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I am trying to understand a solution to a problem. Here's the problem.

Abe will sell his calculator to the person to offer him at least \$130 for it. The offers are independent exponential random variables with mean $100. What's the expected number of offers Abe will receive?

The solution notes that the offers for the calculator are $C_x \sim Exponential(\frac{1}{100})$, which I can understand from the problem. It then says that the number of offers that are below \$130 has a geometric distribution with $p = e^{\frac{-130}{100}}$.

Where did $e^{\frac{-130}{100}}$ come from? I thought that the expected value for an exponential distribution was $\frac{1}{\lambda} = 100$.

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up vote 2 down vote accepted

Imagine the offers come in sequentially and are independent. Call an offer a "success" if it is $\gt 130$.

The probability that an individual offer is $\ge 130$ is $$\int_{130}^\infty \frac{1}{100}e^{-t/100}\,dt.$$ A short calculation shows that this integral is $e^{-130/100}$.

Let $p=e^{-130/100}$. Let $X$ be the number of trials (offers) until the first success. Then $$\Pr(X=n)=(1-p)^{n-1}p.$$ This is straightforward: the first success occurs at the $n$-th trial precisely if there are $n-1$ consecutive failures followed by a success. Thus $X$ has geometric distribution with parameter $p$. By a perhaps familiar formula, $$E(X)=\frac{1}{p}.$$

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Thanks! Why is the lower bound for the integral 100 instead of 0? – John Hoffman Oct 25 '12 at 0:55
@JohnHoffman: Oh no, a typo. Thanks for pointing it out. The lower bound is meant to be $130$. The probability that an offer is $\lt 130$ is the integral from $0$ to $130$. The probability that an offer is $\ge 130$ is the integral over the rest of the world, from $130$ to $\infty$. I went directly to the probability $\ge 130$ because the integral is simpler. The integral from $0$ to $130$ is $1-e^{-130/100}$, which we would have to subtract from $1$. – André Nicolas Oct 25 '12 at 1:04

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