This is exercise 5.29 of Introduction to Probability models by Ross.
Let X and Y be independent exponential random variables with respective rates $\lambda$ and $\mu$, where $\lambda > \mu$. Let $c>0$.
(a) Show that the conditional function of $X$, given that $X+Y=c$, is:
$f_{X|X+Y}(x|c) = \frac{(\lambda - \mu)e^{-(\lambda - \mu)x}}{1-e^{-(\lambda- \mu)c}}$, where $ \ \ \ \ \ 0<x<c$
(b) Use part (a) to find $E[X|X+Y=c]$.
(c) Find $E[Y|X+Y=c]$.
I have solved part (a) and mine matches the distribution given in part (a). However, when I solve for part (b), it is different from the book's solutions manual. Mine is
$E[X|X+Y=c] = \int_0^c f_{X|X+Y}(x|c) \ x \ dx = \frac{1-e^{-(\lambda - \mu)c}(1+(\lambda - \mu)c)}{(\lambda - \mu)(1-e^{-(\lambda - \mu)c})}$
While the solution is
$\frac{1-e^{-(\lambda - \mu)c}(1+(\lambda - \mu)c)}{\lambda(1-e^{-(\lambda - \mu)c})}$
which is slightly different from mine (I have ($\lambda - \mu$) in the denominator while the solution has $\lambda$). Am I doing something wrong or the answer is incorrect?
Thanks.
