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what is the difference between "infinite moment" and "moments don't exist"?

moreover, if I find out the moment generating function for some distribution, and take first derivative set s=1, and if I get $\infty$, what does it mean? infinite mean? or mean doesn't exist?

Thanks in advance.

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    $\begingroup$ If $X$ is a random variable and $X^+$ is the positive part and $X^-$ the negative part, then by definition $E[X]=E[X^+]-E[X^-]$, if at least one of $E[X^+]$ and $E[X^-]$ is finite. If $E[X^+]=\infty$ and $E[X^-]<\infty$ then $E[X]=\infty$. But if both $E[X^+]=E[X^-]=\infty$ then the moment is not well-defined and hence it does not exist. $\endgroup$ – Stefan Hansen Mar 6 '12 at 20:00
  • $\begingroup$ Expanding on Stefan's excellent comment above, basically the central idea to keep on mind is that $\infty - \infty$ is undefined. See e.g. this question: math.stackexchange.com/questions/60766/… So if the moment "is $\infty - \infty$", it is indeterminate, so we say it doesn't exist (the same thing with $1/0$ en.wikipedia.org/wiki/Indeterminate_form). But if the moment is "$\infty$", but still not "$\infty - \infty$", then we can at least say that it exists. $\endgroup$ – Chill2Macht Jun 30 '17 at 17:10
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If $X$ has a Cauchy distribution, then $E(X^2)=\infty$, and one sometimes expresses that by saying the second moment does not exist. But concerning $E(X^3)$, one may say that it does not exist, but one cannot say that it is infinite. If you look at $$ E(X^3) = \int_{-\infty}^\infty x^3 \frac{dx}{\pi(1+x^2)}, $$ what you find is that both the positive and negative parts are infinite: $$ \int_{-\infty}^0 x^3 \frac{dx}{\pi(1+x^2)}=-\infty\text{ and }\int_0^\infty x^3 \frac{dx} {\pi(1+x^2)} = +\infty. $$ When that happens, you can't say the integral has any particular value, because you can actually change the value by rearraning the way in which the bounds approach infinity. For example, if you find $$ \lim_{a\to\infty} \int_{-a}^a \frac{x}{1+x^2}\;dx \text{ and }\lim_{a\to\infty} \int_{-a}^{2a} \frac{x}{1+x^2}\;dx, $$ then you actually get two different finite numbers. (That can happen only if the positive and negative parts are both infinite.)

Later note: Things like the central limit theorem and the law of large numbers fail in cases where the integral that defines the first moment has both the positive and negative parts infinite. For example, suppose $$ X_1,\ldots,X_N\sim\operatorname{i.i.d.}\operatorname{Cauchy}, $$ so the probability density of each is $\displaystyle f(x) = \frac{1}{\pi(1+x^2)}$. Then the probablity distribution of the sample mean $\bar X=(X_1+\cdots+X_n)/n$ is the same Cauchy distribution. So $\bar X$ won't converge to a point as $n\to\infty$, nor will $\sqrt{n}\;\bar X$ converge weakly to a normal distribution.

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  • $\begingroup$ the non-existence or infinite comes from the convergence of the integration, rather than from a probability theory view, am I right saying that? THanks $\endgroup$ – fast tooth Mar 6 '12 at 22:28
  • $\begingroup$ @wenhoujx : Yes. $\endgroup$ – Michael Hardy Mar 7 '12 at 0:17
  • $\begingroup$ But there are probabilistic consequences. Things like the law of large numbers and the central limit theorem fail when you try to apply them to cases where the integral that gives the expected value converges only conditionally. $\endgroup$ – Michael Hardy Mar 7 '12 at 0:18
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When the sum or integral defining $E[|X^n|]$ diverges, $E[X^n]$ might diverge to $+\infty$, so you could say "the $n$'th moment is $+\infty$" as an alternative to "the $n$'th moment does not exist". This would have to be the case if $n$ is even or if $X \ge 0$ almost surely. However, if $n$ is odd and $X$ can be either positive or negative, both positive and negative parts might diverge, so all you could say is "the $n$'th moment does not exist".

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