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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,
$\begin{array}{cc}\text{for}y<0,& P(Y<y)=0\text{and}\\ \text{for}y\ge 0,& P(Y<y)=P(X^2<y)=P(|X|<\sqrt{y})=\\ & =F_X(\sqrt{y})-F_X(-\sqrt{y})=F_X(\sqrt{y})-(1-F_X(\sqrt{y}))=2F_X(\sqrt{y})-1\end{array}$

$\begin{array}{cc}{f}_Y(y)& =2\frac{d}{dy}{F}_X(\sqrt{y})-0=2\frac{d}{dy}\left({\int }_{-\mathrm{\infty }}^{\sqrt{y}}\frac{1}{\sqrt{2\pi }}{e}^{\frac{-t^2}{2}}dt\right)\\ & =2\frac{1}{\sqrt{2\pi }}{e}^{-\frac{y}{2}}(\sqrt{y}){\text{'}}_y=2\frac{1}{\sqrt{2}\sqrt{\pi }}{e}^{-\frac{y}{2}}\left(\frac{1}{2}{y}^{-\frac{1}{2}}\right)=\frac{1}{2^{\frac{1}{2}}\Gamma (\frac{1}{2})}{y}^{\frac{1}{2}-1}{e}^{-\frac{y}{2}}\end{array}$

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-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 $x\sim {\chi }_1^2$ and $y\sim {\chi }_1^2$, 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:

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

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:

$P(Q)dQ={\int }_{𝒱}\prod _{i=1}^k(N(x_i)d{x}_i)={\int }_{𝒱}\frac{e^{-(x_1^2+x_2^2+\cdots +x_k^2)/2}}{(2\pi {)}^{k/2}}d{x}_{1}d{x}_2\cdots d{x}_k$

where $N(x)$ is the standard normal distribution and $𝒱$ 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.

$P(Q)dQ=\frac{e^{-Q/2}}{(2\pi {)}^{k/2}}{\int }_{𝒱}d{x}_{1}d{x}_2\cdots d{x}_k$

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:

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

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

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