Unsure how to calculate mean square error of a variable with a joint distribution I'm working through a homework assignment and I'm having an inordinately difficult time trying to calculate the MSE of two different estimators. The homework is described as this:
Let $X$ and $Y$ be random variables with joint density:
$p(x,y) =  2$ if $x\in A$
It is $0$ otherwise.
$A$ is a set surrounded by $X = 0, Y = 0$, and $X + Y = 1$.
I have to calculate the MSE of 2 estimators: 
\begin{equation}
\hat{X} = E[X|Y]
\end{equation}
and 
\begin{equation}
\hat{X} = Y
\end{equation}
I've seen the proof that $E[X|Y]$ is the better estimator, but I am absolutely stumped on how to actually calculate the MSE $E[(\hat X - X)^2]$ for either estimator. I've spent hours reading and trying to figure this out but I have almost no experience with statistics and it is killing me in this problem. Any help would be appreciated.
Edit:
$E[X|Y] = \int_0^1 \int_0^{1-y} 2xy dx dy = 1/12$
$E[Y] = \int_0^1 2ydy = 1$
 A: In this problem, the conditional density of $X$ given the value of $Y$ is $\alpha$
is a uniform density on $[0,1-\alpha]$ and thus has mean $\frac{1}{2}(1-\alpha)$.
Thus, $\hat{X} = E[X\mid Y] = \frac{1}{2}(1-Y)$. Can you calculate
$E[(\hat{X}-X)^2]$ from this?
Added Note:
$\displaystyle E[(\hat{X}-X)^2] = E\left[\left(\frac{1}{2}(1-Y) - X\right)^2\right] = \int_{y=0}^1 \int_{x=0}^{1-y}\left(\frac{1}{2}(1-y) - x\right)^2\cdot 2
\,\mathrm dx \,\mathrm dy$
Further note in response to the OP's comments:
An easier way of calculating the value of $E[(\hat{X}-X)^2]$ is to note
that conditioned on $Y = \alpha$, the conditional pdf of $X$ is
a uniform density support 
$[0, 1-\alpha]$ and  mean $\hat{X} = \frac{1}{2}(1-\alpha)$, so that
$$E\left[(\hat{X}-X)^2\mid Y = \alpha\right] = \frac{(1-\alpha)^2}{12},$$
from which we get $E\left[(\hat{X}-X)^2\mid Y\right] = \frac{(1-Y)^2}{12}$
and
$$E[(\hat{X}-X)^2] = E\left[E\left[(\hat{X}-X)^2\mid Y\right]\right]
= E\left[\frac{(1-Y)^2}{12}\right]
= \int_0^1 \frac{(1-\alpha)^2}{12}\cdot 2(1-\alpha)\,\mathrm d\alpha
= \frac{1}{24}$$
