# Conditional Expectation for Probability Distribution [duplicate]

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How is the minimum of a set of random variables distributed?

I have two RVs from the same distribution (exponential distribution with parameter λ). How do I calculate E(min|min>x)? I know how to get E(X|X>x) but I specifically need X to be the smallest of the two RVs.

## marked as duplicate by Macro, whuber♦Apr 13 '12 at 14:27

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• If you have the density function for the minimum, then it's exactly the same as doing the calculation for an arbitrary random variable, $X$. Are you asking how to derive the density of the minimum? – Macro Apr 12 '12 at 22:39
• Sorry for confusion - yes, that is basically what I am asking. – Ashal Apr 12 '12 at 22:40
• This is a commonly asked question on here. See, for example, this thread: stats.stackexchange.com/questions/220/… – Macro Apr 12 '12 at 22:43

## 1 Answer

Macro has pointed you at previous discussion of the distribution of the minimum. For your particular problem with exponentially distributed random variables it is even easier.

An exponential distribution is the time for the first occurrence of a Poisson process with rate $\lambda$ so the minimum of $n$ iid exponentially distributed random variables is the time for the first occurrence of a Poisson process with rate $n\lambda$ and so is exponentially distributed with mean $\frac{1}{n\lambda}$.

But an exponential distribution is memoryless, so $E(\min | \min > x) = x+ \frac{1}{n\lambda}$.

In your particular question, $n=2$.