Proofs related to chi-squared distribution

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

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

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

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

Alternative proof directly using the change of variable formula

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

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

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

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

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

So here:

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

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

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

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

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

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

where $\Gamma ({\tfrac {1}{2}})^{2}=\pi$. 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. $x+y$:

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

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

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

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:

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

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

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