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

Theorem: Let $T$ be a random variable following a t-distribution:

$\label{eq:t} T \sim t(\nu) \; .$

Then, the probability density function of $T$ is

$\label{eq:t-pdf} f_T(t) = \frac{\Gamma\left( \frac{\nu+1}{2} \right)}{\Gamma\left( \frac{\nu}{2} \right) \cdot \sqrt{\nu \pi}} \cdot \left( \frac{t^2}{\nu}+1 \right)^{-\frac{\nu+1}{2}} \; .$

Proof: A t-distributed random variable is defined as the ratio of a standard normal random variable and the square root of a chi-squared random variable, divided by its degrees of freedom

$\label{eq:t-def} X \sim \mathcal{N}(0,1), \; Y \sim \chi^2(\nu) \quad \Rightarrow \quad T = \frac{X}{\sqrt{Y/\nu}} \sim t(\nu)$

where $X$ and $Y$ are independent of each other.

$\label{eq:snorm-pdf} f_X(x) = \frac{1}{\sqrt{2 \pi}} \cdot e^{-\frac{x^2}{2}}$ $\label{eq:chi2-pdf} f_Y(y) = \frac{1}{\Gamma\left( \frac{\nu}{2} \right) \cdot 2^{\nu/2}} \cdot y^{\frac{\nu}{2}-1} \cdot e^{-\frac{y}{2}} \; .$

Define the random variables $T$ and $W$ as functions of $X$ and $Y$

$\label{eq:TW-XY} \begin{split} T &= X \cdot \sqrt{\frac{\nu}{Y}} \\ W &= Y \; , \end{split}$

such that the inverse functions $X$ and $Y$ in terms of $T$ and $W$ are

$\label{eq:XY-TW} \begin{split} X &= T \cdot \sqrt{\frac{W}{\nu}} \\ Y &= W \; . \end{split}$

This implies the following Jacobian matrix and determinant:

$\label{eq:XY-TW-jac} \begin{split} J &= \left[ \begin{matrix} \frac{\mathrm{d}X}{\mathrm{d}T} & \frac{\mathrm{d}X}{\mathrm{d}W} \\ \frac{\mathrm{d}Y}{\mathrm{d}T} & \frac{\mathrm{d}Y}{\mathrm{d}W} \end{matrix} \right] = \left[ \begin{matrix} \sqrt{\frac{W}{\nu}} & \frac{T}{2 \nu \sqrt{W/\nu}} \\ 0 & 1 \end{matrix} \right] \\ \lvert J \rvert &= \sqrt{\frac{W}{\nu}} \; . \end{split}$

Because $X$ and $Y$ are independent, the joint density of $X$ and $Y$ is equal to the product of the marginal densities:

$\label{eq:f-XY} f_{X,Y}(x,y) = f_X(x) \cdot f_Y(y) \; .$

With the probability density function of an invertible function, the joint density of $T$ and $W$ can be derived as:

$\label{eq:f-TW-s1} f_{T,W}(t,w) = f_{X,Y}(x,y) \cdot \lvert J \rvert \; .$

Substituting \eqref{eq:XY-TW} into \eqref{eq:snorm-pdf} and \eqref{eq:chi2-pdf}, and then with \eqref{eq:XY-TW-jac} into \eqref{eq:f-TW-s1}, we get:

$\label{eq:f-TW-s2} \begin{split} f_{T,W}(t,w) &= f_X\left( t \cdot \sqrt{\frac{w}{\nu}} \right) \cdot f_Y(w) \cdot \lvert J \rvert \\ &= \frac{1}{\sqrt{2 \pi}} \cdot e^{-\frac{\left( t \cdot \sqrt{\frac{w}{\nu}} \right)^2}{2}} \cdot \frac{1}{\Gamma\left( \frac{\nu}{2} \right) \cdot 2^{\nu/2}} \cdot w^{\frac{\nu}{2}-1} \cdot e^{-\frac{w}{2}} \cdot \sqrt{\frac{w}{\nu}} \\ &= \frac{1}{\sqrt{2 \pi \nu} \cdot \Gamma\left( \frac{\nu}{2} \right) \cdot 2^{\nu/2}} \cdot w^{\frac{\nu+1}{2}-1} \cdot e^{-\frac{w}{2} \left( \frac{t^2}{\nu} + 1 \right)} \; . \end{split}$

The marginal density of $T$ can now be obtained by integrating out $W$:

$\label{eq:f-T-s1} \begin{split} f_T(t) &= \int_{0}^{\infty} f_{T,W}(t,w) \, \mathrm{d}w \\ &= \frac{1}{\sqrt{2 \pi \nu} \cdot \Gamma\left( \frac{\nu}{2} \right) \cdot 2^{\nu/2}} \cdot \int_{0}^{\infty} w^{\frac{\nu+1}{2}-1} \cdot \mathrm{exp}\left[ -\frac{1}{2}\left( \frac{t^2}{\nu}+1 \right) w \right] \, \mathrm{d}w \\ &= \frac{1}{\sqrt{2 \pi \nu} \cdot \Gamma\left( \frac{\nu}{2} \right) \cdot 2^{\nu/2}} \cdot \frac{\Gamma\left( \frac{\nu+1}{2} \right)}{\left[ \frac{1}{2}\left( \frac{t^2}{\nu}+1 \right) \right]^{(\nu+1)/2}} \cdot \int_{0}^{\infty} \frac{\left[ \frac{1}{2}\left( \frac{t^2}{\nu}+1 \right) \right]^{(\nu+1)/2}}{\Gamma\left( \frac{\nu+1}{2} \right)} \cdot w^{\frac{\nu+1}{2}-1} \cdot \mathrm{exp}\left[ -\frac{1}{2}\left( \frac{t^2}{\nu}+1 \right) w \right] \, \mathrm{d}w \; . \end{split}$

At this point, we can recognize that the integrand is equal to the probability density function of a gamma distribution with

$\label{eq:f-W-gam-ab} a = \frac{\nu+1}{2} \quad \text{and} \quad b = \frac{1}{2}\left( \frac{t^2}{\nu}+1 \right) \; ,$

and because a probability density function integrates to one, we finally have:

$\label{eq:f-T-s2} \begin{split} f_T(t) &= \frac{1}{\sqrt{2 \pi \nu} \cdot \Gamma\left( \frac{\nu}{2} \right) \cdot 2^{\nu/2}} \cdot \frac{\Gamma\left( \frac{\nu+1}{2} \right)}{\left[ \frac{1}{2}\left( \frac{t^2}{\nu}+1 \right) \right]^{(\nu+1)/2}} \\ &= \frac{\Gamma\left( \frac{\nu+1}{2} \right)}{\Gamma\left( \frac{\nu}{2} \right) \cdot \sqrt{\nu \pi}} \cdot \left( \frac{t^2}{\nu}+1 \right)^{-\frac{\nu+1}{2}} \; . \end{split}$
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Metadata: ID: P263 | shortcut: t-pdf | author: JoramSoch | date: 2021-10-12, 08:15.