二元正态分布的概率密度函数

二元正态分布随机变量

如果随机变量 X X X Y Y Y的联合PDF为
p X , Y ( x , y ) = 1 2 π σ x σ Y 1 − p 2 exp ⁡ { − ( x − μ X ) 2 σ X 2 + ( y − μ Y ) 2 σ Y 2 − 2 ρ ( x − μ X ) ( y − μ Y ) σ X Σ Y 2 ( 1 − ρ 2 ) } p_{X,Y}(x,y)=\frac{1}{2\pi \sigma_x \sigma_Y \sqrt{1-p^2}}\exp\left\{-\frac{\frac{(x-\mu_X)^2}{\sigma_X^2}+\frac{(y-\mu_Y)^2}{\sigma_Y^2}-\frac{2\rho(x-\mu_X)(y-\mu_Y)}{\sigma_X\Sigma_Y}}{2(1-\rho^2)}\right\} pX,Y(x,y)=2πσxσY1p2 1exp2(1ρ2)σX2(xμX)2+σY2(yμY)2σXΣY2ρ(xμX)(yμY)
则称 X X X Y Y Y满足二元正态分布,其中 X ∼ N ( μ X , σ X 2 ) , Y ∼ N ( μ Y , σ Y 2 ) X \sim N(\mu_X,\sigma_X^2),Y \sim N(\mu_Y,\sigma_Y^2) XN(μX,σX2)YN(μY,σY2),协方差 C X Y = E { X Y } − μ X μ Y C_{XY}={\rm E}\{XY\}-\mu_X\mu_Y CXY=E{XY}μXμY,相关系数 ρ = C X Y σ X σ Y \rho=\frac{C_{XY}}{\sigma_X\sigma_Y} ρ=σXσYCXY

进一步进行归一化处理,设
V = X − μ x σ X ,   W = Y − μ Y σ Y V=\frac{X-\mu_x}{\sigma_X},\ W=\frac{Y-\mu_Y}{\sigma_Y} V=σXXμx, W=σYYμY
则有
p V , W ( v , w ) = 1 2 π 1 − ρ 2 exp ⁡ { − v 2 − 2 ρ v w + w 2 2 ( 1 − ρ 2 ) } . p_{V,W}(v,w)=\frac{1}{2\pi \sqrt{1-\rho^2}}\exp \{-\frac{v^2-2\rho v w+w^2}{2(1-\rho^2)}\}. pV,W(v,w)=2π1ρ2 1exp{2(1ρ2)v22ρvw+w2}.

下面来求 V , W V,W V,W的条件PSD。由于
v 2 + w 2 − 2 ρ v w = ( w − ρ v ) 2 + v 2 ( 1 − ρ 2 ) v^2+w^2-2\rho vw=(w-\rho v)^2+v^2(1-\rho^2) v2+w22ρvw=(wρv)2+v2(1ρ2)
因此
p V , W ( v , w ) = 1 2 π exp ⁡ { − v 2 2 } 1 2 π ( 1 − p 2 ) exp ⁡ { − ( w − ρ v ) 2 2 ( 1 − ρ 2 ) } p_{V,W}(v,w)=\frac{1}{\sqrt{2\pi} }\exp\left\{-\frac{v^2}{2}\right\}\frac{1}{\sqrt{2\pi(1-p^2)}}\exp\left\{-\frac{(w-\rho v)^2}{2(1-\rho^2)}\right\} pV,W(v,w)=2π 1exp{2v2}2π(1p2) 1exp{2(1ρ2)(wρv)2}
进一步,我们有
p V ( v ) = 1 2 π exp ⁡ { − v 2 2 } p_V(v)=\frac{1}{\sqrt{2\pi} } \exp\left\{-\frac{v^2}{2}\right\} pV(v)=2π 1exp{2v2}

