The condition doesn't vanish, you just chose some arbitrary realisation of the random variable N to work with, in this case "n".
So E[S|N=n]= E[X(1) + X(2) +....+ X(n) ] = nE[X] since X1, X2 ... Xn are independent and identically distributed.
But rather than chose an actual number, just condition on knowledge of the random variable by writing an analogous expression to the one above and then average over all it's values. ie. E[S|N] = NE[X]. Now using the unconditional expectation result E = E[E[S|N]] = E[NE[X]] = E[N]E[X].
Last edited by a moderator: Sep 7, 2010