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Homework Help: Finding the joint moment generating function given the joint PDF

  1. Mar 31, 2014 #1
    1. The problem statement, all variables and given/known data


    2. Relevant equations

    3. The attempt at a solution


    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.

    Attached Files:

  2. jcsd
  3. Mar 31, 2014 #2


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    Try something.
    Try integrating over x for example.
    You will be surprised.
  4. Mar 31, 2014 #3
    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...
  5. Mar 31, 2014 #4


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    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:


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

    Other hint: 2bx + x² = (x+b)² - b²
  6. Mar 31, 2014 #5

    Ray Vickson

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    Do the x-integration first; it has the form
    [tex] G(t,y) = \int_{-\infty}^{\infty} e^{-y} e^{tx} e^{-(x-y)^2/2}/\sqrt{2 \pi} \, dx [/tex]
    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 .
  7. Apr 1, 2014 #6
    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 gonna 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...
  8. Apr 1, 2014 #7
    wolfram gives

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

    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: May 6, 2017
  9. Apr 1, 2014 #8


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    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.
  10. Apr 1, 2014 #9


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    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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