Proofs related to chi-squared distribution

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

\text{for} ~ y \geq 0, & ~~ F_Y(y) = P(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 $Y\; =\; X^2\; \backslash sim\; \backslash chi^2\_1.$

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

:$f\_Y(y)\; =\; \backslash sum\_\{i\}\; f\_X(g\_\{i\}^\{-1\}(y))\; \backslash left|\; \backslash frac\{d\; g\_\{i\}^\{-1\}(y)\}\{d\; y\}\; \backslash right|.$

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

:$f\_Y(y)\; =\; 2f\_X(g^\{-1\}(y))\; \backslash left|\; \backslash frac\{d\; g^\{-1\}(y)\}\{d\; y\}\; \backslash right|.$

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

$\backslash frac\{d\; g^\{-1\}(y)\}\{d\; y\}\; =\; \backslash frac\{1\}\{2\backslash sqrt\{y\}\}\; .$

So here:

:$f\_Y(y)\; =\; 2\backslash frac\{1\}\{\backslash sqrt\{2\backslash pi\}\}e^\{-y/2\}\; \backslash frac\{1\}\{2\backslash sqrt\{y\}\}\; =\; \backslash frac\{1\}\{\backslash sqrt\{2\backslash pi\; y\}\}e^\{-y/2\}.$

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

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 $X\backslash sim\backslash chi^2\_1$ and $Y\backslash sim\backslash chi^2\_1$, 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 $\backslash Gamma(\backslash tfrac\{1\}\{2\})^2\; =\; \backslash pi$. Further{{Clarify|reason=Rationale would be helpful.|date=January 2024}}, 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{{Clarify|reason=Reasoning unclear, provide reference.|date=January 2024}}:

: $$

\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 $f(A,B)${{Clarify|reason=The new function in (A,B) needs a new name because it is not the same function only differing in parameters.|date=January 2024}}:

: $$

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 $A${{Clarify|reason=Integration boundaries unclear.|date=January 2024}} 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 $A=\backslash frac\{B^2\}\{4\}\backslash sin^2(t)$ 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}

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 $\backslash mathcal\{V\}$ is that elemental shell volume at Q(x), which is proportional to the (k − 1)-dimensional surface in k-space for which

: $Q=\backslash sum\_\{i=1\}^k\; x\_i^2$

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 $R=\backslash sqrt\{Q\}$, 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

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

The area of a (k − 1)-sphere is:

:$$

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

Substituting, realizing that $\backslash Gamma(z+1)=z\backslash Gamma(z)$, 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

Category:Article proofs