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masu
June 9th 2006, 05:24 AM
Is there anyone can help me to deal with the following proof?

Why Sum (0 to inf) Pr(M1+...+Mn=k | N=n)z^k
=[PM(z)]^n ? where M is a r.v. and PM(z) is the pgf of M

this question comes from a proof of "Loss Models 2nd ed."
(In the top of pg.90)

Thanks a lot^^

ctperng
June 9th 2006, 11:19 PM
Hi, Masu,

This follows from the fact that if X and Y are independent, discrete random variables, taking nonnegative integer values, then writing S = X+Y, one has
P_S(z) = P_X(z)P_Y(z), where P_S, P_X and P_Y represent the corresponding pgfs.
(To prove it, you compare the coefficients of z^n on both sides, using
Pr(S = n) = Sum_i Pr(X = i)Pr(Y = n-i), by independence.)

This generalizes to the sum of more than two random variables.
Since each M_j ( j =1,,n) is of the same type (namely M), the result is clear.

ctperng

masu
June 10th 2006, 01:28 AM

Actually, I know what you mean already:
P_S(z)=E(e^zs)=E(e^z(m1+m2+...))=[E(e^zM)]^n, right?

But my question is about the proof
How to get Sum (0 to inf) Pr(M1+...+Mn=k | N=n)z^k =[PM(z)]^n ?

Sorry, I just feel confused about
1. [PM(z)]^n = [Sum {Pr(M)*z^k}]^n
2. Sum (0 to inf) Pr(M1+...+Mn=k | N=n)z^k =Sum {Pr(M)^n*z^k}

I don't get why they are equal, are I misunderstanding anything?

Once again, thankx for your help

ctperng
June 10th 2006, 03:14 AM
Hi, Masu,

I just showed you the proof.

You are confused about probability generating function and moment generating function: they are not the same thing. Please refer to p.73 for definition.

Regards,
ctperng