两个随机变量的联合分布

hbghlyj

问题2及参考答案 A8_2022 problem 2.pdf (271.29 KB, Downloads: 67)

2023-4-20 23:28 Upload

Click on the file to download the attachment

广告: PDF toolkit extract pages from PDF

(c) Consider the transformation $T:ℝ^2→ℝ^2$ given by $T(x,y)=(x-y,x+y)$. Let $(R,S)$ be a pair of random variables with joint probability density function$$f(r,s)=\begin{cases}\frac{1}{4}e^{-{|s|}},&(r,s)∈[-1,1]×ℝ\\0,&\text { otherwise. }\end{cases}$$ and $(X,Y)$ such that $(R,S)=T(X,Y)$.
(i) Derive the joint probability density function of $(X,Y)$.
(ii) Find the marginal probability density functions of $X$ and $Y$.
(iii) Find the correlation of $X$ and $Y$.

(c)(i) 因 $r=x-y\in[-1,1]$
$$
f_{X, Y}(x, y)={|J(x, y)|} f_{R, S}(T(x, y))= \begin{cases}\frac{1}{2} e^{-{|x+y|}} & \text { if }{|x-y|} \leq 1 \\ 0 & \text { otherwise. }\end{cases}
$$

问题：(c)(ii)为什么分成${|y|} \geq \frac{1}{2},{|y|} \leq \frac{1}{2}$两个区域？

For ${|y|} \geq \frac{1}{2}$,\begin{aligned}
f_Y(y)=f_Y({|y|}) & =\int_{{|y|}-1}^{{|y|}+1} \frac{1}{2} e^{-x-{|y|}} d x=\frac{1}{2} e^{-{|y|}}\left(e^{-{|y|}+1}-e^{-{|y|}-1}\right) \\
& =\frac{1}{2}\left(e-\frac{1}{e}\right) e^{-2{|y|}}
\end{aligned}and for ${|y|} \leq \frac{1}{2}$, we split into two integrals over $({|y|}-1,-{|y|})$ and $(-{|y|},{|y|}+1)$ :\begin{aligned}
f_Y(y)=f_Y({|y|}) & =\left(\frac{1}{2}-\frac{1}{2} e^{2{|y|}-1}\right)+\left(-\frac{1}{2} e^{-2{|y|}-1}+\frac{1}{2}\right) \\
& =1-\frac{1}{e} \cosh (2 y) .
\end{aligned}

hbghlyj

我懂了：区域以水平线（固定 $y$）的截面$x\in[y-1,y+1]$，考虑 $x+y$ 的符号，如果 $\abs y\ge\frac12,x+y$ 的符号不变，如果 $\abs y\le\frac12,x+y$ 的符号在$x\in[y-1,-y],x\in[-y,y+1]$两段上不同

hbghlyj

(c)(iii)参考答案用斜体评论“risks inefficient attempts(考生有可能计算很繁琐)”
在(c)(ii)刚算出X,Y的分布，考生可能直接拿X,Y的分布来计算 var(X),var(Y),cov(X,Y)，这是低效的。
参考答案只用了一行：
以$R,S$表示$X,Y$：$X=\frac{R+S}2,Y=\frac{R-S}2$ 从(b)(ii)算的var(R),var(S)立得var(X),var(Y),cov(X,Y).

这样看来,倒可以跳过(c)(ii)直接答(c)(iii)😉