Expected Value of independent random variables with an ${\rm Exp}(\lambda)$ common distribution Question is:

Jack is selling his house, and has decided to accept the first offer exceeding $k$. Assume that successive offers are independent random variables with common distribution $\mathrm{Exp}(\lambda)$.
a) Find the expected number of offers he will reject before he sells the house;
b) Find the expected value of the offer he will accept.

I have solved Part A as :
$p = e^{-\lambda r}$
$P(N=n) = (1-p)^{n-1}p$
$E(N) = \sum_{n=1}^\infty nP(N=n) = p \sum_{n=1}^\infty n(1-p)^{n-1}$
Eventually this will reduce down to $E(N) = \frac {1}{p} = e^{\lambda r}$
This should be correct.
Part B :
I am not sure how we go about finding the value of the actual offer that he will accept. I feel we have to find some conditional expectation but not sure. Any help would be great!
The exponential distribution is memoryless: the conditional distribution of $X-k$ given $X&gt;k$ coincides with the distribution of $X$ (you may check this property). That said, $$\mathrm{E}[X\mid X&gt;k] = k + \mathrm{E}[X-k\mid X&gt;k] =  k + \mathrm{E}[X].$$
Can you see why this is the calculation?

$$
{\cal E}[x] = \frac{\int\limits_{x = x_0}^\infty x p(x)\ dx }{ \int\limits_{x = x_0}^\infty p(x)\ dx}
$$



The only conditions you must consider are those for $x&gt;x_0$.

The only cases that contribute to your answer are those where the value is greater than some cutoff, $x_0$.  Consider this the relevant distribution.  The numerator is the expected value of $x$ from this distribution.  However, this is not a true distribution because it isn't normalized.  That's what the denominator accomplishes, where of course $p(x) \propto e^{-x}$.