are imposed, meaning that the duration of events on the LHS is equal to the RHS. Thus: ${\displaystyle w_{l}=w_{r},v_{l}=v_{r},x_{a}=x_{0}=0,x_{2}=2,x_{1}=1}$ For the LHS concave curve of form ${\displaystyle f_{l}(x)=(p_{1}-p_{3})x^{2}+p_{3}}$, the formula is used (making reference to original paper): ${\displaystyle \phi _{l}(x_{a})v_{l}(x_{a})={\frac {2v_{0}}{3(p_{1}-p_{2})}}\left(p_{1}-3p_{2}+2p_{3}+2(p_{2}-p_{3}){\sqrt {p_{2}-p_{3} \over p_{1}-p_{3}}}\right)}$ Hence, ${\displaystyle \phi _{l}=2/3}$ (because ${\displaystyle p_{3}=p_{2}}$) And ${\displaystyle a_{l}=(p_{1}+2p_{3})/3}$ The equilibrium equation ${\displaystyle 2(p_{1}+2p_{3})/9=\phi _{r}a_{r}}$ which is solved for d (one for the concave RHS and convex RHS): convex ${\displaystyle d=(3p_{1}+{\sqrt {5p_{1}^{2}-4p_{3}p_{1}+8p_{3}^{2}}})/6}$ concave ${\displaystyle d=(p_{1}+2p_{3})/3}$ These are plugged into their respective ${\displaystyle \phi _{r}}$ formulas to calculate the defect. The goal is to show the convex solution has a greater defect than the concave one: concave ${\displaystyle \phi _{r}=2/3}$ convex ${\displaystyle \phi _{r}={\frac {p_{1}-({\sqrt {5p_{1}^{2}-4p_{3}p_{1}+8p_{3}^{2}}})/3}{p_{1}-p_{3}}}}$ ${\displaystyle 2/3>convex\phi _{r}}$ Because of the scale invariance properties of ${\displaystyle \phi }$, the above formula reduces to: ${\displaystyle \phi _{r}={\frac {x-({\sqrt {5x^{2}-4x+8}})/3}{x-1}}}$ where ${\displaystyle x>1}$ letting ${\displaystyle x=1+\epsilon }$, we have the infinite series expansion about ${\displaystyle \epsilon =0}$: ${\displaystyle 2/3>2/3-(2\epsilon )/9+(2\epsilon ^{2})/27-(4\epsilon ^{4})/243+(2\epsilon ^{5})/243+O(\epsilon ^{6})}$ Because the concave ${\displaystyle \phi _{r}}$ has a smaller defect, the path is shorter (closest to a strait line), and hence the concave path is chosen as minimizing the action. Which completes the proof. This concave-on-concave symmetry agrees with examples in real life of various asset bubbles bursting ### Example 2: (buy order equation) Example: ${\displaystyle v_{l}=v_{r}=10^{6}}$, the x coordinates are the same as the example above, and ${\displaystyle p_{2}=p_{3}=50,p_{1}=53}$, ${\displaystyle f_{l}(x)=3x^{2}+50}$ and ${\displaystyle \eta _{l}=1}$ For this example, the RHS is a concave reflection of the RHS, thus ${\displaystyle \phi _{l},r=2/3}$ and ${\displaystyle w_{l},a_{l}=w_{r},a_{r}}$. To compute the buy and sell orders for the RHS: ${\displaystyle b_{o}={\frac {10^{6}}{2a_{v}}}\left(1+{\frac {2\ln(50/53)}{3(-\ln(50/53)+2)}}\right)}$ ${\displaystyle b_{o}\approx {\frac {490,000}{a_{v}}}}$ For the rest of this summary, ${\displaystyle a_{v}=1}$ The actual choice of ${\displaystyle a_{v}}$ does not matter for non-statistical problems. ### Example 3: (system of equations) What if ${\displaystyle b_{o}}$ is different? Consider a general case where ${\displaystyle b_{o}}$ is only slightly less than ${\displaystyle v_{r}/2}$. Then it becomes more complicated because ${\displaystyle p_{2}}$ is unknown and ${\displaystyle a_{r}}$ cannot be assumed to be equal to ${\displaystyle a_{l}}$ and ${\displaystyle \phi _{l}}$ becomes a function in terms of ${\displaystyle p_{2}}$ instead of just 2/3. The buy order equation and equilibrium equation must be combined for problems where the ${\displaystyle w_{r},w_{l}}$ and or ${\displaystyle v_{r},v_{l}}$ components are not equal, the result being a system of two equations that is solved for ${\displaystyle d}$ (both for the concave and convex) and ${\displaystyle p_{2}}$ (the final price of the stock, for both the concave and convex) The value of ${\displaystyle p_{2}}$ and ${\displaystyle d}$ corresponding to the greatest defect is discarded. As before, ${\displaystyle x_{1}=1,x_{2}=2,x_{0}=0}$ The components are as follows: The inverse of ${\displaystyle f_{l}}$ evaluated at ${\displaystyle p_{2}}$: ${\displaystyle x_{a}={\sqrt {p_{2}-50 \over 3}}}$ ${\displaystyle \phi _{l}v_{l}=\phi _{l}(x_{a})v_{l}(x_{a})={\frac {2v_{o}}{3(53-p_{2})}}\left(153-3p_{2}+2(p_{2}-50){\sqrt {p_{2}-50 \over 3}}\right)}$ Both ${\displaystyle v_{o},v_{r}}$ will be specified later. In the first example, ${\displaystyle v_{o}=v_{l}=v_{r}=10^{6}}$. ${\displaystyle v_{l}}$ is the functional form of LHS volume in terms of ${\displaystyle p_{2}}$, whereas ${\displaystyle v_{o}}$ is the total volume along the interval ${\displaystyle x_{0},x_{1}}$ ${\displaystyle w_{l}={\frac {-\ln(p_{2}/53)}{(-\ln(p_{2}/53)+2)}}}$ ${\displaystyle w_{r}={\frac {-\ln(p_{2}/53)}{-\ln(p_{2}/53)+2/(1-x_{a})}}}$ ${\displaystyle a_{l}={\frac {p_{2}+{\sqrt {3}}{\sqrt {p_{2}-50}}+103}{3}}}$ ${\displaystyle a_{r}=d}$ RHS Convex: ${\displaystyle \phi _{r}={\frac {106-2d}{53-p_{2}}}}$ RHS Concave: ${\displaystyle \phi _{r}={\frac {2d-2p_{2}}{53-p_{2}}}}$ The system of equations are solved for ${\displaystyle d}$ and ${\displaystyle p_{2}}$: ${\displaystyle \Lambda _{l}=\Lambda _{R}}$ ${\displaystyle (v_{r}-2b_{o})d=\Lambda _{l}}$ After some labor, ${\displaystyle \Lambda _{l}}$ has the series approximation about ${\displaystyle p_{2}=53}$: ${\displaystyle \Lambda _{l}\approx v_{o}\left({\frac {(p_{2}-53)^{2}}{12}}-{\frac {11(p_{2}-53)^{3}}{2862}}\right)}$ ${\displaystyle w_{r}\approx {\frac {(p_{2}-53)^{2}}{636}}-{\frac {59(p_{2}-53)^{3}}{404496}}}$ As ${\displaystyle b_{o}\to v_{r}/2}$, the solution is ${\displaystyle p_{2}=53}$. This is because if the number of buy orders is half of the RHS volume, we expect the stock to end unchanged. Consider a small imbalance: ${\displaystyle v_{r}-2b_{o}=1000}$. Let: ${\displaystyle v_{o},v_{r}=10^{6}}$ Solving fox p and d gives six possible solution pairs, but the only one that logically makes sense Convex: ${\displaystyle d\approx 52.62,p_{2}\approx 52.23}$ Concave: ${\displaystyle d\approx 52.60,p_{2}\approx 52.22}$ The convex and concave curves enclose roughly the same area, indicating the that resulting LHS path is very close to being linear. Plugging these solutions into their respective convex and concave ${\displaystyle \phi _{r}}$ shows that the defect is very small, roughly 2.5%. ### Example 4: deriving the Market impact square-root rule A formula very similar to the 'square-root' rule [3] is derived. Consider the instantaneous sale of stock. For simplicity, let the LHS be linear. An instantaneous sale means ${\displaystyle x_{2}-x_{1}=0}$. Therefore, ${\displaystyle \phi _{r}=1,w_{r}=1,a_{r}=(p_{1}+p_{2})/2}$ Define ${\displaystyle p_{3}}$ as the lower-end of the stock range and ${\displaystyle p_{1}}$ as the present price. ${\displaystyle p_{1}>p_{3}}$ The LHS be visualized as a triangle with the vertices ${\displaystyle (0,p_{3}),(1,p_{3}),(1,p_{1})}$ Since the LHS is linear, ${\displaystyle \phi _{l}=1,a_{l}=a_{r}}$ ${\displaystyle p_{2}}$ is the final price of the stock after the instantaneous sale ${\displaystyle v_{r}}$ is rendered. Because ${\displaystyle \eta _{l}=1}$ (the LHS is rising), ${\displaystyle p_{3}\leq p_{2}\leq p_{1}}$ ${\displaystyle v_{l}=v_{o}\left({\frac {p_{1}-p_{2}}{p_{1}-p_{3}}}\right)}$ Where ${\displaystyle v_{o}}$ is the total volume of the LHS between x=0,x=1 (some period of time) This is obtained by taking the inverse of ${\displaystyle x_{a}=f_{l}^{-1}(p_{2})}$ and finding the proportion of ${\displaystyle v_{o}}$ volume that is 'liberated' by the stock falling to ${\displaystyle p_{2}}$. Via the triangle, ${\displaystyle f_{l}=(p_{1}-p_{3})x_{a}+p_{3}}$ Set ${\displaystyle f_{l}=p_{2}}$. Then ${\displaystyle x_{a}=(p_{2}-p_{3})/(p_{1}-p_{3})}$ and ${\displaystyle 1-x_{a}}$ gives the proportion. ${\displaystyle w_{l}}$ can be approximated as ${\displaystyle w_{l}\approx {\frac {1}{2}}\left(1-{\frac {p_{2}}{p_{1}}}\right)}$ Set ${\displaystyle p_{2}=p_{1}-\epsilon }$ where ${\displaystyle \epsilon }$ is the 'impact' Setting up the equilibrium equation and solving for ${\displaystyle \epsilon }$ we have: ${\displaystyle \epsilon ={\sqrt {\frac {2p_{1}(p_{1}-p_{3})v_{r}}{v_{o}}}}}$ The volatility-like variable ${\displaystyle \alpha }$ can be written as: ${\displaystyle \alpha ={\sqrt {\frac {p_{1}-p_{3}}{p_{1}}}}}$. Hence, we have: ${\displaystyle \epsilon =p_{1}{\sqrt {2}}\alpha {\sqrt {\frac {v_{r}}{v_{o}}}}}$ As we would expect, the volatility term is scale invariant, but the impact ${\displaystyle \epsilon }$ is proportional to the initial price ${\displaystyle p_{1}}$. If ${\displaystyle p_{3}}$ is much small than ${\displaystyle p_{1}}$, we have a greater price range (more volatility). The ${\displaystyle {\sqrt {2}}}$ term is somewhat arbitrary, but we still have the volatility and square-root impact relation. ### Example 5: inflows and outflows Inflows and outflows for the RHS chart are calculated. However, this is only an approximation; calculating an exact inflow or outflow using a single, closed-form expression is impossible and impractical, but we can get a ballpark estimate. For small increases or decreases in price (${\displaystyle p_{2}/p_{1}\approx 1}$), we can use the linear approximation of the natural log and ${\displaystyle w_{r}}$. We have: Inflow$ ${\displaystyle \approx \left({\frac {p_{1}+p_{2}}{2}}\right)\left({\frac {p_{2}}{p_{1}}}-1\right){\frac {v_{r}\phi _{r}}{2}}}$

