Proofs related to chi-squared distribution

The following are proofs of several characteristics related to the chi-square 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 if $y<0,~P(Y<y)=0$ and if $y\geq 0,~P(Y<y)=P(X^{2}<y)=P(|X|<{\sqrt {y}})=F_{X}({\sqrt {y}})-F_{X}(-{\sqrt {y}})$

f_{Y}(y)=f_{X}({\sqrt {y}}){\frac {\partial ({\sqrt {y}})}{\partial y}}-f_{X}(-{\sqrt {y}}){\frac {\partial (-{\sqrt {y}})}{\partial y}}

={\frac {1}{\sqrt {2\pi }}}e^{\frac {-y}{2}}{\frac {1}{2y^{1/2}}}+{\frac {1}{\sqrt {2\pi }}}e^{\frac {-y}{2}}{\frac {1}{2y^{1/2}}}

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

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

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

Derivation of the pdf for two degrees of freedom

To derive the chi-square 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.

let $x$ and $y$ are two independent variables and satisfy that $x\sim \chi _{1}^{2}$ and $y\sim \chi _{1}^{2}$ , thus, 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:

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

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:

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

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 {S}}\prod _{i=1}^{k}(N(x_{i})\,dx_{i})=\int _{\mathcal {S}}{\frac {e^{-(x_{1}^{2}+x_{2}^{2}+...+x_{k}^{2})/2}}{(2\pi )^{k/2}}}\,dx_{1}dx_{2}...dx_{k}

Where $N(x)$ is the standard normal distribution and ${\mathcal {S}}$ is that 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.