# Quod Erat Demonstrandum

## 2010/02/26

### 獨立隨機變量

Filed under: Additional / Applied Mathematics,HKALE — johnmayhk @ 7:47 上午

$X, Y, Z$ 是互相獨立（mutually independent）的隨機變數，其平均數及方差分別是 $\mu_1, \mu_2, \mu_3$$\sigma^2_1, \sigma^2_2, \sigma^2_3$，可否尋求 $Var(XY + YZ + ZX)$

$Var(XY + YZ + ZX) = Var(XY) + Var(YZ) + Var(ZX)$

$Var(W) = E(W^2) - E^2(W)$

$Var(XY + YZ + ZX)$
$= E((XY + YZ + ZX)^2) - E^2(XY + YZ + ZX)$
$= E(X^2Y^2 + Y^2Z^2 + Z^2X^2 + 2XYZ(X + Y + Z)) - (E(X)E(Y) + E(Y)E(Z) + E(Z)E(X))^2$

$P(X = x_1$ and $Y = y_1) = a$
$P(X = x_2$ and $Y = y_1) = b$
$P(X = x_1$ and $Y = y_2) = c$
$P(X = x_2$ and $Y = y_2) = d$

$P(X = x_1 | Y = y_1) = P(X = x_1 | Y = y_2)$
$P(X = x_2 | Y = y_1) = P(X = x_2 | Y = y_2)$
$P(Y = y_1 | X = x_1) = P(Y = y_1 | X = x_2)$
$P(Y = y_2 | X = x_1) = P(Y = y_2 | X = x_2)$

$\frac{a}{a + b} = \frac{c}{c + d} \Rightarrow ad = bc$

$M = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$

$\begin{pmatrix} a & b\\ c & d\end{pmatrix}$

（注意：我根本不是在證明，只是驗證某個特例。）

$M = \begin{pmatrix} a & b & c\\ d & e & f\end{pmatrix}$

$\begin{pmatrix} a & b\\ d & e\end{pmatrix}$
$\begin{pmatrix} b & c\\ e & f\end{pmatrix}$
$\begin{pmatrix} c & a\\ f & d\end{pmatrix}$
$\begin{pmatrix} a & b + c\\ d & e + f\end{pmatrix}$
$\begin{pmatrix} b & a + c\\ e & d + f\end{pmatrix}$
$\begin{pmatrix} c & a + b\\ f & d + e\end{pmatrix}$

$\begin{pmatrix} a & b & c\\ d & e & f\end{pmatrix}$

$\begin{pmatrix} a + b & c\\ d + e & f\end{pmatrix}$

$P(X^2 < a$ and $Y^2 < b) = P(X^2 < a)P(Y^2 < b)$ for all $a, b \in \mathbb{R}$

Case 1 : $a < 0$$b < 0$

$P(X^2 < a$ and $Y^2 < b) = 0$
$P(X^2 < a)P(Y^2 < b) = 0$

$P(X^2 < a$ and $Y^2 < b) = P(X^2 < a)P(Y^2 < b)$

Case 2 : $a \ge 0$$b \ge 0$

$P(X^2 < a$ and $Y^2 < b)$
$= P(-\sqrt{a} < X < \sqrt{a}$ and $-\sqrt{b} < Y < \sqrt{b})$
$= P(-\sqrt{a} < X < \sqrt{a})P(-\sqrt{b} < Y < \sqrt{b})$ （因 $X, Y$ 互相獨立）
$= P(X^2 < a)P(Y^2 < b)$

## 2 則迴響 »

1. Hi,

Thanks for an insightful and interesting sharing, which I have learnt much! I’m sure your students will benefit a great deal and be keen to take up mathematics research in future :)

Yuan xiao jie kuai le!

Cheers,
Wen Shih

迴響 由 Wen Shih — 2010/02/27 @ 6:52 上午 | 回應

• Thank you Wen Shih, glad to know you are using WordPress for blog writing!

Yuan xiao jie kuai le!

迴響 由 johnmayhk — 2010/02/28 @ 7:16 下午 | 回應