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CT6 2018 April Q9ii Compound distributions

E

Edward Smith

Member
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
 
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
 
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