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$X,Y$ are independent exponential random variables with respective rates $\lambda,\mu$. Let $M=\text{min}(X,Y)$.

How to calculate $E[MX\;|\;M=X]$.

In general, how to calculate $E[X\;|\;B]$ where $X$ is a continuous RV and $B$ is an event?

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It would be better to post these as separate questions, as some may be able to answer one or the other. – Ross Millikan Dec 1 '12 at 3:07
up vote 1 down vote accepted

$E[M X \mid M = X]$ is the same thing as $E[X^2 \mid X < Y]$, which is:

$$ E[X^2 \mid X < Y] = \int_0^\infty x^2 p(x \mid X < Y) dx. $$

Now, via Bayes' Rule, $p(x \mid X < Y)$ is $$p(x \mid X < Y) = \frac{p(X < Y \mid x) p(x)}{p(X < Y)}.$$

$P(X < Y \mid X = x)$ is just $P(Y > x) = 1 - P(Y \le x) = e^{-\mu x}$, and $p(x) = \lambda e^{-\lambda x}$.

$p(X < Y)$ is a little trickier, but we can do it as $$ p(X < Y) = \int_0^\infty \int_0^\infty I(x<y) \lambda e^{-\lambda x} \mu e^{-\mu y} dy dx \\ = \int_0^\infty \lambda e^{-\lambda x} \left( \int_x^\infty \mu e^{-\mu y} dy \right) dx \\ = \int_0^\infty \lambda e^{-\lambda x} e^{-\mu x} dx \\ = \int_0^\infty \lambda e^{-(\lambda + \mu) x} dx \\ = \frac{\lambda}{\lambda + \mu}. $$

Putting that together, we have $$ p(x \mid X < Y) = e^{-\mu x} \lambda e^{-\lambda x} \frac{\lambda + \mu}{\lambda} = (\lambda + \mu) e^{-(\lambda + \mu) x}, $$ so that $E[X^2 \mid X < Y]$ is just the second moment of an exponential with rate $\lambda + \mu$ (call it $Z$), $$E[Z^2] = Var[Z] + E[Z]^2 = \frac{2}{(\lambda + \mu)^2}.$$

Moderately surprisingly, this means that $E[X^2 \mid X < Y] = E[Y^2 \mid Y < X]$.

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very clear, thanks! – hxhxhx88 Dec 1 '12 at 5:34

For the second question, if by $B$ is an event means it is a discrete distribution, let $p(B_i)$ be the probabilities of each value of each value of $B$. Then $E[X|B]=\sum_i p(B_i)E[X(B_i)]$ where $E[X(B_i)]$ is the expectation of $X$ given that $B=B_i$

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oh..I should post a separate question..anyway, thank you. – hxhxhx88 Dec 1 '12 at 5:35

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