Finding the joint moment generating function given the joint PDF

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Homework Help Overview

The discussion revolves around finding the joint moment generating function given a joint probability density function (PDF). Participants are exploring the challenges posed by a non-elementary integral that arises in the process, particularly in the context of Gaussian integrals.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the difficulty of integrating a non-elementary function and consider various substitutions. Some suggest starting with integration over one variable, while others note the appearance of Gaussian-like terms and express uncertainty about how to handle the additional exponential factor.

Discussion Status

There are multiple lines of reasoning being explored, with some participants suggesting specific integration techniques and others questioning the implications of the integral's form. Hints and references to known results are provided, but no consensus has been reached on a definitive approach.

Contextual Notes

Participants mention constraints related to the course material, specifically the use of the error function, which is noted to be outside the scope of their studies. There is also a reference to the limits of integration spanning from negative to positive infinity, which influences the evaluation of the integral.

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Homework Statement



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Homework Equations





The Attempt at a Solution



attachment.php?attachmentid=68178&stc=1&d=1396258056.jpg


The problem is the integral is non-elementary, so now what? Part (b) follows trivially from part (a). But is there some kind of shortcut I have to take, because no matter what substitution I do, the integral needs to be written in terms of the error function, which is out of the scope of this course. Any hints would be great. Thanks.
 

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Try something.
Try integrating over x for example.
You will be surprised.
 
it seems something will just "vanish" but i can't find what, mainly because there is always the factor of e^x in there that gets in the way. I realize that it looks like a bivariate normal distribution though...
 
Try integrating on x, as a start.
Since the integration spans [-Inf, +Inf], you will end up with a definite integral.
Alternatively, you might re-use known results on Laplace transforms.
See for example:

http://planetmath.org/laplacetransformofagaussianfunction

Proving what you find there is not difficult and could be interesting.

Other hint: 2bx + x² = (x+b)² - b²
 
stripes said:

Homework Statement



attachment.php?attachmentid=68177&stc=1&d=1396257940.jpg


Homework Equations





The Attempt at a Solution



attachment.php?attachmentid=68178&stc=1&d=1396258056.jpg


The problem is the integral is non-elementary, so now what? Part (b) follows trivially from part (a). But is there some kind of shortcut I have to take, because no matter what substitution I do, the integral needs to be written in terms of the error function, which is out of the scope of this course. Any hints would be great. Thanks.

Do the x-integration first; it has the form
G(t,y) = \int_{-\infty}^{\infty} e^{-y} e^{tx} e^{-(x-y)^2/2}/\sqrt{2 \pi} \, dx
Basically, this is a standard Gaussian integral, and it can be done in closed form because the integration limits are ##\pm \infty##. See. eg., http://en.wikipedia.org/wiki/Gaussian_integral .
 
to me, e^-(x-y)^2 looks like a gaussian integral. but with the extra e^(tx), it no longer looks like a "standard" gaussian integral. of course i can see that because the limits are +/- infinity, that the x integration is just going to give root pi, or 1 or something stupid easy, but I'm not terribly familiar with WHAT the actual volume is under the 3d graph of a standard gaussian, let alone a gaussian with an extra e^(tx) in it...
 
wolfram gives

http://www4b.wolframalpha.com/Calculate/MSP/MSP11611geaf4ba3b24ad0800005cf7hi62g6gf6c5i?MSPStoreType=image/gif&s=23&w=230.&h=49.

as the integral wrt x. i can't "see" that. but it looks like it follows from maajdl's suggestion...
 
Last edited by a moderator:
stripes said:
to me, e^-(x-y)^2 looks like a gaussian integral. but with the extra e^(tx), it no longer looks like a "standard" gaussian integral. of course i can see that because the limits are +/- infinity, that the x integration is just going to give root pi, or 1 or something stupid easy, but I'm not terribly familiar with WHAT the actual volume is under the 3d graph of a standard gaussian, let alone a gaussian with an extra e^(tx) in it...

First perform the first integral.
If you want to walk, you need a first step.
And you will see that the second will be much easier than the first step.

That's really a general rule: don't delay the first for the reason that you do not trust the second step.
 
You didn't write here the step where you experience a difficulty.
I can a most guess what is your problem (if any).
Therefore, I add this hint:

Exp(-2bx - (x+c)²) = Exp(-(x+b+c)² + b²) = Exp(-z²+b²) = Exp(b²) Exp(-z²)
where z = x+b+c
 

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