Using a factorial moment generating function to find probability func.

[Fictionarious]

Homework Statement

Hi everyone! Me and my colleague are working our way through Harold J Larson's "Introduction to Probability Theory and Statistical Inference: Third Edition", and we found something interesting. We both have the same edition of the text, but mine is slightly newer?, and one problem is different between our two identical versions, problem 7 of chapter 3.4. His is:

The factorial moment generating function for a discrete random variable Y is

ψ_Y(t) = e^t-1

Find the probability function (cumulative probability function, I presume) for Y.

Mine is the same type of question, except,

ψ_Y(t) = (e^t-1)/(e-1) is what I see.

I suppose my first question is, how could these both possibly have the same answer / be the same? The solution is the same in both our books.

Homework Equations

The book gives a number of relevant equations.

ψ_X(t) = E[t^X] = E[e^Xln(t)] = m_X(ln(t)) This I guess I follow, since E[X] is just expected value or mean, and m_X(t) is the moment generating function.

It also says (d^k/dt^k)ψ_X(t)|_t=0 = k!p_X(n),

that is, that the kth derivative of the factorial moment generating function evaluated at t=0 is just some constant times the probability function for X. This leads me to suspect this is how we'd go about finding an answer to the question.

The Attempt at a Solution

The extent of our attempt to get the answer the back of the book has -

p_X(k) = (1/k!), k = 1,2,3, . . .

has been along these lines:

(using his version of the problem)

Derivative of e^t-1 is just e^t-1. Evaluated at t=0, you get e^-1
Do the same for any subsequent derivatives and the answer is of course the same. Nowhere is k! coming up.

(using my version of the problem)

Derivative of (e^t-1)/(e-1) is just e^t/(e-1). Evaluated at t=0, you get 1/(e-1). Do the same for any subsequent derivatives and the answer is of course the same. Where is the k! coming in?

An explanation of either of our versions of the problem would be much appreciated, and/or an explanation of how they're really the same (I'm hoping that one of our books isn't just plain wrong).

 

[haruspex]

Find the probability function (cumulative probability function, I presume) for Y.

No, it's the PDF.

(d^k/dt^k)ψ_X(t)|_t=0 = k!p_X(n),

(d^k/dt^k)ψ_X(t)|_t=0 = k!p_X(k)

the answer the back of the book has is p_X(k) = (1/k!), k = 1,2,3, . . .

(using his version of the problem)

Derivative of e^t-1 is just e^t-1. Evaluated at t=0, you get e^-1
Do the same for any subsequent derivatives and the answer is of course the same. Nowhere is k! coming up.

Using the (corrected) equation you quoted: e^-1 = k!p_X(k)
What's wrong with that? (But, what is the range of k here?)

(using my version of the problem)

Derivative of (e^t-1)/(e-1) is just e^t/(e-1). Evaluated at t=0, you get 1/(e-1).

Except for one particular value of k . This is the version that matches the answer in the book.

 

[Fictionarious]

k = 0, 1, . . ., n.

yeah, k!p_x(k) is what I meant to say.

So, if 1/(e-1) matches the answer in the back, I guess I just don't see how. Are we summing all the values of (1/k!) for the entire range of k?, because then I would see how Ʃ(1/k!) over the range of k would be 1/(e-1), but only if we excluded k=0 or k=1 (and n was infinity).

And in that case, 1/e would be the (more) correct answer.

I guess what I'm not understanding is this -

p_X(k) is supposed to be 1/k!

k! is clearly . . . k!

Shouldn't that mean that k!p_X(k) is 1? But we're getting 1/e or 1/(e-1), not 1.

We're really just not sure how solve this kind of problem in general.

EDIT: oh, well, in the answer it does specify k starts at 1, so that's something. So could we say?,

in the case of e^t-1, the answer would be 1/k!, k = 0, 1, . . ., n

and in the case of (e^t-1)/(e-1), the answer would be 1/k!, k = 1, 2, . . ., n

 

[haruspex]

p_X(k) is supposed to be 1/k!

No, as I wrote, that is wrong. It doesn't add up to 1. If it's for k = 0, 1... then p_X(k) = e^-1/k! If it's for k = 1, 2... then p_X(k) = (e-1)^-1/k!

Not sure you've understood the difference in the two cases. For ψ_Y(t) = (e^t-1)/(e-1) and k = 0, what is the k^th derivative evaluated at t = 0?