# Homework Help: Moment Generating Functions

1. Feb 15, 2010

### exitwound

I haven't taken math for going on 13 years and I am having a hard time following what this all means. I'm not sure if I'm doing it right.

First of all, can anyone explain to me in English what exactly the Moment Generating Function is? The book does a poor job at it and the teacher wasn't very clear either. In fact, the book *never* says what it is at all, just how it's used.

1. The problem statement, all variables and given/known data

If you toss a fair die with outcome X, p(x) = 1/6 for x=1,2,3,4,5,6. Find Mx(t).

2. Relevant equations

$$M_x(t) = E(e^{tx}) = \sum_{x\in D}e^{tx}p(x)$$

3. The attempt at a solution

If we start with the Expectation equation above:

$$\sum_{x\in D}e^{tx}p(x) = \sum_{x=1}^6 e^{tx}(1/6)$$

$$= e^{t\cdot 0}(1/6) + e^{t\cdot 1}(1/6) + e^{t\cdot 2}(1/6) + ... e^{t\cdot 6}(1/6)$$

$$= 1 + \frac{e^t}{6} + \frac{e^{2t}}{6} + \frac{e^{3t}}{6} + \frac{e^{4t}}{6} + \frac{e^{5t}}{6} + \frac{e^{6t}}{6}$$

First of all, is this correct so far? And if it is, since it's been umteen years since I've had algebra, what happens here? How can this be simplified?

2. Feb 15, 2010

Why do you start your sum with

$$e^{t\cdot 0}$$

when none of the sides have 0 dots? Other than that your initial setup is fine. As for simplifying,

*notice a factor common to the sum of 6 terms, and factor it out
*how would you simplify the geometric sum $$1 + a +a^2 + a^3 + a^4 + a^5$$

I'm not sure the simplified form is much less awkward than the original form.

3. Feb 15, 2010

### vela

Staff Emeritus
The one in the beginning shouldn't be there because x=0 isn't a possible outcome of rolling a die.

The idea of a moment generating function is that if you write it out as a power series in t, the coefficient of [itex]t^n[/tex] is the nth moment divided by n!.

To simplify your expression, note that if [itex]r=e^t[/tex], your moment generating function is

$$M=(1/6)(r+r^2+r^3+r^4+r^5+r^6)$$

Use the formula for the partial sum of a geometric series to get it in closed form.

4. Feb 15, 2010

### exitwound

You're right about the 0 sides on a die. My mistake on that one.

As for "using the formula for the partial sum of a geometric series to get it in closed form", that's speaking MoonMan. I haven't had geometric series in over a decade. I'll have to do some digging. Ugh.

Also, I don't know what you mean by:
To be completely honest, I don't even know what t is. I'm not a math major and the language is mostly foreign to me.

Last edited: Feb 15, 2010
5. Feb 15, 2010

### vela

Staff Emeritus
If you expand the exponential as a Maclaurin series, you get

$$M_x(t) = E(e^{tx}) = E\left[1+tx+\frac{1}{2!}(tx)^2+\frac{1}{3!}(tx)^3+\hdots\right]=E(1)+E(x)t+E(x^2)(t^2/2!)+E(x^3)(t^3/3!)+\hdots$$

So you can each term is proportional to a moment of x. If you have M(t), you can find the nth moment using the formula

$$E(x^n)=\left \frac{d^nM}{dt^n}\right|_{t=0}$$

I don't know of an intuitive interpretation of what t is. I just view it as a bookkeeping device.