p W ∣ V ( w ∣ v ) = 1 2 π ( 1 − ρ 2 ) exp ⁡ { − ( w − ρ v ) 2 2 ( 1 − ρ 2 ) } . p_{W|V}(w|v)=\frac{1}{\sqrt{2\pi(1-\rho^2)}}\exp\left\{-\frac{(w-\rho v)^2}{2(1-\rho^2)}\right\}. pWV(wv)=2π(1ρ2) 1exp{2(1ρ2)(wρv)2}.
所以
E [ W ∣ V ] = ρ v , σ W ∣ V 2 = 1 − ρ 2 . {\rm E}[W|V]=\rho v,\sigma_{W|V}^2=1-\rho^2. E[WV]=ρv,σWV2=1ρ2.

我们再来看X、Y的条件PSD,因为
p X , Y ( x , y ) = 1 2 π σ X σ Y 1 − p 2 exp ⁡ { − ( x − μ X ) 2 σ X 2 + ( y − μ Y ) 2 σ Y 2 − 2 ρ ( x − μ X ) ( y − μ Y ) σ X σ Y 2 ( 1 − ρ 2 ) } = 1 2 π σ X σ Y 1 − p 2 exp ⁡ { − v 2 + w 2 − 2 ρ v w 2 ( 1 − ρ 2 ) } = 1 2 π σ X σ Y 1 − p 2 exp ⁡ { − v 2 2 } exp ⁡ { − ( w − ρ v ) 2 2 ( 1 − ρ 2 ) } = 1 2 π σ X exp ⁡ { − v 2 2 } 1 2 π σ Y 1 − p 2 exp ⁡ { − ( w − ρ v ) 2 2 ( 1 − ρ 2 ) } = 1 σ x p X ( x ) 1 σ y p Y ∣ X ( y ∣ x ) p_{X,Y}(x,y)=\frac{1}{2\pi \sigma_X \sigma_Y \sqrt{1-p^2}}\exp\left\{-\frac{\frac{(x-\mu_X)^2}{\sigma_X^2}+\frac{(y-\mu_Y)^2}{\sigma_Y^2}-\frac{2\rho(x-\mu_X)(y-\mu_Y)}{\sigma_X\sigma_Y}}{2(1-\rho^2)}\right\}\\ =\frac{1}{2\pi \sigma_X \sigma_Y \sqrt{1-p^2}}\exp\left\{-\frac{v^2+w^2-2\rho vw}{2(1-\rho^2)}\right\}\\ =\frac{1}{2\pi \sigma_X \sigma_Y \sqrt{1-p^2}}\exp\left\{-\frac{v^2}{2}\right\}\exp\left\{-\frac{(w-\rho v)^2}{2(1-\rho^2)}\right\}\\ =\frac{1}{\sqrt{2\pi} \sigma_X }\exp\left\{-\frac{v^2}{2}\right\}\frac{1}{\sqrt{2\pi}\sigma_Y \sqrt{1-p^2}}\exp\left\{-\frac{(w-\rho v)^2}{2(1-\rho^2)}\right\}\\ =\frac{1}{\sigma_x}p_X(x)\frac{1}{\sigma_y}p_{Y|X}(y|x) pX,Y(x,y)=2πσXσY1p2 1exp2(1ρ2)σX2(xμX)2+σY2(yμY)2σXσY2ρ(xμX)(yμY)=2πσXσY1p2 1exp{2(1ρ2)v2+w22ρvw}=2πσXσY1p2 1exp{2v2}exp{2(1ρ2)(wρv)2}=2π σX1exp{2v2}2π σY1p2 1exp{2(1ρ2)(wρv)2}=σx1pX(x)σy1pYX(yx)
故有
p Y ∣ X ( y ∣ x ) = σ Y ⋅ p W ∣ X ( w ∣ x ) . p_{Y|X}(y|x)=\sigma_Y\cdot p_{W|X}(w|x). pYX(yx)=σYpWX(wx).

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