If ${\displaystyle p_{2}, we have an outflow. If ${\displaystyle p_{1}=p_{2}}$, there is neither an inflow nor outflow

Here, the average price is ${\displaystyle (p_{1}+p_{2})/2}$, which is a linear approximation of the exact average price of ${\displaystyle f_{r}}$. This is good enough for small changes in price.

${\displaystyle \phi _{r}=2/3}$ can be added if the shape resembles a hyperbola, regardless of concavity

Let's assume we want to calculate the dollar inflow for Microsoft for a single trading day. If on the RHS we observe ${\displaystyle p_{1}=30,p_{2}=p_{3}=31,v_{r}=10^{7},x_{2}=1,x_{1}=0,\eta _{r}=1}$ and the shape of the RHS is concave and resembles the function ${\displaystyle x^{2}+30}$ (noticing this satisfies and endpoints ${\displaystyle P_{2},p_{1}}$ and has ${\displaystyle d/dx=0}$ at ${\displaystyle x=0}$) We have ${\displaystyle \phi _{r}=2/3,w_{r}=.016}$ and ${\displaystyle a_{r}=91/3}$. Putting it all together, we have an inflow of around $32.6 million while around$303 million changed hands.

## References

1. ^ Baggett, L. Scott; Thompson, James; Williams, Edward; Wojciechowski, William (October 2006). "Nobels for nonsense". Journal of Post Keynesian Economics. 29 (1): 3–18.
2. ^ Hull, John; White, Alan (June 1987). "The Pricing of Options on Assets with Stochastic Volatilities". Journal of Finance. 42 (2): 281–300.
3. ^ Gatheral, Jim (October 2011). "Optimal order execution". Text " JOIM Fall Conference, Boston " ignored (help); Text " http://faculty.baruch.cuny.edu/jgatheral/JOIM2011.pdf " ignored (help)