Let X and Y be discrete random variable with joint pdf $f(x,y) = 4/5xy$ if $x = 1,2$ and $y = 2,3$ and zero otherwise. Find:
E(Y)
Basically I found the marginal pdf and summed it up. I'll save you all the trouble of doing the work yourself. I just want to make sure my method is correct
$E(Y) = \sum_{y} yf(y) = \sum_{y=2,3}y6/5y = \sum_{y=2,3} 6/5 = 6/5 + 6/5 + 6/5 - 6/5 = 12/5$
The answer in the book is 12/5 as well, but I wrote it out elaborately like I did above to check. I added and subtracted 6/5 because you have to start at y = 1 to sum up and then you have to subtract what you shouldn't have, i.e. 6/5. Is my method correct?