Find the conditional expectation $\mathbb{E}\left[\left.X_{1}\right|Y\right]$ if $X_1,..., X_n\sim\mathrm{Uniform}\left(0,1\right)$, where $Y=\max\left\{ X_{1},...,X_{n}\right\}$.
MY ATTEMPT:
We have
$\mathbb{E}\left[\left.X_{1}\right|Y\right]=\dfrac{1}{n}\cdot\mathbb{E}\left[\left.X_{1}\right|\left(X_{1}=Y\right)\cap Y\right]+\dfrac{n-1}{n}\cdot\mathbb{E}\left[\left.X_{1}\right|\left(X_{1}<Y\right)\cap Y\right]$.
- If $X_{1}=Y$, then $\mathrm{\mathbb{E}}\left[\left.X_{1}\right|\left(X_{1}=Y\right)\cap Y\right]=Y$.
- If $X_{1}<Y$, then $X_{1}\sim\textrm{Uniform}\left[0,\mathbb{E}\left(Y\right)\right]$, therefore $\mathbb{E}\left[\left(\left.X_{1}\right|X_{1}<Y\right)\cap Y\right]=\dfrac{\mathbb{E}\left[Y\right]}{2}$.
Therefore $\mathbb{E}\left[\left.X_{1}\right|Y\right]=\dfrac{1}{n}\cdot Y+\dfrac{n-1}{n}\cdot\dfrac{\mathbb{E}\left[Y\right]}{2}$.
On the other hand, we have that
$f_{Y}(y)=n\cdot\left[F_{X}(y)\right]^{n-1}\cdot f_{X}(y)=\left\{ \begin{array}{cc} ny^{n-1} & \textrm{if }0\leq y\leq1\\ 0 & \textrm{otherwise} \end{array}\right.$
is the density function of $Y$.
Therefore, $\mathbb{E}\left[Y\right]=\int_{0}^{1}y\cdot f_{Y}(y)dy=\int_{0}^{1}y\cdot ny^{n-1}dy=\dfrac{n}{n+1}$.
Thus, $\mathbb{E}\left[\left.X_{1}\right|Y\right]=\dfrac{1}{n}\cdot Y+\dfrac{n-1}{n}\cdot\dfrac{1}{2}\cdot\dfrac{n}{n+1}=\dfrac{1}{n}\cdot Y+\dfrac{1}{2}\cdot\dfrac{n-1}{n+1}$.
It's ok?
No, it's not OK. Your basic idea of distinguishing the cases $X=Y$ and $X\ne Y$ is good, but you didn't apply it properly; as Did remarked, $\mathbb{E}\left[X_{1}\mid\left(X_{1}=Y\right)\cap Y\right]$ makes no sense.
The probability that two of the $X_i$ are equal is $0$. Thus, by symmetry, $X=Y$ with probability $\frac1n$, and otherwise $X$ is one of $n-1$ points uniformly distributed in $[0,Y]$. Thus
$$ \mathbb E[X_1\mid Y]=\frac1nY+\frac{n-1}n\frac Y2=\frac Y{2n}\left(2+n-1\right)=\frac Y2\left(1+\frac1n\right)\;. $$