Probabilty and Moment Generating Functions

Firstly.....A PGF question:-

The random variable X can take values 1,2,3...and its pgf is G(t). Show that the probability that X is even is give by 0.5[1+G(-1)]

It's the first question in the exercise, so I think I'm missing something pretty fundamental there....

Secondly......I was meant to have read up on MGFs, so if anyone could explain the concept of them, their application, what sorts of questions they are used in etc, that would be hugely appreciated.

Rep to anyone who can help!

Reply 1

RichE

k@tie

The random variable X can take values 1,2,3...and its pgf is G(t). Show that the probability that X is even is give by 0.5[1+G(-1)]

G(t) = p1t + p2t^2 + p3t^3 + ....

where p1+p2+p3+...=1.

P(X is even) = p2+p4+p6+...

1/2[1+G(-1)] = 1/2[1-p1+p2-p3+p4-...] = 1/2[2p2+2p4+...] = p2+p4+...

as required.

PGFs are just a way of efficiently encoding information about a discrete random variable and MGFs are just a different way of doing the same. Do you have a specific question about them?

Reply 2

k@tie

coool cheers :smile:

will rep u in....23 hours :smile:

Reply 3

RichE

k@tie

Sorry, how do you go from one to the other?
1/2[1+G(-1)] = 1/2[p2-p1-p3]

I wrote rubbish originally because I read 1,2,3 and not 1,2,3,.... which makes a world of difference. :eek:

Now edited!

Reply 4

k@tie

RichE

.....MGFs are just a different way of doing the same. Do you have a specific question about them?

No, just need an overview

Reply 5

RichE

k@tie

No, just need an overview

Well:

M(X, t) = E(exp(tX))

For independent X and Y

M(X+Y,t) = M(X,t) M(Y,t)

And

d^nM/dt^n = E(X^n exp(tX))

in particular at t=0 it equals E(X^n)

Also X is recoverable from M: that is

M(X,t) = M(Y,t) means X=Y.

But wouldn't know which of the above is on your syllabus.

And I guess you may need to know the MGFs of standard distributions like Poisson etc.