Derivation of probability density function of t distribution

The derivation process is compiled from https://www.bilibili.com/video/BV1s54y1S7Ji .

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Preliminary knowledge

$\Gamma$ function (gamma function)

定义： $\Gamma(s)=\int_{0}^{+\infty}e^{-t}t^{s-1}\text{d}t$
Supplementary formula: $\Gamma(s+1)=s\Gamma(s),\space(s>0)$
Several important values: $\Gamma(1)=1$ ， $\Gamma(\frac{1}{2})=\sqrt{\pi}$

standard normal distribution

Probability density function: $\varphi(x)=\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2 }}$
Distribution function: $\phi(x)=\int_{-\infty}^{x}\frac{1}{\sqrt{2\pi }}e^{-\frac{t^2}{2}}\text{d}t$

chi-square distribution

Probability density function: $p(x)=\left\{\begin{ array}{cc} \frac{1}{2^{n/2}\Gamma(n/2)}e^{-\frac{x}{2}}x^{\frac{n}{2} -1}, & x>0 \\ 0, & x\leqslant 0 \end{array}\right.$

derivation target

Known $X$ obeys standard normal distribution $N (0, 1)$ ， $Y$ obeys chi-square distribution $\chi^2(n)$ , and $X$ 与 $Y$ are independent of each other.

求 $T=\frac{X}{\sqrt{Y/n}}$ The probability density function of .

Lemma: Distribution of Quotients of Continuous Random Variables

设 $(X, Y)$ is a two-dimensional random variable, and its joint density function is $p (x, y)$ ，则 $Z =$ Density function $p_{Z}(z)$ $of X$ $/$ $Y$ $p_{Z} (z)$ satisfyp
$p_{Z}(z) = \int_{-\infty}^{+\infty}p(zy , y)|y|\text{d}y \thinspace。$
prove

Using the distribution function method, first find $The distribution function of Z$ , and then derive the distribution function to get $p_{Z}(z)$ 。

againstzz $The positive and negative nature of z$ is discussed, and the integral area is obtained as shown in the shaded part in the figure below.

We found that these two cases can be combined, and the integration area is $\{(y,x) \space|\space y<0 \space\wedge\space x \geqslant zy\}$ $\space\cup\space$ $\{(y,x) \space|\space y>0 \space\wedge\space x\leqslant zy\}$ . So there is
$\begin{aligned} F_{Z}(z) &\space\space=\space\space P(Z<z) \\ &\space\space=\space\space P(\frac{X}{Y}<z) \\ &\space\space=\space\space \int_{-\infty}^{0}\text{d}y\int_{yz}^{+\infty}p(x,y)\text{d}x + \int_{0}^{+\infty}\text{d}y\int_{-\infty}^{yz}p(x,y)\text{d}x \\ &\overset{x=uy}{=} \int_{-\infty}^{0}\text{d}y\int_{z}^{+\infty}p(uy,y)y\text{d}u + \int_{0}^{+\infty}\text{d}y\int_{-\infty}^{z}p(uy,y)y\text{d}u \\ &\space\space=\space\space \int_{z}^{+\infty}\text{d}u\int_{-\infty}^{0}p(uy,y)y\text{d}y + \int_{-\infty}^{z}\text{d}u\int_{0}^{+\infty}p(uy,y)y\text{d}y \thinspace。 \end{aligned}$ 因此
$\begin{aligned} p_{Z}(z) &= F_{Z}'(z) \\ &= -\int_{-\infty}^{0}p(zy,y)y\text{d}y + \int_{0}^{+\infty}p(zy,y)y\text{d}y \\ &= \int_{-\infty}^{+\infty}p(zy,y)|y|\text{d}y \thinspace。 \end{aligned}$

