# Proving the conditional probability $E(y|x) = \operatorname{corr}(y,x) x/(\operatorname{stdev}(x)^2)$

In my statistical signal modelling class we were given the following task:

For simplicity, let's assume $E(y) = E(x) = 0$. Show that if the real variables $x$ and $y$ are normally distributed, then the equation $E(y|x) = \operatorname{corr}(y,x) \cdot x/\operatorname{var}(x)$ holds.

We were also given a hint: write the assisting variable $s = y - \operatorname{corr}(x,y)\cdot x/\operatorname{var}(x)$ and show that it's uncorrelated with $x$, meaning that $E(sx) = E(s)E(x) = 0$. Because $s$ and $x$ together are normally distributed, it follows that $s$ and $x$ are uncorrelated, ie. the conditional probability of $s$ doesn't depend on $x$. Now you can easily write $y$ using $s$, and calculate $E(y|x)$.

I've been hacking at this for 3 hours without much progress so any help is appreciated

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I believe the crucial observation is that normality implies s and x are independent, not just uncorrelated: that is what is needed to conclude that the conditional probability of s is independent of x. –  whuber Sep 23 '10 at 19:00

Note that $corr(x,y) = E(xy)-E(x)E(y) = E(xy)$ and $var(x) = E(x^2)-E(x)^2=E(x^2)$

Hence,

$E(sx) = E(yx) - \frac{corr(y,x)}{var(x)}E(x^2) = E(yx) -\frac{E(xy)}{E(x^2)}E(x^2) = 0$

Then also $y=s+\frac{corr(x,y)x}{var(x)}$ so

$E(y|x) = E(s+\frac{corr(x,y)x}{var(x)}|x) = E(s|x) + E(\frac{corr(x,y)x}{var(x)}|x) = E(s) + \frac{corr(x,y)x}{var(x)}$

but $E(s)=0$ giving the result.

You will need to justify all these formulas and steps.

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Thanks, I didn't realize to look at any alternative formulas for correlation. –  tsiki Sep 25 '10 at 12:41
Cheers. No problem. –  alext87 Sep 25 '10 at 13:36