# Proofs related to chi-squared distribution

The following are proofs of several characteristics related to the chi-squared distribution.

## Derivations of the pdf

### Derivation of the pdf for one degree of freedom

Let random variable Y be defined as Y = X2 where X has normal distribution with mean 0 and variance 1 (that is X ~ N(0,1)).

Then,
\begin{alignat}{2} \text{for} ~ y < 0, & ~~ P(Y

\begin{align} f_Y(y) & = 2 \frac{d}{dy} F_X(\sqrt{y}) - 0 = 2 \frac{d}{dy} \left( \int_{-\infty}^\sqrt{y} \frac{1}{\sqrt{2\pi}} e^{\frac{-t^2}{2}} dt \right) \\ & = 2 \frac{1}{\sqrt{2 \pi}} e^{-\frac{y}{2}} (\sqrt{y})'_y = 2 \frac{1}{\sqrt{2}\sqrt{\pi}} e^{-\frac{y}{2}} \left( \frac{1}{2} y^{-\frac{1}{2}} \right) = \frac{1}{2^{\frac{1}{2}} \Gamma(\frac{1}{2})}y^{\frac{1}{2}-1}e^{-\frac{y}{2}} \end{align}

Where $F$ and $f$ are the cdf and pdf of the corresponding random variables.

Then $Y = X^2 \sim \chi^2_1.$

### Derivation of the pdf for two degrees of freedom

To derive the chi-squared distribution with 2 degrees of freedom, there could be several methods. Here presented is one of them which is based on the distribution with 1 degree of freedom.

Suppose that $x$ and $y$ are two independent variables satisfying $x\sim\chi^2_1$ and $y\sim\chi^2_1$, so that the probability density functions of $x$ and $y$ are respectively:

$f(x)=\frac{1}{2^{\frac{1}{2}}\Gamma(\frac{1}{2})}x^{-\frac{1}{2}}e^{-\frac{x}{2}}$

and

$f(y)=\frac{1}{2^{\frac{1}{2}}\Gamma(\frac{1}{2})}y^{-\frac{1}{2}}e^{-\frac{y}{2}}$

Simply, we can derive the joint distribution of $x$ and $y$:

$f(x,y)=\frac{1}{2\pi}(xy)^{-\frac{1}{2}}e^{-\frac{x+y}{2}}$

where $\Gamma(\frac{1}{2})^2$ is replaced by $\pi$. Further, let $A=xy$ and $B=x+y$, we can get that:

$x = \frac{B+\sqrt{B^2-4A}}{2}$

and

$y = \frac{B-\sqrt{B^2-4A}}{2}$

or, inversely

$x = \frac{B-\sqrt{B^2-4A}}{2}$

and

$y = \frac{B+\sqrt{B^2-4A}}{2}$

Since the two variable change policies are symmetric, we take the upper one and multiply the result by 2. The Jacobian determinant can be calculated as:

$\operatorname{Jacobian}\left( \frac{x, y}{A, B} \right) =\begin{vmatrix} -(B^2-4A)^{-\frac{1}{2}} & \frac{1+B(B^2-4A)^{-\frac{1}{2}}}{2} \\ (B^2-4A)^{-\frac{1}{2}} & \frac{1-B(B^2-4A)^{-\frac{1}{2}}}{2} \\ \end{vmatrix} = (B^2-4A)^{-\frac{1}{2}}$

Now we can change $f(x,y)$ to $f(A,B)$:

$f(A,B)=2\times\frac{1}{2\pi}A^{-\frac{1}{2}}e^{-\frac{B}{2}}(B^2-4A)^{-\frac{1}{2}}$

where the leading constant 2 is to take both the two variable change policies into account. Finally, we integrate out $A$ to get the distribution of $B$, i.e. $x+y$:

$f(B)=2\times\frac{e^{-\frac{B}{2}}}{2\pi}\int_0^{\frac{B^2}{4}}A^{-\frac{1}{2}}(B^2-4A)^{-\frac{1}{2}}dA$

Let $A=\frac{B^2}{4}\sin^2(t)$, the equation can be changed to:

$f(B)=2\times\frac{e^{-\frac{B}{2}}}{2\pi}\int_0^{\frac{\pi}{2}} \, dt$

So the result is:

$f(B)=\frac{e^{-\frac{B}{2}}}{2}$

### Derivation of the pdf for k degrees of freedom

Consider the k samples $x_i$ to represent a single point in a k-dimensional space. The chi square distribution for k degrees of freedom will then be given by:

$P(Q) \, dQ = \int_\mathcal{V} \prod_{i=1}^k (N(x_i)\,dx_i) = \int_\mathcal{V} \frac{e^{-(x_1^2 + x_2^2 + \cdots +x_k^2)/2}}{(2\pi)^{k/2}}\,dx_1\,dx_2 \cdots dx_k$

where $N(x)$ is the standard normal distribution and $\mathcal{V}$ is that elemental shell volume at Q(x), which is proportional to the (k − 1)-dimensional surface in k-space for which

$Q=\sum_{i=1}^k x_i^2$

It can be seen that this surface is the surface of a k-dimensional ball or, alternatively, an n-sphere where n = k - 1 with radius $R=\sqrt{Q}$, and that the term in the exponent is simply expressed in terms of Q. Since it is a constant, it may be removed from inside the integral.

$P(Q) \, dQ = \frac{e^{-Q/2}}{(2\pi)^{k/2}} \int_\mathcal{V} dx_1\,dx_2\cdots dx_k$

The integral is now simply the surface area A of the (k − 1)-sphere times the infinitesimal thickness of the sphere which is

$dR=\frac{dQ}{2Q^{1/2}}.$

The area of a (k − 1)-sphere is:

$A=\frac{kR^{k-1}\pi^{k/2}}{\Gamma(k/2+1)}$

Substituting, realizing that $\Gamma(z+1)=z\Gamma(z)$, and cancelling terms yields:

$P(Q) \, dQ = \frac{e^{-Q/2}}{(2\pi)^{k/2}}A\,dR= \frac{1}{2^{k/2}\Gamma(k/2)}Q^{k/2-1}e^{-Q/2}\,dQ$