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Joram Soch |
BCCN Berlin |
joram.soch@bccn-berlin.de |
2021-10-12 01:15:00 -0700 |
Probability density function of the t-distribution |
Probability Distributions |
Univariate continuous distributions |
t-distribution |
Probability density function |
| authors |
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Computation Empire |
2021 |
Student's t Distribution: Derivation of PDF |
YouTube |
retrieved on 2021-10-11 |
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P263 |
t-pdf |
JoramSoch |
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.
The probability density function of the standard normal distribution is
$$ \label{eq:snorm-pdf}
f_X(x) = \frac{1}{\sqrt{2 \pi}} \cdot e^{-\frac{x^2}{2}}
$$
and the probability density function of the chi-squared distribution is
$$ \label{eq:chi2-pdf}
f_Y(y) = \frac{1}{\Gamma\left( \frac{\nu}{2} \right) \cdot 2^{\nu/2}} \cdot y^{\frac{\nu}{2}-1} \cdot e^{-\frac{y}{2}} ; .
$$
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:
$$ \label{eq:f-T-s2}
\begin{split}
f_T(t) &= \frac{1}{\sqrt{2 \pi \nu} \cdot \Gamma\left( \frac{\nu}{2} \right) \cdot 2^{\nu/2}} \cdot \frac{\Gamma\left( \frac{\nu+1}{2} \right)}{\left[ \frac{1}{2}\left( \frac{t^2}{\nu}+1 \right) \right]^{(\nu+1)/2}} \\
&= \frac{\Gamma\left( \frac{\nu+1}{2} \right)}{\Gamma\left( \frac{\nu}{2} \right) \cdot \sqrt{\nu \pi}} \cdot \left( \frac{t^2}{\nu}+1 \right)^{-\frac{\nu+1}{2}} ; .
\end{split}
$$