The annual profit Y(in $100,000) can be expressed as a continuous function of drug demand x(in 1,000): Y(x) = 2(1-e^(-2x)). Suppose the demand for their drug has the probability function: f(x)= 6e^(-6x), x>0. Find the company's expected annual profit. So do I have to start by doing ? I'm sure I have the upper bound wrong... 2. The way I see this problem, it is a mixed distribution where the given profit function Y is conditional on X:$\displaystyle f_{Y|X}(y|x)=2(1-e^{-2x})$and therefore to find E(Y) you need first to find marginal distribution$\displaystyle f_Y(y)$and then go from there. But I may be totally wrong! 3. Hello,$\displaystyle E[Y]=E[E[Y|X]]=E[E[2(1-e^{-2X})|X]]=E[2(1-e^{-2X})]$, which is the formula alakaboom1 wrote. As for the boundaries, it should rather be on the whole set of real numbers. Then since f(x)=0 if x<0, the boundaries will indeed go from 0 to infinity. Volga : there's something disturbing in what you wrote.$\displaystyle 2(1-e^{-2x})$is not a pdf. We just have that$\displaystyle Y=2(1-e^{-2X})\$, so in order to find Y's pdf, there's a change of variable to make in X's pdf