평균과 분산 성질

$Var(X)=E[(X-\mu {)}^2]$

$=\underset{x}{\mathrm{\Sigma }}(x-\mu {)}^{2}p(x)$

$=\underset{x}{\mathrm{\Sigma }}(x^2-2\mu x+{\mu }^2)p(x)$

$=\underset{x}{\mathrm{\Sigma }}{x}^{2}p(x)-2\mu \underset{x}{\mathrm{\Sigma }}xp(x)+{\mu }^2\underset{x}{\mathrm{\Sigma }}p(x)$

$=E[X^2]-2{\mu }^2+{\mu }^2$

$=E[X^2]-{\mu }^2$

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$E[X+Y]=E[X]+E[Y]$

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$E[aX+b]=aE[X]+b=a\mu +b$

 

$Var(aX+b)=E[(aX+b-a\mu -b{)}^2]=a^{2}E[(X-\mu {)}^2]=a^{2}Var(X)$

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Theorem: If $X$ & $Y$ are independent, then $E[XY]=E[X]E[Y]$

Proof:

Let $x_i,y_i,i=1,2,\dots$ be the possible values of $X,Y$.

 

$E[XY]=\underset{i}{\mathrm{\Sigma }}\underset{j}{\mathrm{\Sigma }}{x}_i{y}_{j}P(X=x_i\wedge Y=y_j)$ $=\underset{i}{\mathrm{\Sigma }}\underset{j}{\mathrm{\Sigma }}{x}_i{y}_{j}P(X=x_i)P(Y=y_j)$ $=\underset{i}{\mathrm{\Sigma }}{x}_{i}P(X=x_i)\underset{j}{\mathrm{\Sigma }}{y}_{j}P(Y=y_j)$ $=E[X]E[Y]$

 

Note: Not true in general;

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In general: $Var[X+Y]\ne Var[X]+Var[Y]$

Theorem: If $X$ & $Y$ are independent, then $Var[X+Y]=Var[x]+Var[y]$

Proof:

$\hat{X}=X-E[X],\hat{Y}=Y-E[Y]$ 
$E[\hat{X}]=0,E[\hat{Y}]=0$
$Var[\hat{X}]=Var[X],Var[\hat{Y}]=Var[Y]$

 

$Var[X+Y]=Var[\hat{X}+\hat{Y}]$ 
$=E[(\hat{X}+\hat{Y}{)}^2]-(E[\hat{X}+\hat{Y}]{)}^2$
$=E[{\hat{X}}^2+2\hat{X})\hat{Y}+{\hat{Y}}^2]-0$
$=E[{\hat{X}}^2]+2E[\hat{X})\hat{Y}]+E[{\hat{Y}}^2]$
$=Var[\hat{X}]+0+Var[\hat{Y}]$
$=Var[X]+Var[Y]$

 

$Std[X+Y]=\sqrt{Var[X]+Var[Y]}$