正态分布
对于x ∼ N ( μ , σ 2 ) x\sim N(\mu,\sigma^2) x ∼ N ( μ , σ 2 )
f ( x ) = 1 2 π σ e − ( x − μ ) 2 2 σ 2 f(x)=\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x-\mu)^2}{2\sigma^2}}
f ( x ) = 2 π σ 1 e − 2 σ 2 ( x − μ ) 2
二维正态分布
对于( X , Y ) ∼ N ( μ 1 , μ 2 , σ 1 2 , σ 2 2 , r ) (X,Y)\sim N(\mu_1,\mu_2,\sigma_1^2,\sigma_2^2,r) ( X , Y ) ∼ N ( μ 1 , μ 2 , σ 1 2 , σ 2 2 , r )
f ( x , y ) = 1 2 π σ 1 σ 2 1 − r 2 exp { − 1 2 ( 1 − r 2 ) [ ( x − μ 1 ) 2 σ 1 2 − 2 r ( x − μ 1 ) ( y − μ 2 ) σ 1 σ 2 + ( y − μ 2 ) 2 σ 2 2 ] } f(x,y)=\frac{1}{2\pi\sigma_1\sigma_2\sqrt{1-r^2}}\exp\left\{-\frac{1}{2(1-r^2)} \left[\frac{(x-\mu_1)^2}{\sigma_1^2}-2r\frac{(x-\mu_1)(y-\mu_2)}{\sigma_1\sigma_2}+\frac{(y-\mu_2)^2}{\sigma_2^2} \right]\right\}
f ( x , y ) = 2 π σ 1 σ 2 1 − r 2 1 exp { − 2 ( 1 − r 2 ) 1 [ σ 1 2 ( x − μ 1 ) 2 − 2 r σ 1 σ 2 ( x − μ 1 ) ( y − μ 2 ) + σ 2 2 ( y − μ 2 ) 2 ] }
有如下结论：
若Z = a X + b Y Z=aX+bY Z = a X + bY Z ∼ N ( E ( Z ) , D ( Z ) ) Z\sim N(E(Z),D(Z)) Z ∼ N ( E ( Z ) , D ( Z )) X , Y X,Y X , Y
例题
Let X ∼ N ( 0 , 1 ) X\sim N(0,1) X ∼ N ( 0 , 1 ) X = x X = x X = x Y ∼ N ( x , 4 ) Y\sim N(x,4) Y ∼ N ( x , 4 ) Y Y Y ‾ \underline{\quad}
Solve: First, calculate E ( Y ) = E X ( E Y ( Y ∣ X ) ) = E ( X ) = 0 note: we view X in E(Y|X) as a constant, and get a expression dependent on X Then, calculate E ( Y 2 ) = E X ( E Y ( Y 2 ∣ X ) ) = E X ( D Y ( Y 2 ∣ X ) + E Y 2 ( Y ∣ X ) ) = E X ( 4 + X 2 ) = 4 + E X ( X 2 ) = 4 + D X ( X ) + E X 2 ( X ) = 5 Therefore, D ( Y ) = E ( Y 2 ) − E 2 ( Y ) = 5 − 0 = 5 \begin{aligned}
\text{Solve:}\\
&\text{First, calculate }E(Y)=E_X(E_Y(Y|X))=E(X)=0\\
&\text{note: we view $X$ in E(Y|X) as a constant, and get a expression dependent on $X$}\\
&\text{Then, calculate }E(Y^2)\\
&\qquad =E_X(E_Y(Y^2|X))\\
&\qquad =E_X(D_Y(Y^2|X)+E_Y^2(Y|X))\\
&\qquad =E_X(4+X^2)\\
&\qquad =4+E_X(X^2)\\
&\qquad =4+D_X(X)+E_X^2(X)\\
&\qquad =5\\
&\text{Therefore, }D(Y)=E(Y^2)-E^2(Y)=5-0=5
\end{aligned}
Solve: First, calculate E ( Y ) = E X ( E Y ( Y ∣ X )) = E ( X ) = 0 note: we view X in E(Y|X) as a constant, and get a expression dependent on X Then, calculate E ( Y 2 ) = E X ( E Y ( Y 2 ∣ X )) = E X ( D Y ( Y 2 ∣ X ) + E Y 2 ( Y ∣ X )) = E X ( 4 + X 2 ) = 4 + E X ( X 2 ) = 4 + D X ( X ) + E X 2 ( X ) = 5 Therefore, D ( Y ) = E ( Y 2 ) − E 2 ( Y ) = 5 − 0 = 5
*结论：全方差公式
D ( Y ) = E [ D ( Y ∣ X ) ] + D ( E [ Y ∣ X ] ) D(Y)=E[D(Y∣X)]+D(E[Y∣X])
D ( Y ) = E [ D ( Y ∣ X )] + D ( E [ Y ∣ X ])
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