Great Expectations!

Came across this cool question the other day.

If X and Y are Uniform(0,1) random variables, what is E[X/(X+Y)]?

Direct calculation of the expectation as a double integral turns out to be a bit nasty.

But there is a trick.

X/(X+Y) + Y/(X+Y) = 1

=> E[X/(X+Y)] + E[Y/(X+Y)] = 1

Since X~Y, both the expectations above are equal since they are symmetrical in X and Y.

Thus each equals 1/2.

And we are done!

Corollaries:

Now it is easy to calculate the following (X, Y, Z ~ U(0,1) )


E[X/(X+Y+Z)] = 1/3
E[XY / (XY + YZ + ZX)] = 1/3

and so on!

Question:
Would the results be the same if X, Y, Z were iid but from another distribution?