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My problem:

Let $Y_1$ and $Y_2$ be two random variables and let $f(y_1,y_2)=e^{-y_2}$ from $0\leq y_1\leq y_2 < \infty$ and $0$ otherwise be the joint density function.

(a) Calculate $E(Y_1|Y_2 = y_2)$ and $E(Y_2 |Y_1 = y_1)$.

(b) Calculate the density functions for $E(Y_1|Y_2 )$ and $E(Y_2 |Y_1 )$.

My solution in (a) were easy. I got $E(Y_1|Y_2 = y_2)=y_2/2$ and $E(Y_2 |Y_1 = y_1)=y_1+1$. For (b) I believe I should use transformation method, i.e. I define two random variables as \begin{equation} U=Y_2/2 \hspace 0.5cm and \hspace 0.5cm V=Y_1+1 \end{equation} With the transformation method I obtain for $U$ the following density function \begin{equation} f_U(u)=2e^{-2u} \end{equation}

My problem is now to find $V=Y_1+1$ because I don't have $Y_1$ in my joint density function...

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Your formula for $f_U$ is not correct. You first calculate $f_{Y_1}$ and $f_{Y_2}$ as follows: $f_{Y_1}(y_1)=\int f(y_1,y_2)dy_2 =\int_{y_1}^{\infty} e^{-y_2} dy_2=e^{-y_1}$ for $0<y_1<\infty$; $f_{Y_2}(y_2)=\int f(y_1,y_2)dy_1 =\int_{0}^{y_2} e^{-y_2} dy_1=y_2e^{-y_2}$ for $0<y_2<\infty$. Using $f_{Y_2}$ we get $f_U(y)=4ye^{-2y}$ for $0<y<\infty$.

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  • $\begingroup$ So if I use $f_{Y_2}(y_2)=y_2e^{-y_2}$ then I get $f_U(u)=4ue^{-2u}$ and not the same as you. So I'm supposed to use my marginals instead my joint density function. $\endgroup$ – Joey Adams Jan 10 at 0:05
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    $\begingroup$ @JoeyAdams There was a typo. Indeed $f_U(u)=4ue^{-2u}$ $\endgroup$ – Kavi Rama Murthy Jan 10 at 0:36
  • $\begingroup$ Okay, I see! And for the second one, i.e. $E(Y_2|Y_1)$ I get $f_V(v)=e^{v-1}$ $\endgroup$ – Joey Adams Jan 10 at 0:39

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