Expected number of offers until selling a house and expected selling price A man puts his house for sale, and decides to accept the first offer that exceeds the reserve price of $£r$. Let $X_1,X_2,...$ represent the sequence of offers received, and suppose that the $X_i$ are independent and identically distributed random variables, each having exponential distribution with rate parameter $\lambda$. 
(1) What is the expected number of offers received before the house is sold? 
(2) What is the expected selling price of the house?  
I'm assuming that part 2 is just whatever the value of $X_n$ is where $n$ is the value you get from part 1? Not sure how I would calculate that exactly. 
For part 1 I did: 
Let F denote the common CDF of the $X_i$. 
By independence we know that $P(X_1\le r)=P(X_2\le r)=P(X_i\le r) = F(r)$ 
$P(X_n$ offer is accepted$)=P(X_1\le r)P(X_2\le r)...P(X_{n-1}\le r)P(X_n\gt r)=$ $F(r)^{n-1}(1-F(r))$ 
This is a geometric distribution so the expected number of offers is just $\dfrac{1}{1-F(r)}$.
 A: You are correct in your expected number of offers, but are probably expected to express it in terms of $r$ and $\lambda$.  Do you know the CDF of an exponential distribution?  Wikipedia does.  For part 2, you want the expected value for a bid that exceeds $r$, which is $$\frac {\int_r^\infty xP(x)\ dx}{\int_r^\infty P(x)\ dx}$$
A: Let $p=e^{-\lambda r}$ and represent the number of offers with the random variable $N$. The number of offers before one succeeds is $P(N=n)=(1-p)^{n-1}p$. The number of offers we expect to get is
$$
E(N)=\sum_{n=1}^\infty nP(N=n)=\sum_{n=1}^\infty n(1-p)^{n-1}p=p\sum_{n=1}^\infty n(1-p)^{n-1}
$$
Let $q=1-p$ then 
$$
\sum_{n=0}^\infty nq^{n-1}={d\over dq}\sum_{n=1}^\infty q^n={d\over dq}{q\over 1-q}={1\over (1-q)^2}={1\over p^2}
$$
Thus
$$
E(N)={1\over p} = e^{\lambda r} 
$$
This probably has a name but I worked it all out myself from first principles so I don't know what the name is.
For the second part of the question you just need the conditional probability
$$
P(X|X\geq r)
$$
I believe @RossMillikan's answer is correct for that.
