$X$ is a random variable which has exponential distribution. We define the event $A=\{a<X<b\}$. Then how is the conditional expectation $E(X\mid A)$ and conditional density $f_{X\mid A}(x)$ defined?

Is $f_{X\mid A}=\frac{f_X(x)}{\Pr(A)}$ for $a<X<b$ and $0$ else?

Is $E(X\mid A)$ same as $E(1_AX)$ where $1_A$ is the indicator function. The latter I suppose is $E(1_AX)=\int_a^bf_X(t)tdt$, where $f_X(x)$ is the density of $X$.

Or is $E(X\mid A)=\int_0^\infty f_{X\mid A}(t)tdt$ which, assuming the above definition of conditional density is correct, is equal to $\frac{E(1_AX)}{\Pr(A)}$


1 Answer 1


The expression for $f_{X\mid A}$ is correct.

But $E(X\mid A)$ is not the same as $E(1_AX)$. To understand why, first think about the expression $E(X\mid A)$. When $a$ tends to $b$ we expect it to lay in the interval $(a,b)$ and thus to be roughly equal to $a$ and $b$. On the other hand the $E(1_AX)$ will go to 0 in that case because (intuitively) the random variable $1_AX$ will be almost everywhere equal to zero.

The rest of your approach is correct and in particular the result that $E(X\mid A)=\frac{E(1_AX)}{\Pr(A)}$ is correct too.


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