I have the following: $X$ and $Y$ are exponentially distributed with parameter $\lambda$. They are independent of each other. I also have $U = X+Y$ and $V = X-2Y$. I am asked to find $E(V|U=1)$. What I tried was: I found the combined distribution of $U$ and $V$, by finding $f(x,y)$ which is
$f(x,y) = \lambda^{2}\cdot \exp(-\lambda(x+y))$, and since $U = X+Y$ and using the Jacobian for the transformation I got
$f(U,V) = \frac{\lambda^{2}}{3}\exp(-λu)$
also since $U=X+Y$ I can find its distribution function using convolution which came out to:
$f(U) = \lambda^{2}u\cdot exp(-\lambda u)$ so
$f(V|U) = f(U,V)/f(U) =\frac{1}{3u}$
therefore $E(V|U=1)$ is the integral of $\frac{v}{u}\,dv$ from $v=0$ to infinity right? but I get infinity which is not the answer
The correct answer is $-0.5$
Please tell me where I went wrong and if there is an easier way to solve this question.
Write out the domain of the pdfs as well. i.e. where is the pdf zero and where is it non-zero?. The region where it is $0$, you don't need to integrate. It is one theme that I see always among people starting probability theory, that they just blindly do a variable change without worrying about the domain. For example, can $U-V=3Y$ be negative? If so, then you should include that in the joint pdf. Also $2X=2U+V$ must be positive.
So the pdf $f(u,v)=\begin{cases}\frac{1}{3}\lambda^{2}\exp(-\lambda u)\,,\, u>v>-2u,\\0\,,\text{otherwise}\end{cases}$
Similar to what you have done, you will get $f(V|U=1)(v)=\frac{1}{3u}=\frac{1}{3}\,$ but again, you need to mention the domain .
i.e. you still need to have $1>v>-2$.
So $f(V|U=1)(v)=\frac{1}{3}\,,-2<v<1$ and $0$ elsewhere.
So now compute $\int_{\Bbb{R}}vf(V|U=1)(v)\,dv$ to find the expectation.
i.e.
$$\frac{1}{3}\int_{-2}^{1}\,v\,dv=\frac{-1}{2}$$