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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?

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up vote 1 down vote accepted

It's generally a good idea to avoid notation like $4/5xy$ because it's ambiguous with respect to the order of operations; my tendency would have been to interpret it as $(4/5)xy$, whereas you apparently intended $4/(5xy)$.

The notation $y=2,3$ is also suboptimal, both because $y$ is equated to two different values in a single equation and because many people around the world use decimal commas and might interpret this as $2.3$.

Yes, your method is correct; it would have been OK though to write $\sum_{y\in\{2,3\}}6/5=6/5+6/5$ without adding and subtracting a term for $y=1$.

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