Moment Generating Function

1. Apr 24, 2007

Glass

I'm given the probability density function:
f(x) = 3x^2 for x in [0, 1]
f(x) = 0 elsewhere
I want to find E[X^2] which is easy if I use the integral definition (I got 3/5). Yet, when I try and do this using Moment Generating Function (mgf) I cannot seem to get the same answer (in fact I get an unbounded answer). I'm wondering why this is the case. Any help?

2. Apr 24, 2007

quasar987

Well as you guessed, that's not normal. So there has to be something quite not right with one of your calculations. But nobody can help you find what is the problem if we can't see your calculations.

3. Apr 24, 2007

Glass

From the definition of mgf I have:
$$M(t) = \int_{0}^{1}e^{xt}3x^2dx$$
Then after some integration by parts (or using a symbolic integrator) I get:
$$M(t) = e^t(\frac{3}{t} - \frac{6}{t^2} + \frac{6}{t^3}) - \frac{6}{t^3}$$
Then if we differentiate that twice (or even once or zero times) there are still 1/t (or some power of 1/t) terms in there.

4. Apr 24, 2007

quasar987

Actually, I did the calculations, and I get what you're saying when you say it's unbounded. At first I imagined that by "unbounded" you meant that the integral blew up, because you forgot that the bounds were 0 and 1 and left them to ±infinity or something. But it's actually that M(0) is not defined.

This is nothing to be so surprised about though. There are many cases when you can't use the mgf because it simply is not defined on any nbhd of 0.

A trivial exemple of when the mgf does not exist is for the uniform distribution. We then have

$$M_X(t)=\int_{0}^{1}e^{tx}dx=\frac{1}{t}(e^t-1)$$

Undefined at t=0.