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I'm studying the distributional properties of a laplace distribution, and I'm trying to get some intuition beyond plotting the distribution of what it means to have an undefined moment.

In wikipedia you can see that the mgf is only defined for $|t| < 1/b$ so as the variance of the laplace distribution increases to 1, you lose all moments including the mean. Does this matter? What is the intuition? For example the fourth moment can blow up, but the distribution will still look generally okay. What is the benefit of having a defined moment if you have the distribution?

If I have some data that fits the laplace distribution fits well with a very high b, should I be concerned? If for two data sets where b is close to 1 in one data set, but smaller in another, am I more confident in the fit to the data set with a b that generates more moments?

Any thoughts would be greatly appreciated. And if I'm thinking about this the wrong way let me know.

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    $\begingroup$ "so as the variance of the laplace distribution increases to 1, you lose all moments including the mean" --- can you please explain why you think this follows? I'm pretty sure all moments of the Laplace exist no matter what the value of the scale parameter. $\endgroup$ – Glen_b -Reinstate Monica Sep 20 '13 at 1:39
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    $\begingroup$ I'm quite familiar with that specific page, thanks. I believe you've mistaken in your understanding of how mgfs and moments are related. You statements above imply that you think that a Laplace with $b=1$ would not have a mean; that's certainly false, directly from the definition of the mean. Ergo, the manner in which you arrive at your conclusion is flawed. You think MGFs are telling you something about moments (and they do), but what you think the mgf is telling you isn't what its telling you. Can you be specific about how you arrived at your conclusion, so I can help you find the error? $\endgroup$ – Glen_b -Reinstate Monica Sep 20 '13 at 21:34
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    $\begingroup$ Specifically: all $k^\text{th}$ moments exist, $k=0,1,2,\ldots$. The odd central moments are all 0. The central moments for $k$ even are $\mu_k = k!b^k$. (You only need investigate $b=1$; the fact that $b$ is a scale parameter immediately leads to the factor of $b^k$. Note that the central moments of the Laplace are the raw moments of the corresponding exponential, and it's trivial to show that the raw $k^\text{th}$ moment of a standard exponential is $k!$.) $\endgroup$ – Glen_b -Reinstate Monica Sep 20 '13 at 21:40
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    $\begingroup$ hey Glen_B you are totally right. I misunderstood how to use the moment generating function. Great that I asked this question and thanks for being persistent. Digging into this more closely I see that form utdallas.edu/~jjue/cs6352/probability/node2.html I was using the moment generating function incorrectly to arrive at the moment. Now when I'm calculating the moments I don't end up with moments exploding. $\endgroup$ – phubaba Sep 21 '13 at 14:47
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    $\begingroup$ Hi phubaba, could you please generate a short explanation of how moments relate to the MGF based on your new understanding, including how to get say the first moment of the above Laplace from that and post it as an answer? $\endgroup$ – Glen_b -Reinstate Monica Sep 22 '13 at 1:10
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I was incorrectly using the moment generating function which led to my misunderstanding of the Laplace distribution.

The moment generating function is $M_X(\theta) = \text{E}(e^{\theta X})$.

When you use that to find the $n^{\text{th}}$ moment, you take the $n^{\text{th}}$ derivative at $\theta=0$:

$$\frac{d^{n}(M_X(\theta))}{d(\theta)^{n}} |_{\theta=0}\quad\text{.}$$

If you see my note above there is a proof using Taylor series expansion of $E(e^{\theta X})$. When you take the n-th derivative the leading term will not have a $\theta$ in it, but higher order terms will have $\theta$. This allows you to set $\theta=0$ and use the moment generating function to produce moments.

So for the Laplace we have $E(e^{\theta X}) = e^{\mu\theta}/(1-b^{2}\theta^{2})$ (from Wikipedia)

$E(X) = d^{1}(M_X(\theta))/d(\theta)^{1} = (e^{\theta\mu} (\mu + b^2 \theta (2 - \theta \mu)))/(-1 + b^2 \theta^2)^2$

if you evaluate this for $\theta=0$, then you get $E(X) = \mu$ as expected.

Now the second part of my question is trying to understand undefined moments. The implication of an undefined moment means that trying to estimate the parameters of the distribution by matching moments will generally require more advanced techniques (such as maximizing log-likelihood).

There is a good discussion about this for the Cauchy distribution which does not have defined moments see http://en.wikipedia.org/wiki/Cauchy_distribution

As an added thought, in Python there is a symbolic algebra package called sympy that makes evaluating moments very simple using symbolic algebra. There are simple formulas to convert non-central moments to central moments, allowing you to calculate skewness and kurtosis fairly easily for many distributions.

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    $\begingroup$ I've made a few edits to clean the exposition up a little. I plan to come back and do a more detailed check of the calculations at the end. $\endgroup$ – Glen_b -Reinstate Monica Sep 24 '13 at 2:55

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