Proofs related to chi-squared distribution

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The following are proofs of several characteristics related to the chi-squared distribution.

Derivations of the pdf[edit]

Derivation of the pdf for one degree of freedom[edit]

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)).


\text{for} ~ y < 0, & ~~ P(Y<y) = 0 ~~ \text{and} \\
\text{for} ~ y \geq 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}))= 2 F_X(\sqrt{y})-1

f_Y(y) & = 2 \frac{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}-1}e^{-\frac{y}{2}}

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

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

Derivation of the pdf for two degrees of freedom[edit]

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^2_1 and y\sim\chi^2_1, so that the probability density functions of x and y are respectively:




Simply, we can derive the joint distribution of x and y:


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}


y = \frac{B-\sqrt{B^2-4A}}{2}

or, inversely

x = \frac{B-\sqrt{B^2-4A}}{2}


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)
                 -(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}             \\
       = (B^2-4A)^{-\frac{1}{2}}

Now we can change f(x,y) to f(A,B):


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:


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:


Derivation of the pdf for k degrees of freedom[edit]

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


The area of a (k − 1)-sphere is:


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