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.


$E[X|Y] = \int_0^1 \int_0^{1-y} 2xy dx dy = 1/12$

$E[Y] = \int_0^1 2ydy = 1$

  • $\begingroup$ Have you calculated $E[X|Y]$? $\endgroup$ Nov 9, 2013 at 0:31
  • $\begingroup$ I think so: \begin{equation}\int_0^1 \int_0_(1-Y) 2xy dx dy\end{equation} = 1/12 It turns out I'm not great at using LaTeX. The value is Integral(0->1){Integral(0->1-Y){ 2xy dx}dy} = 1/12 (I've done the math on paper and it worked out to this) $\endgroup$ Nov 9, 2013 at 0:44
  • $\begingroup$ Did you mean $\int_0^1 \int_0^{(1-y)} 2xy \,dx \,dy\,\,$? $\endgroup$
    – Glen_b
    Nov 9, 2013 at 0:52
  • $\begingroup$ That is what I meant, yes. $\endgroup$ Nov 9, 2013 at 0:54

1 Answer 1


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 $$E[(\hat{X}-X)^2] = E\left[E\left[(\hat{X}-X)^2\mid Y\right]\right] = E\left[\frac{(1-Y)^2}{12}\right] = \int_0^1 \frac{(1-\alpha)^2}{12}\cdot 2(1-\alpha)\,\mathrm d\alpha = \frac{1}{24}$$

  • $\begingroup$ Okay, so $E[(\hat X - X)^2] = Var(\hat X) + Bias(\hat X, X)^2$ Because $\hat X = E[X|Y]$, and $Bias(\hat X, X) = E[\hat X] - E[X]$, then $Bias(\hat X, X) = E[E[X|Y]] - E[X]$, and by the law of iterated expectations this becomes $E[X] - E[X] = 0$. It follows, then, that $E[(\hat X - X)^2] = Var(\hat X)$. $Var(\hat X) = E[\hat {X}^2] - (E[\hat X])^2$. By my calculations $E[\hat {X}^2] = 1/48$ and $(E[\hat X])^2 = 1/144$, which means $E[(\hat X - X)^2] = Var(\hat X) = 1/72$. That's what I have, is it wrong? $\endgroup$ Nov 9, 2013 at 20:42
  • $\begingroup$ Also, as a part of the problem I (think) I calculated $E[(\hat X - X)^2]$ for $\hat X = Y$. The estimator is unbiased according to my work, and this makes sense because X and Y are uniformly distributed over the same set which means that the error is based on the variance. Then, $Var(Y) = E[Y^2] - (E[Y])^2$ where $E[Y^2] = \int_0^1 y^2p_{y}(y)dy = \int_0^1 y^2(2-2y)dy = \int_0^1 2y^2 - 2y^3 dy = [(2/3)y^3-(1/2)y^4]_0^1 = 1/6$ and $(E[Y])^2 = ((1/2)\int_0^1 1-x dx)^2 = ((1/2)[x-(1/2)x^2]_0^1)^2 = ([1-1/2]*(1/2))^2 = (1/4)^2 = 1/16$ $\endgroup$ Nov 9, 2013 at 22:28
  • $\begingroup$ $(E[Y])^2 = (\frac{1}{2}\int_0^1 1-x dx)^2 = (\frac{1}{2}[x-\frac{x^2}{2}]_0^1)^2 = ([1-\frac{1}{2}]*\frac{1}{2})^2 = (\frac{1}{4})^2 = \frac{1}{16}$ So, $Var(Y) = \frac{1}{6} - \frac{1}{16} = \frac{5}{48}$. As a result, we can compare the MSE for the estimators $\hat X_1 = E[X|Y]$ and $\hat X_2 = Y$ and see that $\hat X_1 < \hat X_2$ which makes sense given that the conditional expectation is the best predictor. $\endgroup$ Nov 9, 2013 at 22:39
  • $\begingroup$ I evaluated the integral (using a bunch of substitutions) and I got that $E[(\hat X - X)^2] = \frac{1}{24}$ $\endgroup$ Nov 9, 2013 at 23:42
  • $\begingroup$ $\frac{1}{24}$ is the correct answer. $\endgroup$ Nov 9, 2013 at 23:49

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