Index: The Book of Statistical ProofsProbability Distributions ▷ Univariate continuous distributions ▷ Chi-squared distribution ▷ Probability density function

Theorem: Let $Y$ be a random variable following a chi-squared distribution:

$\label{eq:chi2} Y \sim \chi^{2}(k) \; .$

Then, the probability density function of $Y$ is

$\label{eq:chi2-pdf} f_Y(y) = \frac{1}{2^{k/2} \, \Gamma (k/2)} \, y^{k/2-1} \, e^{-y/2} \; .$

Proof: A chi-square-distributed random variable with $k$ degrees of freedom is defined as the sum of $k$ squared standard normal random variables:

$\label{eq:chi2-def} X_1, \ldots, X_k \sim \mathcal{N}(0,1) \quad \Rightarrow \quad Y = \sum_{i=1}^{k} X_i^2 \sim \chi^{2}(k) \; .$

Let $x_1, \ldots, x_k$ be values of $X_1, \ldots, X_k$ and consider $x = \left( x_1, \ldots, x_k \right)$ to be a point in $k$-dimensional space. Define

$\label{eq:y-x} y = \sum_{i=1}^{k} x_i^2$

and let $f_Y(y)$ and $F_Y(y)$ be the probability density function and cumulative distribution function of $Y$. Because the PDF is the first derivative of the CDF, we can write:

$\label{eq:y-pdf-s0} F_Y(y) = \frac{F_Y(y)}{\mathrm{d}y} \, \mathrm{d}y = f_Y(y) \, \mathrm{d}y \; .$

Then, the cumulative distribution function of $Y$ can be expressed as

$\label{eq:y-cdf-s1} f_Y(y) \, \mathrm{d}y = \int_{V} \prod_{i=1}^{k} \left( \mathcal{N}(x_i; 0, 1) \, \mathrm{d}x_i \right)$

where $\mathcal{N}(x_i; 0, 1)$ is the probability density function of the standard normal distribution and $V$ is the elemental shell volume at $y(x)$, which is proportional to the $(k-1)$-dimensional surface in $k$-space for which equation \eqref{eq:y-x} is fulfilled. Using the probability density function of the normal distribution, equation \eqref{eq:y-cdf-s1} can be developed as follows:

$\label{eq:y-cdf-s2} \begin{split} f_Y(y) \, \mathrm{d}y &= \int_{V} \prod_{i=1}^{k} \left( \frac{1}{\sqrt{2 \pi}} \cdot \exp \left[ -\frac{1}{2} x_i^2 \right] \, \mathrm{d}x_i \right) \\ &= \int_{V} \frac{\exp \left[ -\frac{1}{2} \left( x_1^2 + \ldots + x_k^2 \right) \right]}{(2 \pi)^{k/2}} \; \mathrm{d}x_1 \, \ldots \, \mathrm{d}x_k \\ &= \frac{1}{(2 \pi)^{k/2}} \int_{V} \exp \left[ -\frac{y}{2} \right] \; \mathrm{d}x_1 \, \ldots \, \mathrm{d}x_k \; . \end{split}$

Because $y$ is constant within the set $V$, it can be moved out of the integral:

$\label{eq:y-cdf-s3} f_Y(y) \, \mathrm{d}y = \frac{\exp \left[ -y/2 \right]}{(2 \pi)^{k/2}} \int_{V} \; \mathrm{d}x_1 \, \ldots \, \mathrm{d}x_k \; .$

Now, the integral is simply the surface area of the $(k-1)$-dimensional sphere with radius $r = \sqrt{y}$, which is

$\label{eq:A} A = 2 r^{k-1} \, \frac{\pi^{k/2}}{\Gamma(k/2)} \; ,$

times the infinitesimal thickness of the sphere, which is

$\label{eq:dR} \frac{\mathrm{d}r}{\mathrm{d}y} = \frac{1}{2} y^{-1/2} \quad \Leftrightarrow \quad \mathrm{d}r = \frac{\mathrm{d}y}{2 y^{1/2}} \; .$

Substituting \eqref{eq:A} and \eqref{eq:dR} into \eqref{eq:y-cdf-s3}, we have:

$\label{eq:y-cdf-s4} \begin{split} f_Y(y) \, \mathrm{d}y &= \frac{\exp \left[ -y/2 \right]}{(2 \pi)^{k/2}} \cdot A \, \mathrm{d}r \\ &= \frac{\exp \left[ -y/2 \right]}{(2 \pi)^{k/2}} \cdot 2 r^{k-1} \, \frac{\pi^{k/2}}{\Gamma(k/2)} \cdot \frac{\mathrm{d}y}{2 y^{1/2}} \\ &= \frac{1}{2^{k/2} \, \Gamma(k/2)} \cdot \frac{2 \sqrt{y}^{k-1}}{2 \sqrt{y}} \cdot \exp \left[ -y/2 \right] \, \mathrm{d}y \\ &= \frac{1}{2^{k/2} \, \Gamma(k/2)} \cdot y^{\frac{k}{2}-1} \cdot \exp \left[ -\frac{y}{2} \right] \, \mathrm{d}y \; . \end{split}$

From this, we get the final result in \eqref{eq:chi2-pdf}:

$\label{eq:y-cdf-s5} f_Y(y) = \frac{1}{2^{k/2} \, \Gamma (k/2)} \, y^{k/2-1} \, e^{-y/2} \; .$
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Metadata: ID: P197 | shortcut: chi2-pdf | author: JoramSoch | date: 2020-11-25, 05:56.