Transformation method of two random variables

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...

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 $$0; $$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. Using $$f_{Y_2}$$ we get $$f_U(y)=4ye^{-2y}$$ for $$0.
• 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. – Joey Adams Jan 10 at 0:05
• @JoeyAdams There was a typo. Indeed $f_U(u)=4ue^{-2u}$ – Kavi Rama Murthy Jan 10 at 0:36
• Okay, I see! And for the second one, i.e. $E(Y_2|Y_1)$ I get $f_V(v)=e^{v-1}$ – Joey Adams Jan 10 at 0:39