The derivation process

First calculate $W=\sqrt{Y/n}$ The probability density function $p_{W}(w)$

Again, we use the distribution function approach. When $w\leqslant 0$ , notice that $W=\sqrt{Y/n}$ Constantly non-negative, so $P (W < w) = 0$ , so $p_{W}(w)=0$ . Below for $w > 0$ is calculated.
$\begin{aligned} F_{W}(w) &= P(W<w) \\ &= P(Y<nw^2) \\ &= \int_{0}^{nw^2}\frac{ 1}{2^{n/2}\Gamma(n/2)}e^{-\frac{x}{2}}x^{\frac{n}{2}-1}\text{d} x \thinspace. \end{aligned}$ Form
$\begin{aligned} p_{W}(w) &= F_{W}'(w) \\ &= \frac{1}{2^ {n/2}\Gamma(n/2)}e^{-\frac{nw^2}{2}}(nw^2)^{\frac{n}{2}-1}\cdot 2nw \ \ &= \frac{1}{\Gamma(n/2)}2^{1-\frac{n}{2}}n^{\frac{n}{2}} e^{-\frac{ nw^2}{2}}w^{n-1} \thinspace。\end{aligned}$ 综上，有
$p_{W}(w) = \left\{\begin{array}{cc} \frac{1}{\Gamma(n/2)}2^{1-\frac{n}{2}}n^{\frac{n}{2}} e^{-\frac{nw^2}{2}}w^{n-1}, & w>0 \\ 0, & w\leqslant 0 \end{array}\right. \thinspace。$

Then calculate $XWT=\frac{X}{W}$ Probability density function $p_{T}(t)$

because $X$ 与 $Y$ are independent of each other, so $X$ given $W=\sqrt{Y/n}$ are also independent of each other, so $(X, W)$ joint probability density function $p (x, w)$ 满足
$p_{X}(x)p_{W}(w) \thinspace。$ Then by the lemma, we have
$\begin{aligned} p_{T}(t) &\quad\space\,=\quad\space\, \int_{-\infty}^{+\infty}p(tw, w)|w|\text{d}w \\ &\quad\space\,=\quad\space\, \int_{-\infty}^{+\infty}p_{X}(tw)p_{W}(w)|w|\text{d}w \\ &\quad\space\,=\quad\space\, \int_{0}^{+\infty}p_{X}(tw)p_{W}(w)w\text{d}w \\ &\quad\space\,=\quad\space\, \int_{0}^{+\infty} \frac{1}{\sqrt{2\pi}}e^{-\frac{t^2w^2}{2}}\cdot \frac{1}{\Gamma(\frac{n}{2})}2^{1-\frac{n}{2}}n^{\frac{n}{2}} e^{-\frac{nw^2}{2}}w^{n-1}\cdot w\text{d}w \\ &\quad\space\,=\quad\space\, \frac{1}{\sqrt{\pi}\cdot \Gamma(\frac{n}{2})}2^{\frac{1-n}{2}}n^{\frac{n}{2}} \int_{0}^{+\infty}e^{-\frac{n+t^2}{2}w^2}w^n\text{d}w \\ &\overset{w=\sqrt{\frac{2z}{n+t^2}}}{=} \frac{1}{\sqrt{\pi}\cdot\Gamma(\frac{n}{2})}2^{\frac{1-n}{2}}n^{\frac{n}{2}}\cdot \frac{1}{2}(\frac{2}{n+t^2})^{\frac{n+1}{2}} \int_{0}^{+\infty}e^{-z}z^{\frac{n-1}{2}}\text{d}z \\ &\quad\space\,=\quad\space\, \frac{\Gamma(\frac{n+1}{2})}{\sqrt{\pi}\cdot \Gamma(\frac{n}{2})}\cdot n^{\frac{n}{2}}(\frac{1}{n+t^2})^{\frac{n+1}{2}} \\ &\quad\space\,=\quad\space\, \frac{\Gamma(\frac{n+1}{2})}{\sqrt{n\pi}\cdot \Gamma(\frac{n}{2})} (1+\frac{t^2}{n})^{-\frac{n+1}{2}} \thinspace。 \end{aligned}$ The derivation is complete.