Derivation of the probability density function of F distribution

The derivation process is compiled from https://www.bilibili.com/video/BV1qf4y1R7FA .

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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

已知 $U$ obeys chi-square distribution $\chi^2(n_1)$ ， $V$ obeys chi-square distribution $\chi^2(n_2)$ , and $UV$ and $V$ are independent of each other.

求 $F=\frac{U/n_1}{V/n_2}$ 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 $X=U/n_1$ Sum $Y=V/n_2$ The probability density function of

Note that $X=U/n_1$ Sum $Y=V/n_2$ have a similar structure. We just need to calculate $The probability density function of X$ , you can use the same method to getThe probability density function of $Y.$

with $n$ independent and identically distributed random variables $Z_1,Z_2,\cdots,Z_n$ , they all obey the standard normal distribution $N (0, 1)$ ，则 $Z=\sum\limits_{i=1}^{n}Z_i^2$ Subject to chi-square distribution $\chi^2(n)$ . We next use the distribution function method to calculate $W =$ Probability density function $p_{W}(w)$ $of Z$ $/$ $n$ $p_{W} (w)$ 。

当 $w\leqslant 0$ , because $W=Z/n=(\sum\limits_{i=1}^{n}Z_i^2)/n$ is always non-negative, so $P (W < w) = 0$ , so $p_{W}(w)=0$ . Below for $w > 0$ is calculated.
$\begin{aligned} F_{W}(w) &= P(W<w) \\ &= P(Z<nw) \\ &= \int_{0}^{nw}\frac{1}{2 ^{n/2}\Gamma(n/2)}e^{-\frac{x}{2}}x^{\frac{n}{2}-1}\text{d}x \thinspace. \end{aligned}$ Form
$\begin {aligned} p_{W}(w) &= F_{W}'(w) \\ &= \frac{1}{2^{n/2 }\Gamma(n/2)}e^{-\frac{nw}{2}}(nw)^{\frac{n}{2}-1}\cdot n \\ &= \frac{1} {2^{n/2}\Gamma(n/2)}n^{\frac{n}{2}} e^{-\frac{nw}{2}}w^{\frac{n}{ 2}-1} \thinspace。 \end{aligned}$
综上，有
$p_{W}(w) = \left\{\begin{array}{cc} \frac{1}{2^{n/2}\Gamma(n/2)}n^{\frac{n}{2}} e^{-\frac{nw}{2}}w^{\frac{n}{2}-1}, & w>0 \\ 0, & w\leqslant 0 \end{array}\right. \thinspace。$

因此
$p_{X}(x) = \left\{\begin{array}{cc} \frac{1}{2^{n_1/2}\Gamma(n_1/2)}n_1^{\frac{n_1}{2}} e^{-\frac{n_1x}{2}}x^{\frac{n_1}{2}-1}, & x>0 \\ 0, & x\leqslant 0 \end{array}\right. \thinspace，$

$p_{Y} (y) = \left\{\begin{array}{cc} \frac{1}{2^{n_2/2}\Gamma(n_2/2)}n_2^{\frac{n_2}{2}}e ^{-\frac{n_2y}{2}}y^{\frac{n_2}{2}-1}, & y>0 \\ 0, & y\leqslant 0 \end{array}\right. \thinspace。$

Then calculate $XYF=\frac{X}{Y}$ Probability density function $p_{F}(t)$

当 $t\leqslant 0$ , because $X=U/n_1$ Sum $Y=V/n_2$ Constantly non-negative, so $P(F<t)=P(\frac{X}{Y}<t)=0$ , so $p_{F}(t)=0$ . Below for $t > 0$ is calculated.

Cause $UV$ and $V$ are independent of each other, so $X=U/n_1$ Given $Y=V/n_2$ are also independent of each other, so $(X, Y)$ joint probability density function $p (x, y)$ 满足
$p_{X}(x)p_{Y}(y) \thinspace。$ Then by the lemma, we have
$\begin{aligned} p_{F}(t) &\quad\,=\quad\, \int_{-\infty}^{+\infty}p(ty, y)|y|\text{d}y \\ &\quad\,=\quad\, \int_{-\infty}^{+\infty}p_{X}(ty)p_{Y}(y)|y|\text{d}y \\ &\quad\,=\quad\, \int_{0}^{+\infty}p_{X}(ty)p_{Y}(y)y\text{d}y \\ &\quad\,=\quad\, \int_{0}^{+\infty}\frac{1}{2^{n_1/2}\Gamma(n_1/2)}n_1^{\frac{n_1}{2}} e^{-\frac{n_1ty}{2}}(ty)^{\frac{n_1}{2}-1}\cdot \frac{1}{2^{n_2/2}\Gamma(n_2/2)}n_2^{\frac{n_2}{2}} e^{-\frac{n_2y}{2}}y^{\frac{n_2}{2}-1}\cdot y\text{d}y \\ &\quad\,=\quad\, \frac{1}{2^{(n_1+n_2)/2}\Gamma(n_1/2)\Gamma(n_2/2)}n_1^{\frac{n_1}{2}}n_2^{\frac{n_2}{2}}t^{\frac{n_1}{2}-1} \int_{0}^{+\infty}e^{-\frac{n_1t+n_2}{2}y}y^{\frac{n_1+n_2}{2}-1}\text{d}y \\ &\overset{y=\frac{2z}{n_1t+n_2}}{=} \frac{1}{2^{(n_1+n_2)/2}\Gamma(n_1/2)\Gamma(n_2/2)} n_1^{\frac{n_1}{2}}n_2^{\frac{n_2}{2}} t^{\frac{n_1}{2}-1}\cdot (\frac{2}{n_1t+n_2})^{\frac{n_1+n_2}{2}} \int_{0}^{+\infty}e^{-z}z^{\frac{n_1+n_2}{2}-1}\text{d}z \\ &\quad\,=\quad\, \frac{\Gamma[(n_1+n_2)/2]}{\Gamma(n_1/2)\Gamma(n_2/2)}n_1^{\frac{n_1}{2}}n_2^{\frac{n_2}{2}} t^{\frac{n_1}{2}-1} (n_1t+n_2)^{-\frac{n_1+n_2}{2}} \\ &\quad\,=\quad\, \frac{\Gamma[(n_1+n_2)/2]}{\Gamma(n_1/2)\Gamma(n_2/2)}(\frac{n_1}{n_2})^{\frac{n_1}{2}} t^{\frac{n_1}{2}-1} (1+\frac{n_1}{n_2}t)^{-\frac{n_1+n_2}{2}} \thinspace。 \end{aligned}$ 综上，我们得到
$p_{F}(t) = \left\{\begin{array}{cc} \frac{\Gamma[(n_1+n_2)/2]}{\Gamma(n_1/2)\Gamma(n_2/2)}(\frac{n_1}{n_2})^{\frac{n_1}{2}} t^{\frac{n_1}{2}-1} (1+\frac{n_1}{n_2}t)^{-\frac{n_1+n_2}{2}}, & t>0 \\ 0, & t\leqslant 0 \end{array}\right. \thinspace。$