Question X2.3 Does anyone understand the answer? I'm totally lost! Question X2.10 (iii) Does anyone know why E[I - E(I)]^3 = E[I^3 - 3I^2 E(I) + 2(E(I))^3] ? (iv) The solution shos that for independent variables skewness is additive too, just like variance? Thanks in advance!
X2.10 E[I-E(I)]^3 = E[I^3-3I^2E(I)+3IE(I)^2-E(I)^3] (I think the assignment solution is incorrect as it looks like its cancelled some term prematurely??) = E(I^3)-3E(I^2)E(I)+3E(I)E(I)^2-E(I)^3 = E(I^3)-3qE(I^2)+2q^3 (as per solution) Haven't looked at the other question, sorry.
Thanks for your reply! I had the same explanation in mind. Anyone knows about the other two questions? Thanks!
It's looking at the proof that the sum of compound Poissons is also a compound Poisson from the end of Chapter 6. Suppose that S1 is compound Poisson with (Poisson) parameter 5 and claim distribution f1(x) and S2 is compound Poisson with parameter 3 and claim distribution f2(x) then: S is compound Poisson with parameter 5 + 3 = 8 and claim distribution: 5/8 f1(x) + 3/8 f2(x) Does this shed enough light to make the solution understandable? Yes if X and Y are independent then: var(X + Y) = var(X) + var(Y) skew(X + Y) = skew(X) + skew(Y)
