For part (c), given that we are dealing with a compound poisson distribution, I have calculated the variance = lambda*E(X^2) where lambda = 0.3 and E(X^2) = 200 + 250^2 so variance = 18810 Where is the mark scheme getting 23810 from?
Hi Edward This isn't explained in the Examiner's Report, but the reason is that we aren't dealing with a single Poisson process. This is because the parameter lambda is not fixed but a random variable. This is what part (i) is getting at. Therefore we need to use the formula for "Moments of a compound distribution" from p16, not the "Compound Poisson distribution" formulae. Using this we get: var(Si) = E[N] * var(X) + var(N) * (E[X])^2 = 0.3 * (200 + 250^2) + 0.08 * 250^2 = 23810 Hope that helps. Dave
Thanks Dave... Are you not using E(X^2) instead of V(X)? I would have thought to use the V(X)=200 given in the question
Hi Edward, My apologies - I was too hasty answering your question and I set up the right formula but with the wrong justification! You need to use the formula for the "Conditional variance" on p16 of the Tables. var(S) = var(E[S|lambda]) + E[var(S|lambda)] From (ii)(b) and (c) we can substitute in: var(S) = var(lambda * m1) + E[lambda * m2] = m1^2 * var(lambda) + m2 * E[lambda] Then using results from earlier in the question, and what we know about X: var(S) = 250^2 * 0.08 + (200 + 250^2) * 0.3 = 23810 Sorry about that error. This solution should replace the one above. Dave
