I´d like to calculate the expectation of an algorithm. The input to the algorithm are n customers with values $v_1$,..$v_n$(i.i.d) drawn from a normal distribution. Now the algorithm sells an item with a fixed price p and offers this price to the customers sequentially. The first customer i with value $v_i > p$ gets the item.
What is the expectation of the value of the customers we sold to($v$) achieved by the algorithm? I tried to solve it by calculating the expectation of the normal distribution from p as the lower bound to infinity, but it didn´t match the result I was getting when solving this empirically.
Many thanks in advance!
Suppose $v_i\in N(\mu,\sigma^2)$. Let $z_i=(v_i-\mu)/\sigma$, so the $z_i$ are standard normal, and let $q=(p-\mu)/\sigma$ be the standardized price. Let $\Phi(x)$ be the standard normal cdf, with pdf $\phi(x)$.
Conditioned on the existence at least one customer whose value is above the price, then the first customer whose value is larger than $p$ will have a value which is normally distributed, conditional on being greater than $p$. The conditional expected value is $$ E[v\mid v>p]=E[\sigma z+\mu|z>q]=\mu+\sigma E[z|z>q] =\mu+\sigma\cdot\frac{\int_q^\infty x\,\phi(x)\,dx}{P(z>q)} $$ The normal cdf satisfies $x\phi(x)=-\phi'(x)$, so using the fundamental theorem of calculus this is $$ \mu+\sigma\frac{\phi(q)}{1-\Phi(q)} $$ To make this unconditional, we have to multiply by the probability that at least one customer's value is above the price, resulting in $$ \boxed{\Big(1-\Phi(q)^n\Big)\left(\mu+\sigma\frac{\phi(q)}{1-\Phi(q)}\right)} $$