91影视

$Theprobabilityof{X}_1....X_2$

Given: $X_1....Xn$

Formula to used: ${\mathrm{F}}_{\mathrm{X}}\left(\mathrm{X}\right)=\mathrm{P}(\mathrm{X}鈮�\mathrm{x})$

Calculation :

X be a continuous random variable

$\mathrm{X}~Uni(1,2,3,鈥�n)$

The total combination${\mathrm{n}}^{\mathrm{n}}$

The random vector

${\mathrm{X}}_1,{\mathrm{X}}_2,鈥�鈥�鈥�{\mathrm{X}}_{\mathrm{n}}$

Every random variable assumes one and only one variable The probability is

$p=\frac{n!}{n^n}$

The probability of mass function of$X_1,X_2....X_n$

Given: X be a continuous random variable

$\mathrm{X}~Uni(1,2,3,鈥�n)$

The total combination ${\mathrm{n}}^{\mathrm{n}}$

The random vector ${\mathrm{X}}_1,{\mathrm{X}}_2,鈥�鈥�鈥�{\mathrm{X}}_{\mathrm{n}}$

Every random variable assumes one and only one variable The probability is

$\begin{array}{r}P\left(\underset{蟽\{1,2,3,鈥�n\}}{鈭�}\underset{i=1}{\overset{n}{鈭�}}鈥�X_i=蟽\left(i\right)\right)=\underset{蟽\{1,2,3,鈥�n\}}{鈭�}鈥�\underset{i=1}{\overset{n}{鈭�}}鈥�P\left(X_i=蟽\left(i\right)\right)\\ \underset{i=1}{\overset{n}{鈭�}}鈥�P\left(X_i=蟽\left(i\right)\right)=\underset{i=1}{\overset{n}{鈭�}}鈥�p_{蟽\left(i\right)}=p_1{p}_2鈥�鈥�.p_{n鈭�1}{p}_n\\ \underset{蟽\{1,2,3,鈥�n\}}{鈭�}鈥�\underset{i=1}{\overset{n}{鈭�}}鈥�P\left(X_i=蟽\left(i\right)\right)=\underset{蟽\{1,2,3,鈥�n\}}{鈭�}鈥�p_1{p}_2鈥�鈥�.p_{n鈭�1}{p}_n\\ \underset{蟽\{1,2,3,鈥�n\}}{鈭�}鈥�\underset{i=1}{\overset{n}{鈭�}}鈥�P\left(X_i=j\right)=n!p_1{p}_2鈥�鈥�.p_{n鈭�1}{p}_n\\ P\left(X_i=j\right)=n!p_1{p}_2鈥�鈥�.p_{n鈭�1}{p}_n\end{array}$

Therefore the probability of mass function of$P\left(X_i=j\right)=n!p_1{p}_2鈥�鈥�.p_{n-1}{p}_n$