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I only find some materials relating to conditional expectation like $E[X|Y=y]$. How to calculate conditional expectation $E[X|X \geq 0]$?

user462102
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2 Answers2

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Let $Y = I(X \ge 0)$. Then, the density of $Y$ is given by:

$$\frac{f(y)}{\int_0^{\infty}f(y) dy}$$

You can then compute $E(X|X\ge 0)$ by computing $E(Y)$.

Srikant
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  • Thank you for such a prompt reply. You mean the density of $X$? Since $Y = I(X \geq 0)$, we have $E[Y] = E[I(X \geq 0) = P(X \geq 0)]$, right?\ How to calculate $E(X|X \geq 0)$ by computing $E(Y)$? Could you please expain in detail? – user462102 Oct 13 '17 at 13:43
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In general: if $A$ is an event with positive probability, then $$E[X | A] = \frac{E[X \cdot 1_{A}]}{P(A)}$$ where the symbol $X \cdot 1_A$ denotes the random variable that is $X$ on the event $A$ and is $0$ otherwise.

One fact that may be helpful: if $A$ and $B$ are a partition of $\Omega$ (that is, $A \cap B = \emptyset$ and $A \cup B = \Omega$) then: \begin{align*} E[X] &= E[X \cdot 1_A + X \cdot 1_B] \\ &= E[X \cdot 1_A] + E[X \cdot 1_B] \\ &= E[X | A] P(A) + E[X | B] P(B). \end{align*}

Aaron Montgomery
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  • Here, $A$ is the event $X \geq 0$. Then follow the formula given, $$ E[X|X \geq 0] = \frac{E[X \geq 0]}{P(X \geq 0)} . $$ Is this right ? – user462102 Oct 13 '17 at 13:49
  • I don't think the symbol $E[X \geq 0]$ has any meaning. Instead, you want $E[X \cdot 1_{X \geq 0}]$, which means "the random variable that is $X$ if $X$ is nonnegative and is $0$ otherwise". – Aaron Montgomery Oct 13 '17 at 14:03