Proofs related to chi-squared distribution

The following are proofs of several characteristics related to the 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, $$ \begin{alignat}{2} \text{for} ~ y < 0, & F_Y(y) = P(Y<y) = 0  \text{and} \\ \text{for} ~ y \geq 0, & F_Y(y) = P(Y<y) = P(X^2<y) = P(|X|<\sqrt{y}) = P(-\sqrt{y} < X <\sqrt{y})\\ & = F_X(\sqrt{y})-F_X(-\sqrt{y})= F_X(\sqrt{y})-(1-F_X(\sqrt{y}))= 2 F_X(\sqrt{y})-1 \end{alignat} $$



\begin{align} 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{align} $$

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

Then $$Y = X^2 \sim \chi^2_1.$$

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:
 * $$f_Y(y) = \sum_{i} f_X(g_{i}^{-1}(y)) \left| \frac{d g_{i}^{-1}(y)}{d y} \right|. $$

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.


 * $$f_Y(y) = 2f_X(g^{-1}(y)) \left| \frac{d g^{-1}(y)}{d y} \right|.$$

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

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 $$X\sim\chi^2_1$$ and $$Y\sim\chi^2_1$$, 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 &minus; 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 &minus; 1)-sphere times the infinitesimal thickness of the sphere which is


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

The area of a (k &minus; 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 $$