# Expected Profit

• Mar 21st 2011, 06:24 PM
alakaboom1
Expected Profit
Sorry, I already asked a question today, but it would be awesome if someone could help me with this one, I feel like I'm close:

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 http://s3.amazonaws.com/answer-board...9239628497.gif? I'm sure I have the upper bound wrong...
• Mar 22nd 2011, 12:30 AM
Volga
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!
• Mar 23rd 2011, 01:22 AM
Moo
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 :)
• Mar 23rd 2011, 04:28 AM
Volga
Sorry! (Itwasntme)