Proofs related to chi-squared distribution

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,
${\displaystyle {\begin{alignedat}{2}{\text{for}}~y<0,&~~F_{Y}(y)=P(Y<y)=0~~{\text{and}}\\{\text{for}}~y\geq 0,&~~F_{Y}(y)=P(Y<y)=P(X^{2}<y)=P(|X|<{\sqrt {y}})=P(-{\sqrt {y}}<X<{\sqrt {y}})\\~~&=F_{X}({\sqrt {y}})-F_{X}(-{\sqrt {y}})=F_{X}({\sqrt {y}})-(1-F_{X}({\sqrt {y}}))=2F_{X}({\sqrt {y}})-1\end{alignedat}}}$

${\displaystyle {\begin{aligned}f_{Y}(y)&={\tfrac {d}{dy}}F_{Y}(y)=2{\tfrac {d}{dy}}F_{X}({\sqrt {y}})-0=2{\frac {d}{dy}}\left(\int _{-\infty }^{\sqrt {y}}{\frac {1}{\sqrt {2\pi }}}e^{\frac {-t^{2}}{2}}dt\right)\\&=2{\frac {1}{\sqrt {2\pi }}}e^{-{\frac {y}{2}}}({\sqrt {y}})'_{y}=2{\frac {1}{{\sqrt {2}}{\sqrt {\pi }}}}e^{-{\frac {y}{2}}}\left({\frac {1}{2}}y^{-{\frac {1}{2}}}\right)\\&={\frac {1}{2^{\frac {1}{2}}\Gamma ({\frac {1}{2}})}}y^{-{\frac {1}{2}}}e^{-{\frac {y}{2}}}\end{aligned}}}$

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

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

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:

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

and of course ${\displaystyle f_{Y}(y)=f_{X}(y)}$. Then, we can derive the joint distribution of ${\displaystyle (X,Y)}$:

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

where ${\displaystyle \Gamma ({\tfrac {1}{2}})^{2}=\pi }$. Further^{[clarification needed]}, 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^{[clarification needed]}:

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

Now we can change ${\displaystyle f(x,y)}$ to ${\displaystyle f(A,B)}$^{[clarification needed]}:

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

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

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

Substituting ${\displaystyle A={\frac {B^{2}}{4}}\sin ^{2}(t)}$ gives:

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

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:

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