Moment Generating Function problem

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Homework Help Overview

The discussion revolves around the moment generating function (mgf) of a probability density function defined as f(x) = 3x^2 for x in [0, 1] and f(x) = 0 elsewhere. The original poster attempts to find E[X^2] using both the integral definition and the mgf, noting a discrepancy in the results.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of the mgf, with the original poster providing their integral setup and results. Questions arise regarding the validity of the calculations and the nature of the unbounded results encountered.

Discussion Status

The discussion is ongoing, with some participants questioning the calculations and the implications of the mgf not being defined at certain points. There is recognition that the mgf may not be applicable in this case, but no consensus has been reached on the specific issues with the calculations.

Contextual Notes

Participants note that the mgf can be undefined in certain scenarios, which may contribute to the confusion in this problem. The original poster's calculations are under scrutiny, and there is an acknowledgment of the need for clarity in the setup and bounds of the integrals involved.

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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?
 
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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.
 
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.
 
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.
 

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