This is for (a) and (b)

The joint PDF, fxy(x,y) = 4/3 because P[area is filled] = 3/4. We want the volume of the 2-RV PDF to be 1, and (4/3)*(3/4) = 1. To find fx(x), you need to consider 2 cases: when 0 < x < 0.5, and when 0.5 < x < 1. For the first case, where is y filled in? It is filled in from y = 0.5 to y = 1. So we integrate the joint PDF from 0.5 to 1. Same goes for the 2nd case, but this time y is filled all the way from 0 to 1, so we integrate the joint PDF from 0 to 1.

You can just use the formulas given in class to evaluate E[X] and Var(X).

And for part (b), because of the symmetry of the graph, the PDFs, E, and Var of Y should be exactly the same of those of X.

Hope this is correct. Let me know if I did anything wrong....

--- I got the E[x] = 7/12 and Var(x) = -37/144 and tell me if I am wrong because I think I am... (Jonathan Morales)

--- I got the fx(x)=2, E[x]=1 and Var(x) = -1/3 and tell me if i am wrong, because i think i am...(Zhongtian Wang)

--- The E[X] = 7/12. Both of your variances are wrong, remember that you can't have a negative variance. Use the Var[X] = E[X^2] - (E[X])^2 formula. (Gregory Pajot)

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood