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I am slightly confused about the different between

$\mathrm{E}[Y|X = x]$

and

$\mathrm{E}[Y|X]$

and similarly for Variance.

It seems to me the first should be a scalar, because we first pick a specific $X = x$ and then get the expected value of $Y$ within that set whereas the second one is a random variable that depends on the random variable $X$. Is that correct?

Any definition using the probabilities $\mathrm{P}(X)$, $\mathrm{P}(Y)$, $\mathrm{P}(Y|X)$ and $\mathrm{P}(Y, X)$ is appreciated.

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1  
Your interpretation is correct. $\Bbb E[Y|X]$ is a random variable whose value at $X=x$ is $\Bbb E[Y|X=x]$. $\Bbb E[Y|X]$ is called the conditional expectation of $Y$ given $X$. A similar statement holds for the conditional variance of $Y$ given $X$ (symbolized by $\text{Var}[Y|X]$). –  David Mitra Feb 19 '12 at 17:10
    
You need to be clear whether $X$ and $Y$ are random variables or events. If they are random variables then $E[Y|X]$ may be a function of a random variable and so a random variable, but then so too is $P(X)$ which may not be what you intend as you describe it as a probability. –  Henry Feb 19 '12 at 21:36
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1 Answer 1

Your surmise is correct.

Let's say with probability $1/2$ I pick a biased coin, and toss it, getting $Y=1$ with probability $1/3$ and $Y=0$ with probability $2/3$. And with probability $1/2$ I pick an unbiased coin. Let $X$ be $0$ or $1$ according as I pick the biased or unbiased coin. Then $$ \begin{align} E(Y \mid X=0) & = \frac 1 3 \\ \\ E(Y \mid X=1) & = \frac 1 2 \\ \\ \\ E(Y \mid X) & = \begin{cases} \frac 1 3 & \text{with probability }1/2, \\ \\ \frac 1 2 & \text{with probability }1/2. \end{cases} \end{align} $$ And similarly for conditional variances.

Having done that, one can write things like $$ E(E(Y \mid X)) = E(Y) $$ (the law of total expectation) and $$ E(\operatorname{var}(Y \mid X)) + \operatorname{var}(E(Y \mid X)) = \operatorname{var}(Y) $$ (the law of total variance, which breaks the variance into an "explained" part and an "unexplained" part). (Now I notice that I wrote the "unexplained" part first, so don't add "respectively".)

In a similar way, one has $$ E(\Pr(A \mid X)) = \Pr(A) $$ (the law of total probability).

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In words, the variance of $Y$ is $E[\text{var}(Y|X)]$, the mean of the conditional variance plus $\text{var}(E[Y|X])$, the variance of the conditional mean. –  Dilip Sarwate Feb 19 '12 at 23:02
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