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I am selling my house, and have decided to accept the first offering exceeding $K$ dollars. Assuming that offers are independent rv with common distribution $F$, find the expected number of offers received before I sell the house.

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I call $X$ to be number of offers receiver before house is sold. Suppose we have $n$ such offers and call them $X_1,X_2,...,X_n$. Therefore, $X = \sum X_i$. We have that

$$ E(X) = E(X_1) + .. + E(X_n) $$

Since all $X_i$ have common distribution $F$, then

$$ E(X_i) = \int\limits_0^K x f(x) $$

where $f = F' $. So,

$$ E(X) = n \int\limits_0^K x f(x) $$

is this correct?

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James
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1 Answers1

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This is completely the wrong way to do it. $F(K)$ gives the probability that one offer is less than $K$, i.e. the offer fails, so $1-F(K)$ is the probability of an offer succeeding. Now the number of offers received follows a geometric distribution with success rate $1-F(K)$, so the expected number of offers is its reciprocal, $\frac1{1-F(K)}$.

Parcly Taxel
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