欢迎访问 生活随笔！
TAG聚合

生活随笔

当前位置： 首页 > 编程资源 > 编程问答 >内容正文

编程问答

UA MATH564 概率论 多元随机变量的变换 理论与应用2

发布时间：2025/4/14 编程问答 37 豆豆
生活随笔 收集整理的这篇文章主要介绍了 UA MATH564 概率论 多元随机变量的变换 理论与应用2 小编觉得挺不错的,现在分享给大家,帮大家做个参考.

UA MATH564 概率论 多元随机变量的变换 几个例题

例5 X1,X2,X3∼iidEXP(λ)X_1,X_2,X_3 \sim_{iid} EXP(\lambda)X1​,X2​,X3​∼iid​EXP(λ)，Y1=X1+X2+X3,Y2=X1X1+X2,Y3=X1+X2X1+X2+X3Y_1 = X_1+X_2+X_3,Y_2 = \frac{X_1}{X_1+X_2},Y_3 = \frac{X_1+X_2}{X_1+X_2+X_3}Y1​=X1​+X2​+X3​,Y2​=X1​+X2​X1​​,Y3​=X1​+X2​+X3​X1​+X2​​，求Y1,Y2,Y3Y_1,Y_2,Y_3Y1​,Y2​,Y3​的联合概率密度，并判断他们是否互相独立。
先把X1,X2,X3X_1,X_2,X_3X1​,X2​,X3​表示成Y1,Y2,Y3Y_1,Y_2,Y_3Y1​,Y2​,Y3​的函数，
Y3=X1+X2X1+X2+X3=X1+X2Y1⇒X1+X2=Y1Y3Y2=X1X1+X2=X1Y1Y3⇒X1=Y1Y2Y3Y_3 = \frac{X_1+X_2}{X_1+X_2+X_3} = \frac{X_1+X_2}{Y_1} \Rightarrow X_1+X_2 = Y_1Y_3 \\ Y_2 = \frac{X_1}{X_1+X_2} = \frac{X_1}{Y_1Y_3} \Rightarrow X_1 = Y_1Y_2Y_3Y3​=X1​+X2​+X3​X1​+X2​​=Y1​X1​+X2​​⇒X1​+X2​=Y1​Y3​Y2​=X1​+X2​X1​​=Y1​Y3​X1​​⇒X1​=Y1​Y2​Y3​

因此X2=Y1Y3(1−Y2),X3=Y1(1−Y3)X_2 = Y_1Y_3(1-Y_2),X_3 = Y_1(1-Y_3)X2​=Y1​Y3​(1−Y2​),X3​=Y1​(1−Y3​)，
∂(X1,X2,X3)∂(Y1,Y2,Y3)=∣Y2Y3Y3(1−Y2)1−Y3Y1Y3−Y1Y30Y1Y2Y1(1−Y2)−Y1∣=(1−Y3)[Y12(1−Y2)Y3+Y12Y2Y3]−Y1[−Y1Y2Y32−Y1(1−Y2)Y32]=(1−Y3)Y12Y3+Y12Y32=Y12\frac{\partial (X_1,X_2,X_3)}{\partial (Y_1,Y_2,Y_3)} = \left| \begin{matrix}Y_2Y_3 & Y_3(1-Y_2) & 1-Y_3 \\ Y_1Y_3 & -Y_1Y_3 & 0 \\ Y_1Y_2 & Y_1(1-Y_2) &-Y_1 \end{matrix}\right| \\ = (1-Y_3)[Y_1^2(1-Y_2)Y_3+Y_1^2Y_2Y_3]-Y_1[-Y_1Y_2Y_3^2-Y_1(1-Y_2)Y_3^2] \\ = (1-Y_3)Y_1^2Y_3+Y_1^2Y_3^2 = Y_1^2∂(Y1​,Y2​,Y3​)∂(X1​,X2​,X3​)​=∣∣∣∣∣∣​Y2​Y3​Y1​Y3​Y1​Y2​​Y3​(1−Y2​)−Y1​Y3​Y1​(1−Y2​)​1−Y3​0−Y1​​∣∣∣∣∣∣​=(1−Y3​)[Y12​(1−Y2​)Y3​+Y12​Y2​Y3​]−Y1​[−Y1​Y2​Y32​−Y1​(1−Y2​)Y32​]=(1−Y3​)Y12​Y3​+Y12​Y32​=Y12​

X1,X2,X3X_1,X_2,X_3X1​,X2​,X3​的联合概率密度是
fX1,X2,X3(x1,x2,x3)=λ3e−λ(x1+x2+x3),x1,x2,x3>0f_{X_1,X_2,X_3}(x_1,x_2,x_3) = \lambda^3e^{-\lambda(x_1+x_2+x_3)},x_1,x_2,x_3>0fX1​,X2​,X3​​(x1​,x2​,x3​)=λ3e−λ(x1​+x2​+x3​),x1​,x2​,x3​>0

因此Y1,Y2,Y3Y_1,Y_2,Y_3Y1​,Y2​,Y3​的联合概率密度是
fY1,Y2,Y3(y1,y2,y3)=λ3y12e−λy1,y1>0f_{Y_1,Y_2,Y_3}(y_1,y_2,y_3) = \lambda^3 y_1^2 e^{-\lambda y_1},y_1>0fY1​,Y2​,Y3​​(y1​,y2​,y3​)=λ3y12​e−λy1​,y1​>0

