3
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – SPRajagopal Aug 1 '14 at 10:40
  • $\begingroup$ @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 $x_1$. $\endgroup$ – Alecos Papadopoulos Aug 1 '14 at 18:46
  • $\begingroup$ @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. $\endgroup$ – SPRajagopal Aug 2 '14 at 8:24
  • 1
    $\begingroup$ @Glen_b, You're right. I have corrected it. $\endgroup$ – SPRajagopal Aug 3 '14 at 9:27
3
$\begingroup$

Regarding the 2nd question

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.

(Note that when one random variable can be written as a function of the other, without any third random variable present, then strictly speaking, the concept of "dependence" can not be used, because it requires the existence of two distinct random variables. When the one is just a function of the other, we have essentially only one "source of randomness").

The above reflects the "link" between the two random variables -not which causes which. The general rule that links conditional probabilities

$$P(A\mid B) P(B) = P(B\mid A) P(A)$$ applies here -and one can see that both conditional probabilities exist, without implying anything about causality.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.