Finding the joint moment generating function given the joint PDF

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SUMMARY

The discussion centers on finding the joint moment generating function given a joint probability density function (PDF) involving a Gaussian integral. Participants highlight that the integral is non-elementary and suggest integrating over the variable x first to simplify the problem. Key insights include recognizing the integral's resemblance to a bivariate normal distribution and utilizing known results from Laplace transforms. The integration limits spanning from negative to positive infinity allow for closed-form solutions, leveraging properties of Gaussian integrals.

PREREQUISITES
  • Understanding of joint probability density functions (PDFs)
  • Familiarity with Gaussian integrals and their properties
  • Knowledge of Laplace transforms
  • Basic calculus, particularly integration techniques
NEXT STEPS
  • Study the properties of bivariate normal distributions
  • Learn about Gaussian integrals and their applications
  • Explore Laplace transforms in probability theory
  • Review techniques for integrating exponential functions over infinite limits
USEFUL FOR

Mathematicians, statisticians, and students in advanced probability and statistics courses who are working with joint distributions and moment generating functions.

stripes
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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.
 

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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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