I am reading a paper on contracts and there was an expected value calculation which got me confused.
Consider the following primitives
- $\delta \in (0,1)$, an exogenous parameter
- $D$ a positive random variable with distribution $F$
- $S(q) = E[\min(q,D)]$ expected sales. The idea is that you are a retailer, order a quantity $q$ and can sell the min of what you ordered and what demand was. In the paper, it is shown that $S(q) = q(1 - F(q)) + \int_{0}^{q} y f(y) dy = q - \int_0^q F(y) dy$.
- $I(q)$ is expected leftover inventory $ = E[(q-D)^{+}] = q - S(q)$. Clearly the expected leftover inventory is $< q$, and probably $ > 0 $.
In the paper, they state that $$E[\min[I, \delta q]] = \int_{(1 - \delta)q}^{q} F(y) dy.$$
I am getting stuck understanding that relationship. Clearly on the function there is going to be some weight on $\delta q$ and some weight on $S(q)$, but I am getting stuck seeing exactly how this is derived.
EDIT: What I am trying to prove is the expectation of the $\min$. Also, $S(\cdot)$ and $I(\cdot)$ are both expectations. My apologies if this was not clear earlier.
"In the paper", "they" are probably trying to compute $E(X)$, where the random variable $X$ is $$X=\min((q-D)^+,\delta q).$$ Recall that, for every nonnegative random variable $\xi$, $$E(\xi)=\int_0^\infty P(\xi\geqslant x)\mathrm dx.$$ If $x\leqslant\delta q$, $[X\geqslant x]=[(q-D)^+\geqslant x]=[D\leqslant q-x]$. If $x\gt\delta q$, $[X\geqslant x]$ is empty. Hence, $$E(X)=\int_0^{\delta q}P(D\leqslant q-x)\mathrm dx=\int_{(1-\delta) q}^qP(D\leqslant x)\mathrm dx=\int_{(1-\delta) q}^qF(x)\mathrm dx.$$