# MMSE decision rule for gaussian

If $X=Y+Z$ where $Y$ and $Z$ are independent gaussians with zero means and known variances, what is the MMSE decision rule to estimate $Y$ for $X=x$?

I know that $h_{MMSE}(x)=$ posterior mean $= E[Y|X=x]=\int yf(y|x)dy$
and $f(y|x) = \frac{f(x,y)}{f(x)}$, but I don't have a good enough grasp of the material to put this information together. I found here a formula that shows it as:

However I can't follow the steps and don't know how this changes for my case of a linear combination of independent gaussians.

edit: Here's the original question I'm looking at, but I figured I'd just ask about the first part of it (MMSE).

If $Y$ and $Z$ are independent Gaussian random variables, then $(X,Y) = (Y+Z,Y)$ enjoys a bivariate Gaussian distribution. Indeed, we nave that $$E[X] = E[Y]+E[Z], ~ \operatorname{var}(X) = \operatorname{var}(Y) + \operatorname{var}(Z), \\\operatorname{cov}(X,Y) = \operatorname{cov}(Y+Z,Y) = \operatorname{cov}(Y,Y) = \operatorname{var}(Y)$$ and we can use this to write down the joint density function explicitly if we choose to do so.
Now, the MMSE estimate of $Y$ given that $X = x_0$ is, as you know, $E[Y\mid X = x_0]$, that is, the mean of the conditional density of $Y$ conditioned on $X = x_0$. This is a general result that does not require any special conditions on $X$ and $Y$. But when $(X,Y)$ has a bivariate Gaussian distribution, the MMSE estimator coincides with the linear MMSE estimator of $Y$ given that $X = x_0$. The linear MMSE estimator is, of course, the estimator achieves the minimum mean-square error among all estimators that are constrained to be linear functions (actually, affine functions) of $x_0$: $\hat{Y} = ax_0+b$ where $a$ and $b$ are real numbers. So, what is the linear MMSE estimator of $Y$ given that the value of $X$ is $x_0$? If you have never seen this calculation done before, we want to find $a$ and $b$ such that $E[(Y=ax_0+b)^2]$ is as small as possible, and the usual calculus methods for finding minima tell us that $a$ and $b$ must be such that $$\hat{Y} =ax_0+b = E[Y] + \left.\left.\frac{\operatorname{cov}(X,Y)}{\operatorname{var}(X)}\right(x_0 - E[X]\right) = E[Y] + \left.\left.\frac{\rho\cdot\sigma_Y}{\sigma_X}\right(x_0 - E[X]\right)$$ where $\rho$ is the correlation coefficient of $X$ and $Y$, and $\sigma_Y$ and $\sigma_X$ are the standard deviations of $Y$ and $X$ respectively. Note that this result not require that $XY$ and $Y$ are jointly Gaussian; it is perfectly general. But the linear MMSE estimator is also, in general, not the MMSE estimator which is usually a nonlinear function of $x_0$. For jointly Gaussian random variables, we have the happy result that the linear MMSE estimator coincides with the MMSE estimator.
Truth be told, I can never remember the details this formula and prefer to keep it in mind in the more elegant form $$\left(\frac{\hat{Y}-E[Y]}{\sigma_Y}\right) = \rho \left(\frac{X-E[X]}{\sigma_X}\right)$$