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Moment generating function DNE

  1. Aug 8, 2014 #1
    1. The problem statement, all variables and given/known data

    Write the integral that would define the mgf of the pdf,
    [itex]f(x) = \frac 1{\pi} \frac 1{1+x^2} dx [/itex]


    2. Relevant equations

    The moment generating function (mgf) is [itex] E e^{tX}[\itex].

    3. The attempt at a solution
    My question really has to do with improper integrals. I must show the improper integral diverges:

    [itex]\int_0^{\infty} e^{tx} \frac 1{\pi} \frac 1{1+x^2} dx [/itex].

    Now if do integration by parts and let [itex]u=e^{tx}[/itex] and [itex] dV = \frac 1{1+x^2}dx [/itex], then I have:

    [itex] \frac 1{\pi} e^{tx} arctan(x) |_0^{\infty} - \int_0^{\infty} \frac 1{\pi} t arctan(x) e^{tx} dx [/itex].

    However I can see that [itex] arctan(x) e^{tx} \frac 1{\pi} |_0^{\infty} [/itex], will be ∞. So is this enough to show that the improper integral diverges?
     
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  3. Aug 8, 2014 #2

    jbunniii

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    No, it's not enough unless you establish that the second integral (the one being subtracted) is not also ##\infty##. If it is, then you have an indeterminate form ##\infty - \infty## and more work is required to determine convergence.

    By the way, if the moment generating function is ##E[e^{tX}]##, shouldn't you be integrating over all of ##\mathbb{R}## instead of ##[0,\infty)##? As you have written it, your integral does in fact converge for ##t \leq 0##. (Granted, in order for it to be of any use in generating moments, you need convergence on some open interval containing ##t=0##.)
     
  4. Aug 8, 2014 #3
    I was thinking a comparison test would work, except that (I think) t can be any real number.

    The support wasn't stated in the problem, so I just guessed. Looking back, I think the author meant this to be the Cauchy distribution, in which case the support is all reals.
     
  5. Aug 8, 2014 #4

    jbunniii

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    For positive ##t##, you can simply show (using L'Hospital's rule, for example) that
    $$\lim_{x \rightarrow \infty}\frac{e^{tx}}{1 + x^2} = \infty$$
    Similarly, for negative ##t##, show that
    $$\lim_{x \rightarrow -\infty}\frac{e^{tx}}{1 + x^2} = \infty$$
    These two facts imply that the integral does not converge for any nonzero ##t##. (Of course, it converges to 1 for ##t=0##.)
     
  6. Aug 8, 2014 #5
    I maybe (figuratively) beating a dead horse here, but is it always the case that:

    If a function fails to converge at a finite value as it approaches infinity, then the improper integral of said function from an arbritrary value to infinity must diverge

    ?
     
  7. Aug 8, 2014 #6

    jbunniii

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    No, that's not true. Consider the function defined as follows:
    $$f(x) = \begin{cases}
    1 & \text{if }n \leq x \leq n + 1/n^2 \text{ and }n \geq 1 \text{ is an integer}\\
    0 & \text{elsewhere} \\
    \end{cases}$$
    Then this function oscillates between 0 and 1 as ##x \rightarrow \infty##, so it doesn't converge to any value. But the intervals where the function value is 1 become narrower and narrower, and the integral does converge:
    $$\int_0^\infty f(x) dx = \sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6}$$
    However, your function does something more severe: it diverges to ##\infty## as ##x \rightarrow \infty##. In this case it is true that the integral diverges to infinity. Proof: given any positive integer ##N##, there is some ##X > 0## such that ##f(x) \geq N## for all ##x \geq X## (definition of divergence to ##\infty##). Therefore, since the integrand is nonnegative, we have
    $$\int_{0}^{\infty} f(x) dx \geq \int_{X}^{\infty} f(x) dx \geq \int_{X}^{\infty} N dx = N \cdot \infty = \infty$$
     
    Last edited: Aug 8, 2014
  8. Aug 8, 2014 #7

    jbunniii

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    Fixed some typos, sorry.
     
  9. Aug 8, 2014 #8
    What is capital X above?

    Note: the only "proofy" math I know is from a discrete math course I took.
     
  10. Aug 8, 2014 #9

    jbunniii

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    OK, let me state the argument a bit more simply. We established that if ##t > 0##,
    $$\lim_{x \rightarrow \infty}\frac{e^tx}{1+x^2} = \infty$$
    by L'Hospital's rule. If the function goes to infinity, then it certainly must exceed 1 beyond some point (the exact value of the point depends on ##t##). In other words, if you draw a horizontal line at ##y=1##, then the function will eventually always stay above that line if ##x## is large enough. Therefore the area under the function is larger than the area under the line, which is infinite.
     
  11. Aug 8, 2014 #10

    jbunniii

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    Here is a picture which might help. In the picture, ##X## indicates the point beyond which the function stays above the line ##y=1##. So the area of the part of the function to the right of ##X## must be even larger than the area of the rectangle to the right of ##X##, which is infinite (height = 1, width = ##\infty##).
     

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  12. Aug 9, 2014 #11

    statdad

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    Another way to see this: remember that
    [tex]
    e^{tx} = \sum_{n=0}^\infty \frac{(tx)^n}{n!}
    [/tex]

    so the integral for your mgf is really
    [tex]
    \frac 1 {\pi} \, \int_{-\infty}^{\infty} \sum_{n=0}^\infty \frac{t^n x^n}{1+x^2} \, dx
    [/tex]

    If the mgf existed then these integrals would all exist (since they are the moments)

    [tex]
    \int_{-\infty}^{\infty} \frac{x^n}{1+x^2} \, dx \quad n \ge 0
    [/tex]

    but for [itex] n \ge 1 [/itex] none of those integrals converge, therefore the mgf doesn't exist.

    Summary: If an mgf exists then all integral moments exist. So, if some integral moment does not exist the mgf does not exist.
     
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