Definition – (Student’s) t-distribution.

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1. Definition

Definition

Let $T$ be a random variable with pdf

$p_T(t) = \frac{\Gamma(\frac{n+1}{2}) }{\sqrt{n\pi} \Gamma(\frac{n}{2})} (1+\frac{t^2}{n})^{-\frac{n+1}{2}}$

In this instance, we say that $T$ follows a t-distribution with degrees of freedom $n\in\mathbb{N}\setminus{0}$ .

Derivation

$Z \sim N(0,1)$ be normally distributed and let $U \sim \chi_n^2$ follow a Chi-square distribution with $n$ degrees of freedom. We will derive the distribution of

$X = \frac{Z}{\sqrt{\frac{U}{n}}}$ .

Firstly, as very simple application of the change of variable formula for $U \sim \chi_n^2$ , with $g(U) = \sqrt{\frac{U}{n}} = Y_1$ gives

$p_{Y_1}(y_1) \sim \frac{2}{2^{\frac{n}{2}}\cdot \Gamma(\frac{n}{2})} \cdot n^{\frac{n}{2}} x^{n-1} e^{-\frac{x^2}{2}}$

Then, we can make use of the formula for the ratio of two random variables,

$\begin{aligned} p(T = \frac{Z}{Y_1}) &= \int\limits_{-\infty}^{\infty} |y_1| p(y_1)p(y_1)dy_1 \\\\ &= \int\limits_{-\infty}^{\infty} y_1 (\frac{2n^{\frac{n}{2}}}{2^{\frac{n}{2}} \Gamma(\frac{n}{2})} y_1^{n-1} e^{-\frac{n}{2} y_1^2}) (\frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}}y_1^2)dy_1 \\\\ &= \frac{n^{\frac{n}{2}}}{2^{\frac{n-1}{2}} \Gamma(\frac{n}{2}) \sqrt{n\pi}} \int\limits_{-\infty}^{\infty} y_1^n \cdot e^{-\frac{1}{2} (n+t^2)y_1^2}dy_1 \end{aligned}$

But the integral can be worked out via substitution. Let $y_1 = \sqrt{v}$ , then, the integral becomes

$\frac{1}{2} \int\limits_{-\infty}^{\infty} v^{\frac{n-1}{2}} e^{-\frac{1}{2} (n+t^2)v}dv$

Which is just a gamma distribution (without it’s normalising constant), with $\alpha = \frac{n+1}{2}$ and $\beta = \frac{n+t}{2}$ . This means that

$\frac{1}{2} \int\limits_{-\infty}^{\infty} v^{\frac{n-1}{2}} e^{-\frac{1}{2} (n+t^2)v}dv = \frac{\Gamma(\frac{n+1}{2})}{2} (\frac{n+t^2}{2})^{-\frac{n+1}{2}}$

$\frac{n^{\frac{n}{2}}}{2^{\frac{n-1}{2}} \Gamma(\frac{n}{2}) \sqrt{n\pi}} \int\limits_{-\infty}^{\infty} y_1^n \cdot e^{-\frac{1}{2} (n+t^2)y_1^2}dy_1 \\\\ = \frac{n^{\frac{n}{2}}}{2^{\frac{n-1}{2}} \Gamma(\frac{n}{2}) \sqrt{n\pi}} \cdot \frac{\Gamma(\frac{n+1}{2})}{2} (\frac{n+t^2}{2})^{-\frac{n+1}{2}}$

Which, after cancelling out some of the factors, gives the pdf we are looking for,

$p_T(t) = \frac{\Gamma(\frac{n+1}{2}) }{\sqrt{n\pi} \Gamma(\frac{n}{2})} (1+\frac{t^2}{n})^{-\frac{n+1}{2}} \quad \square$