# Proofs related to chi-squared distribution

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

{\displaystyle {\begin{aligned}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}}}e^{-{\frac {y}{2}}}\end{aligned}}}

Where ${\displaystyle F}$ and ${\displaystyle f}$ are the cdf and pdf of the corresponding random variables.

Then ${\displaystyle Y=X^{2}\sim \chi _{1}^{2}.}$

#### Alternative proof directly using the change of variable formula

The change of variable formula (implicitly derived above), for a monotonic transformation ${\displaystyle y=g(x)}$, is:

${\displaystyle f_{Y}(y)=\sum _{i}f_{X}(g_{i}^{-1}(y))\left|{\frac {dg_{i}^{-1}(y)}{dy}}\right|.}$

In this case the change is not monotonic, because every value of ${\displaystyle \scriptstyle Y}$ has two corresponding values of ${\displaystyle \scriptstyle X}$ (one positive and negative). However, because of symmetry, both halves will transform identically, i.e.

${\displaystyle f_{Y}(y)=2f_{X}(g^{-1}(y))\left|{\frac {dg^{-1}(y)}{dy}}\right|.}$

In this case, the transformation is: ${\displaystyle x=g^{-1}(y)={\sqrt {y}}}$, and its derivative is ${\displaystyle {\frac {dg^{-1}(y)}{dy}}={\frac {1}{2{\sqrt {y}}}}.}$

So here:

${\displaystyle f_{Y}(y)=2{\frac {1}{\sqrt {2\pi }}}e^{-y/2}{\frac {1}{2{\sqrt {y}}}}={\frac {1}{\sqrt {2\pi y}}}e^{-y/2}.}$

And one gets the chi-squared distribution, noting the property of the gamma function: ${\displaystyle \Gamma (1/2)={\sqrt {\pi }}}$.

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

There are several methods to derive chi-squared distribution with 2 degrees of freedom. Here is one based on the distribution with 1 degree of freedom.

Suppose that ${\displaystyle x}$ and ${\displaystyle y}$ are two independent variables satisfying ${\displaystyle x\sim \chi _{1}^{2}}$ and ${\displaystyle y\sim \chi _{1}^{2}}$, so that the probability density functions of ${\displaystyle x}$ and ${\displaystyle y}$ are respectively:

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

and

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

Simply, we can derive the joint distribution of ${\displaystyle x}$ and ${\displaystyle y}$:

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

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

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

and

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

or, inversely

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

and

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

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:

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

Now we can change ${\displaystyle f(x,y)}$ to ${\displaystyle f(A,B)}$:

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

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

${\displaystyle f(B)=2\times {\frac {e^{-{\frac {B}{2}}}}{2\pi }}\int _{0}^{\frac {B^{2}}{4}}A^{-{\frac {1}{2}}}(B^{2}-4A)^{-{\frac {1}{2}}}dA}$

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

${\displaystyle f(B)=2\times {\frac {e^{-{\frac {B}{2}}}}{2\pi }}\int _{0}^{\frac {\pi }{2}}\,dt}$

So the result is:

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

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

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

${\displaystyle P(Q)\,dQ=\int _{\mathcal {V}}\prod _{i=1}^{k}(N(x_{i})\,dx_{i})=\int _{\mathcal {V}}{\frac {e^{-(x_{1}^{2}+x_{2}^{2}+\cdots +x_{k}^{2})/2}}{(2\pi )^{k/2}}}\,dx_{1}\,dx_{2}\cdots dx_{k}}$

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

${\displaystyle Q=\sum _{i=1}^{k}x_{i}^{2}}$

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 ${\displaystyle R={\sqrt {Q}}}$, 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.

${\displaystyle P(Q)\,dQ={\frac {e^{-Q/2}}{(2\pi )^{k/2}}}\int _{\mathcal {V}}dx_{1}\,dx_{2}\cdots dx_{k}}$

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

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

The area of a (k − 1)-sphere is:

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

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

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