Hello,

I'll think on this one later ^^

Like for any problem of this kind, take a partition of the set of events.(ii) Let be a sequence of i.i.d random variables with values in and a random variable with values in which is independent of . Define

and

Show that if we denote by and the probability the generating functions of and respectively, then the probability generating function of S is given by

.

Thank you (:

Namely

({.} denotes an event... if you prefer, it's the set of such that the even happens)

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Thus

Hence :

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We have :

But what is ?

Hence

Since the Xi's are independent, this is equal to the product of the expectations :

Since the Xi's follow the same distribution as X, this is :

Now finally, we have :

Can you recognize that it's the pgf of T, taken at a particular value ?

Spoiler:

And you're done... I hope you like it, because it's been quite loooooooong to type lol!

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is obtained just by using the distributive law of intersection and union. It shouldn't be difficult to prove it.

because for , {T=k} and {T=m} are disjoint (intersection = empty set). And hence { =n , T=k} and { =n , T=m} are disjoint.

The equality follows from the third axiom of Kolmogorov.

is actually Bayes' formula

Edit : lol...that's my second 5th post XD