Q&A Bank 2 Q24

[Annie79]

Please can someone help with the final stage of this question?

The question is about a Poisson claims process with given claim size density function which it asks you to derive the MGF for. I am happy with the solution up to the last step where it seems to put limits into the integrated expression.
What happens to the two infinity limits and why do the denominators change from (t-3) and (t-7) to (3-t) and (7-t)? Is there a quick trick to this?

The solution jumps from:
3/2[e^(t-3)x / (t-3)] upper limit infinity and lower limit of zero
+ 7/2[e^(t-7)x / (t-7)] with same limits and integrated wrt x

to the MGF of:
3/2(3-t) + 7/2(7-t)

Thanks!!

 

[Ian Senator]

Dealing just with the first integral (same logic applies to second integral):

The integral converges only if t is less than 3 (otherwise you'll get e to the infinity). If t is less than 3 then the numerator of the integrand is e to the minus something. At x equals infinity this will tend to zero. So the answer becomes 3/2(t-3) times (e to the minus infinity minus 1), which is 3/2(t-3) times -1 which is 3/2(3-t).