# Homework Help: Moment generating function DNE

1. Aug 8, 2014

### Mogarrr

1. The problem statement, all variables and given/known data

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

2. Relevant equations

The moment generating function (mgf) is $E e^{tX}[\itex]. 3. The attempt at a solution My question really has to do with improper integrals. I must show the improper integral diverges: [itex]\int_0^{\infty} e^{tx} \frac 1{\pi} \frac 1{1+x^2} dx$.

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

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

However I can see that $arctan(x) e^{tx} \frac 1{\pi} |_0^{\infty}$, will be ∞. So is this enough to show that the improper integral diverges?

2. Aug 8, 2014

### jbunniii

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 $t=0$.)

3. Aug 8, 2014

### Mogarrr

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.

4. Aug 8, 2014

### jbunniii

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 $t=0$.)

5. Aug 8, 2014

### Mogarrr

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

?

6. Aug 8, 2014

### jbunniii

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
7. Aug 8, 2014

### jbunniii

Fixed some typos, sorry.

8. Aug 8, 2014

### Mogarrr

What is capital X above?

Note: the only "proofy" math I know is from a discrete math course I took.

9. Aug 8, 2014

### jbunniii

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.

10. Aug 8, 2014

### jbunniii

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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11. Aug 9, 2014

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

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

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

$$\int_{-\infty}^{\infty} \frac{x^n}{1+x^2} \, dx \quad n \ge 0$$

but for $n \ge 1$ 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.