MSE estimator of MMSE, E[Y|X] I have the following joint pdf $f(x,y)=2\exp(-x-y)$ ,  for $0<x<y< \infty$.
I found these quantaties:


*

*$f(x)=2\exp(-2x)$

*$E[Y|X]=x+1$

*$E[X]=1/2$

*$E[Y]=3/2$

*$E[XY]=1$

*$E[Y^2]=7/2$

*MSE estimator equals to $E[Var[Y|X]]=E[Y^2]-E[E[Y|X]^2]$
$E[E[Y|X]^2]=E[(x+1)^2]=\int\limits_0^\infty2(x+1)^2\exp(-2x)\,dx = - 7/2$ 
I have doubt on $E[E[Y|X]^2]$ value. Am I right here? 
 A: $$\int\limits_0^\infty2(x+1)^2\exp(-2x)\,dx = 5/2$$
A: There really is no need to evaluate all those integrals in this case.  Remember that if $g(x)$ is a nonnegative function with finite integral, then 
$h(x) = g(x)/\int_{-\infty}^{\infty} g(x) \mathrm dx$ is a probability density function
(because $h(x)$ is non-negative and the "area under the curve $h(x)$" is $1$).  (Exercise: show that this idea works for nonpositive functions too!).  Armed with this and visualizing
the joint density function $f(x,y)$ as a surface above the $x$-$y$ plane, we see that
given $X = x_0$, 
the conditional density of $Y$ is proportional to
$f(x_0, y) = \exp(-x_0 - y),~~ x_0 < y < \infty$, where the constant of 
proportionality is 
$$\frac{1}{\int_{-\infty}^{\infty} f(x_0, y) \mathrm dy}
= \frac{1}{\int_{x_0}^{\infty} f(x_0, y) \mathrm dy} = 
\frac{1}{f_X(x_0)}$$ 
and "normalizes" (unitizes?) the area under the curveto $1$.  Now, instead of
evaluating this integral to get the exact conditional density function,
we can argue that the "shape" of the conditional density is an exponentially decaying
function of $y$, and so the conditional density of $Y$ given $X$ is just
an exponential density (with mean and variance equal to $1$) that has been displaced 
$x_0$ to the right.
Hence, $E[Y\mid X = x_0] = 1 + x_0$, and so $E[Y \mid X] = 1 + X$.  The conditional
mean-square error (MSE) is just the variance ($1$) of this conditional density, and since
this does not depend on the value of $X$, the unconditional MSE is also $1$.  
Note that in this instance, the MMSE estimator is the same as the linear MMSE estimator.
