Vtyjkyuk

Let $Y$ a uniform random variable in $[0,1]$, and let $X$ a uniform variable in $[1,e^Y]$

My work

By definition

$E(Xmid Y)=int_xin Axp(xmid y),dx$

Now, I need the function $f(xmid y)$.

By definition, $f(xmid y)=fracf(x,y)f_y(y)$

I'm stuck trying to finding $f(x,y)$ and $f_y(y)$. Can someone help me?

You don't need $f_Y$ nor $f_XY$ here because $f(xmid y)$ is given straightfowardly.

You know that $Xsim mathcalU[1,e^Y]$, 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 $Xsim mathcalU[1,e^y]$.

On the other hand, if you want $f(ymid x)$, then you should calculate both $f_X$ and $f_XY$.

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 $Y$.

$$E[X|Y=k] = int_1^e^k x f_Y(x|y=k) dx$$

We know that $X|Y=k sim Uniform(1,e^k)$ which means $f_Y(x|y=k) =frac1e^k-1$

So with this knowledge,

$$=int_1^e^k fracxe^k-1 dx =frac1e^k-1 int_1^e^k xdx$$

$$=frace^2k-12(e^k-1)$$

Finally to get $E[X|Y]$ just replace $k$ with $Y$.

$$E[X|Y]=frace^2Y-12(e^Y-1)$$