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X is a random variable with mean 100 and standard deviation 30. Y is a random variable with mean of 50 and standard deviation of 25. The correlation between X and Y is 0.5. What is Var(Y|X)?

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Perhaps you are missing some assumption - eg. X and Y are jointly normal ? Besides, I guess that it's the correllation coefficient which is 0.5 –  leonbloy May 23 '11 at 13:17

2 Answers 2

up vote 2 down vote accepted

There is no unique answer to your question without some additional information on the joint distribution of $(X,Y)$. Here are two examples of couples $(X,Y)$ which both fulfill all the conditions in your post and such that $\mathrm{Var}(Y|X)$ is different.

Your hypothesis are that $E(X)=x$, $\mathrm{Var}(X)=u$, $E(Y)=y$, $\mathrm{Var}(Y)=v$ and $\mathrm{Corr}(X,Y)=c$, with $x=100$, $u=900$, $y=50$, $v=625$ and $c=1/2$. In both situations, an auxiliary random variable $Z$ is introduced, independent on $X$ and such that $E(Z)=z$ and $\mathrm{Var}(Z)=w$, where the values of $z$ and $w$ will be specified later on.

First example: Assume that $Y=aX+Z$ for a given $a$, then $y=ax+z$, $v=a^2u+w$ and $c=a\sqrt{u/v}$. The third equation yields $a=c\sqrt{v/u}$, the second one yields $w=(1-c^2)v$, the first one yields $z=y-cx\sqrt{v/u}$ and this $(a,z,w)$ yields an admissible couple $(X,Y)$. Furthermore, $$ \mathrm{Var}(Y|X)=\mathrm{Var}(Z)=w=(1-c^2)v. $$ Second example: Let $S$ denote a random variable, independent on $X$ and $Z$, with $P(S=1)=p$ and $P(S=0)=q=1-p$. Assume that $Y=SX+(1-S)Z$. In words, $Y=X$ or $Y=Z$ with probability $p$ and $q$ respectively. Then $y=px+qz$, $v=pu+qw$ and $c=p\sqrt{u/v}$. Hence $p=c\sqrt{v/u}$ is imposed, $w=(v-pu)/q$ as well (and $w>0$ for the suggested numerical values) and $z=(y-px)/q$ as well. Furthermore, $$ \mathrm{Var}(Y|X)=pq(X-z)^2+qw. $$

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Note that $$\text{Var}(Y) = E(\text{Var}(Y|X))+ \text{Var}(E(Y|X))$$ so that $$E(\text{Var}(Y|X)) = \text{Var}(Y)-\text{Var}(E(Y|X))$$

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I don't know what E[y|x] is. How do I find that? –  icobes Apr 20 '11 at 21:10
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PEV: I would appreciate if you could make precise the ways in which your post is relevant to the question asked. –  Did May 21 '11 at 10:22

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