# How to find $E[x|y]$ when distributions of y and x are separately known,(p.s. they are both Gaussian)?

In detail, I have these relations (in order of causality):

$u_1 = ax_0$

$x_1 = u_1 + x_0$

$y = x_1 + w$

where $w = N(0,1), x_0 = N(0,\sigma^2)$.

This was my approach: I know the distribution of $x_1 = N(0,(1+a)^2\sigma^2)$ and I know $y = N(0,(1+a)^2\sigma^2 + 1)$ The expectation to find is: $E[x_1|y]$. What I'm unable to figure out is how to go about computing this expectation, and the inter-dependence of the two variables: $x_1$ and $y$. I know that $y$ depends on $x_1$ but is the converse also true?

• No, the information is correct. I checked. I also followed the "answer" given by @Sid below and I was able to find out the linear MMSE. I got the way to do it. – SPRajagopal Aug 1 '14 at 10:40
• @Glen_n Glen, your calculation has a typo: the variance of $u_1$ is $a^2\sigma^2$, not $1$, so we end up with the expression the OP gave for the distribution of $x_1$. – Alecos Papadopoulos Aug 1 '14 at 18:46
• @Glen_b, No, it's not for a subject. It's for my research thesis. Also, I checked. There are no typos. Alecos has made that clear, I hope. – SPRajagopal Aug 2 '14 at 8:24
• @Glen_b, You're right. I have corrected it. – SPRajagopal Aug 3 '14 at 9:27

I know that $y$ depends on $x_1$ but is the converse also true?
Stochastic dependence (say, between two random variables) is not a relation that has anything to do with causality. It is a situation where the realizations of one random variable "affect probabilistically" the realizations of the other. Strictly speaking, this means that we can in principle express their relation as a two-way correspondence: To a value of $X$ correspond many values of $Y$, and to different values of $Y$ correspond different sets (not necessarily disjoint) of values of $X$ (hence a correspondence and not a function). And vice-versa.
$$P(A\mid B) P(B) = P(B\mid A) P(A)$$ applies here -and one can see that both conditional probabilities exist, without implying anything about causality.