Let $Y$ a uniform random variable in $[0,1]$, and let $X$ a uniform variable in $[1,e^Y]$
My work
By definition
$E(X\mid Y)=\int_{x\in A}xp(x\mid y)\,dx$
Now, I need the function $f(x\mid y)$.
By definition, $f(x\mid y)=\frac{f(x,y)}{f_y(y)}$
I'm stuck trying to finding $f(x,y)$ and $f_y(y)$. Can someone help me?
On
My 2 cents... and I'm not an expert so see if this makes sense to you.
$E[X|Y]$ is a random variable and not a constant. So for my attack I'll first get $E[X|Y=k]$ and then once I get that switch $k$ for $Y$.
$$E[X|Y=k] = \int_{1}^{e^k} x f_{X|Y}(x|y=k) dx$$
We know that $X|Y=k \sim Uniform(1,e^k)$ which means $f_{X|Y}(x|y=k) =\frac{1}{e^k-1}$
So with this knowledge,
$$=\int_{1}^{e^k} \frac{x}{e^k-1} dx =\frac{1}{e^k-1} \int_{1}^{e^k} xdx$$
$$=\frac{e^{2k}-1}{2(e^k-1)}$$
Finally to get $E[X|Y]$ just replace $k$ with $Y$.
$$E[X|Y]=\frac{e^{2Y}-1}{2(e^Y-1)}$$
You don't need $f_Y$ nor $f_{XY}$ here because $f(x\mid y)$ is given straightfowardly.
You know that $X\sim \mathcal{U}[1,e^Y]$, what (being $Y$ a r.v.) gives actually the conditional density of $X$ given $Y$. In particular, what we know is that if you are aware of the fact that $Y$ gets the value $y$, then $X\sim \mathcal{U}[1,e^y]$.
On the other hand, if you want $f(y\mid x)$, then you should calculate both $f_X$ and $f_{XY}$.