显然Y1,Y2,Y3Y_1,Y_2,Y_3Y1​,Y2​,Y3​独立。

例5 续 Z1=X1/Y1Z_1 = X_1/Y_1Z1​=X1​/Y1​，Z2=X2/Y1Z_2 = X_2/Y_1Z2​=X2​/Y1​，求Z1,Z2Z_1,Z_2Z1​,Z2​的联合概率密度。
先对Z1,Z2Z_1,Z_2Z1​,Z2​变形一下，
Z1=X1Y1=Y1Y2Y3Y1=Y2Y3Z2=X2Y1=Y1Y3(1−Y2)Y1=Y3(1−Y2)Z_1 = \frac{X_1}{Y_1} = \frac{Y_1Y_2Y_3}{Y_1} = Y_2Y_3 \\ Z_2 = \frac{X_2}{Y_1} = \frac{Y_1Y_3(1-Y_2)}{Y_1} = Y_3(1 - Y_2)Z1​=Y1​X1​​=Y1​Y1​Y2​Y3​​=Y2​Y3​Z2​=Y1​X2​​=Y1​Y1​Y3​(1−Y2​)​=Y3​(1−Y2​)

再把Y2,Y3Y_2,Y_3Y2​,Y3​表示为Z1,Z2Z_1,Z_2Z1​,Z2​的函数
1−Y2Y2=1Y2−1=Z2Z1⇒Y2=Z1Z1+Z2\frac{1-Y_2}{Y_2} = \frac{1}{Y_2}-1= \frac{Z_2}{Z_1} \Rightarrow Y_2 = \frac{Z_1}{Z_1+Z_2}Y2​1−Y2​​=Y2​1​−1=Z1​Z2​​⇒Y2​=Z1​+Z2​Z1​​

所以Y3=Z1+Z2Y_3 = Z_1+Z_2Y3​=Z1​+Z2​，计算Jacobi行列式
∂(Y2,Y3)∂(Z1,Z2)=∣Z2(Z1+Z2)21−Z1(Z1+Z2)21∣=1Z1+Z2\frac{\partial (Y_2,Y_3)}{\partial (Z_1,Z_2)} = \left| \begin{matrix} \frac{Z_2}{(Z_1+Z_2)^2}&1\\-\frac{Z_1}{(Z_1+Z_2)^2} & 1 \end{matrix} \right| = \frac{1}{Z_1+Z_2}∂(Z1​,Z2​)∂(Y2​,Y3​)​=∣∣∣∣∣​(Z1​+Z2​)2Z2​​−(Z1​+Z2​)2Z1​​​11​∣∣∣∣∣​=Z1​+Z2​1​

Y2,Y3Y_2,Y_3Y2​,Y3​的联合概率密度为
fY2,Y3(y2,y3)=∫0∞λ3y12e−λy1dy1=−∫0∞λ2y12e−λy1d(−λy1)=−λ2y12e−λy1∣0∞+∫0∞2λ2y1e−λy1dy1=−∫0∞2λy1e−λy1d(−λy1)=−2λy1e−λy1∣0∞+∫0∞2λe−λy1dy1=−2e−λy1∣0∞=2f_{Y_2,Y_3}(y_2,y_3) = \int_{0}^{\infty} \lambda^3 y_1^2 e^{-\lambda y_1}dy_1 =-\int_{0}^{\infty} \lambda^2 y_1^2 e^{-\lambda y_1}d(-\lambda y_1) \\ = -\lambda^2 y_1^2 e^{-\lambda y_1}|_0^{\infty} + \int_0^{\infty} 2\lambda^2 y_1 e^{-\lambda y_1} dy_1 = -\int_0^{\infty} 2\lambda y_1 e^{-\lambda y_1} d(-\lambda y_1) \\ = -2\lambda y_1e^{-\lambda y_1}|_0^{\infty} + \int_{0}^{\infty} 2\lambda e^{-\lambda y_1}dy_1 = -2e^{-\lambda y_1}|_0^{\infty} = 2fY2​,Y3​​(y2​,y3​)=∫0∞​λ3y12​e−λy1​dy1​=−∫0∞​λ2y12​e−λy1​d(−λy1​)=−λ2y12​e−λy1​∣0∞​+∫0∞​2λ2y1​e−λy1​dy1​=−∫0∞​2λy1​e−λy1​d(−λy1​)=−2λy1​e−λy1​∣0∞​+∫0∞​2λe−λy1​dy1​=−2e−λy1​∣0∞​=2

因此Z1,Z2Z_1,Z_2Z1​,Z2​的联合概率密度是
fZ1,Z2(z1,z2)=2z1+z2,z1,z2∈(0,1)f_{Z_1,Z_2}(z_1,z_2) = \frac{2}{z_1+z_2},z_1,z_2 \in (0,1)fZ1​,Z2​​(z1​,z2​)=z1​+z2​2​,z1​,z2​∈(0,1)

总结

以上是生活随笔为你收集整理的UA MATH564 概率论 多元随机变量的变换 理论与应用2的全部内容，希望文章能够帮你解决所遇到的问题。

如果觉得生活随笔网站内容还不错，欢迎将生活随笔推荐给好友。