View source for 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'' = ''X''<sup>2</sup> where ''X'' has [[normal distribution]] with mean 0 and variance 1 (that is ''X''&nbsp;~&nbsp;''N''(0,1)).

Then,<br>
<math>
\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}
</math>

: <math>
\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}
</math>

Where <math>F</math> and <math>f</math> are the cdf and pdf of the corresponding random variables.

Then <math>Y = X^2 \sim \chi^2_1.</math>

====Alternative proof directly using the change of variable formula====
The [[random variable#Functions of random variables|change of variable formula]] (implicitly derived above), for a [[monotonic]] transformation <math>y=g(x)</math>, is: 
:<math>f_Y(y) = \sum_{i} f_X(g_{i}^{-1}(y)) \left| \frac{d g_{i}^{-1}(y)}{d y} \right|. </math>

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

:<math>f_Y(y) = 2f_X(g^{-1}(y)) \left| \frac{d g^{-1}(y)}{d y} \right|.</math>

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

So here:

:<math> 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}. </math>

And one gets the chi-squared distribution, noting the property of the [[gamma function#Particular values|gamma function]]: <math> \Gamma(1/2)=\sqrt{\pi}</math>.

===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 <math>X</math> and <math>Y</math> are two independent variables satisfying <math>X\sim\chi^2_1</math> and <math>Y\sim\chi^2_1</math>, so that the probability density functions of <math>X</math> and <math>Y</math> are respectively:

: <math>
f_X(x)=\frac{1}{2^{\frac{1}{2}}\Gamma(\frac{1}{2})}x^{-\frac{1}{2}}e^{-\frac{x}{2}}
</math>
and of course <math>f_Y(y) = f_X(y)</math>. Then, we can derive the joint distribution of <math>(X,Y)</math>:

: <math>
f(x,y)=f_X(x)\,f_Y(y) = \frac{1}{2\pi}(xy)^{-\frac{1}{2}}e^{-\frac{x+y}{2}}
</math>

where <math>\Gamma(\tfrac{1}{2})^2 = \pi</math>. Further{{Clarify|reason=Rationale would be helpful.|date=January 2024}}, let <math>A=xy</math> and <math>B=x+y</math>, we can get that:

: <math>
x = \frac{B+\sqrt{B^2-4A}}{2}
</math>
and
: <math>
y = \frac{B-\sqrt{B^2-4A}}{2}
</math>

or, inversely

: <math>
x = \frac{B-\sqrt{B^2-4A}}{2}
</math>
and
: <math>
y = \frac{B+\sqrt{B^2-4A}}{2}
</math>

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}}:

: <math>
\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}}
</math>


Now we can change <math>f(x,y)</math> to <math>f(A,B)</math>{{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}}:

: <math>
f(A,B)=2\times\frac{1}{2\pi}A^{-\frac{1}{2}}e^{-\frac{B}{2}}(B^2-4A)^{-\frac{1}{2}}
</math>

where the leading constant 2 is to take both the two variable change policies into account. Finally, we integrate out <math>A</math>{{Clarify|reason=Integration boundaries unclear.|date=January 2024}} to get the distribution of <math>B</math>, i.e. <math>x+y</math>:

: <math>
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
</math>

Substituting <math>A=\frac{B^2}{4}\sin^2(t)</math> gives:

: <math>
f(B)=2\times\frac{e^{-\frac{B}{2}}}{2\pi}\int_0^{\frac{\pi}{2}} \, dt
</math>

So, the result is:

: <math>
f(B)=\frac{e^{-\frac{B}{2}}}{2}
</math>

=== Derivation of the pdf for ''k'' degrees of freedom ===

Consider the ''k'' samples <math>x_i</math> to represent a single point in a ''k''-dimensional space. The chi square distribution for ''k'' degrees of freedom will then be given by:

:<math>
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
</math>

where <math>N(x)</math> is the standard [[normal distribution]] and <math>\mathcal{V}</math> is that elemental shell volume at ''Q''(''x''), which is proportional to the (''k''&nbsp;&minus;&nbsp;1)-dimensional surface in ''k''-space for which

: <math>Q=\sum_{i=1}^k x_i^2</math>

It can be seen that this surface is the surface of a ''k''-dimensional ball or, alternatively, an [[n-sphere]] where ''n''&nbsp;=&nbsp;''k''&nbsp;-&nbsp;1 with radius <math>R=\sqrt{Q}</math>, 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.

:<math>
P(Q) \, dQ = \frac{e^{-Q/2}}{(2\pi)^{k/2}} \int_\mathcal{V} dx_1\,dx_2\cdots dx_k
</math>

The integral is now simply the surface area ''A'' of the (''k''&nbsp;&minus;&nbsp;1)-sphere times the infinitesimal thickness of the sphere which is

:<math>dR=\frac{dQ}{2Q^{1/2}}.</math>

The area of a [[n-sphere|(''k''&nbsp;&minus;&nbsp;1)-sphere]] is:

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

Substituting, realizing that <math>\Gamma(z+1)=z\Gamma(z)</math>, and cancelling terms yields:

:<math>
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
</math>

[[Category:Article proofs]]