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For 2 numbers which are exponentially distributed with rate 3, I need to calculate: the expected value of the lowest number given that both numbers are above 1/3

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Because of the memorylessness of this distribution, you can just find the expected value of the minimum without the condition that they are greater than $1/3$, and then add $1/3$ to it. (If you don't know about this, then (1) Google the word memorylessness, and (2) if you're taking a course on this material, look for it in your notes.)

$\min\{X_1,X_2\} > x$ if and only if both $X_1>x$ and $X_2>x$. If the two random variables $X_1$ and $X_2$ are independent, then the probability that both of these events occur is just the product of their two probabilities: $\Pr(X_1>x\ \&\ X_2>x)= \Pr(X_1>x)\Pr(X_2>x) = e^{-3x}e^{-3x} = e^{-6x}$. So the CDF of the minimum is $F(x) = 1-e^{-6x}$. Differentiate that to get the density function $f(x)$, then find $\mathbb{E}(\min) = \displaystyle\int_0^\infty x f(x) \; dx$.

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    +1 from me. Your answer helped me think through a question I've had for some time. Thanks.2013-07-10