Suppose $F~$ is a probability distribution symmetrical about 0, for which all moments exist. Let $\mu_i~$be the $i$-th moment (of course $\mu_i=0$ if $i~$ is odd).

We know there are some conditions under which $\{\mu_i\}$ determines $F$, and in that case $\{\mu_i\}$ must determine the absolute moments of $F$.

Q1. How can we write the absolute moments directly in terms of $\{\mu_i\}$? (I have seen a formula of von Bahr (1965) quoted here, but I wonder if there is something simpler than an integral like that.)

Q2. In the case that $\{\mu_i\}$ doesn't determine $F$, does it still determine the absolute moments?


Here is the result of Bengt von Bahr, Ann. Math. Stat. 36 (1965) 808-818. I changed it in the fashion of Ushakov, Statistics & Probability Letters Volume 81, Issue 12, December 2011, Pages 2011-2015, which seems to correct a misplaced bracket in the original. All the integrals are over $(-\infty,\infty)$.

Let $H(x)$ be a function of bounded variation on $(-\infty,\infty)$ with finite absolute moment $ \int |x|^\nu |dH(x)|$ for $\nu>0$ not an even integer. Define the ordinary moments $\mu_j=\int x^j dH(x)$ for $j=0,1,\ldots,\lfloor \nu\rfloor$. Let $\phi(t)$ be the characteristic function of $H(x)$. Then $$ \int |x|^\nu dH(x) = (\Gamma(\nu+1)/\pi)\cos((\nu+1)\pi/2) \int \frac{\Re\phi(t) -\sum_{j=0}^m (-1)^j\mu_{2j}t^{2j}/(2j)!}{|t|^{\nu+1}} dt,$$ where $\Re$ stands for real part and $m=\lfloor \nu/2\rfloor$.

  • $\begingroup$ Specializing the display equation to your particular case of interest, since the distribution is symmetric, $\phi(t) = \Re \phi(t)$ and since an infinite number of moments are assumed to exist, then for any nonnegative integer $n$, $\phi(t) = \sum_{k=0}^n (-1)^k \mu_{2k} t^{2k} / (2k)! + o(t^{2n})$. $\endgroup$ – cardinal Oct 14 '11 at 16:44

This response attempts to address your second question.

I believe the following (a modification of a classical counterexample) is a counterexample, but it would be helpful to have it checked by others.

It is well-known that the lognormal distribution is not determined by its moments. The density of the (standard) lognormal is $$ f_0(x) = \frac{1}{x \sqrt{2\pi}} e^{-(\log x)^2/2} , $$ for $x > 0$ and is $0$ otherwise.

We can construct an indexed family of distributions with the same moments, as follows. Let the density parameterized by $a \in [-1,1]$ be $$ f_a(x) = f_0(x) (1 + a \sin(2\pi\log x)) . $$

Note that $$ \sqrt{2\pi} \int_0^\infty x^r f_0(x) \sin(2\pi\log x) = \int_{-\infty}^{\infty} e^{yr + r^2} e^{-(y+r)^2/2} \sin(2\pi(y+r)) \mathrm{d}y , $$ by making the change of variables $\log x \mapsto y + r$. Simplifying the second integral, we get $$ e^{r^2/2} \int_{-\infty}^{\infty} e^{-y^2/2} \sin(2\pi y + 2\pi r) \mathrm{d}y , $$ and, in particular, the integral is equal to zero for all $r = k/2$ for $k \in \mathbb Z$.

Hence, the $f_a$ are all densities and they all have the same moments and "half-moments".

Counterexample (claimed): Let $\{X_a\}$ be a set of random variables indexed by $a$ having density $f_a$. Let $\epsilon \in \{-1,+1\}$ be a random variable such that $\mathbb P(\epsilon = 1) = 1/2$ and independent of $\{X_a\}$. Set $Y_a = \epsilon X_a^{1/4}$. Then, the $Y_a$ are symmetric and have the same moments, but the absolute moments differ.

Proof: The $Y_a$ are symmetric by construction and $\mathbb E Y_a^{2n} = \mathbb E X_a^{n/2}$ and so they all share the same moments. But, for odd $n$, $$ \mathbb E |Y_a|^n = \mathbb E X_a^{n/4} = \mathbb E X_0^{n/4} + a \frac{e^{(n/4)^2/2}}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-y^2/2} \sin(2\pi y + \pi n / 2) \mathrm{d}y. $$

Taking $n = 1$, we see that $$ \mathbb E |Y_a| = e^{1/32} + a e^{\frac{1}{32}-2\pi^2} = e^{1/32}(1 + a e^{-2 \pi^2}). $$

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  • $\begingroup$ Yes, that is convincing, thanks. A question: can I click on the tick for two different answers, or change it if someone gives a great answer for Q1? $\endgroup$ – Brendan McKay Oct 13 '11 at 15:45
  • $\begingroup$ If you're asking if it's possible to "accept" two different answers, I don't believe so, but I could be wrong. But, please accept whatever answer that appears that you feel is most satisfying. $\endgroup$ – cardinal Oct 13 '11 at 16:21
  • $\begingroup$ I can't see the paper you link to, so maybe I'm a little bit unclear about what you're asking for Q1 when you say you want to write the absolute moments directly (i.e., explicitly?) in terms of (as a function of?) the raw moments. Assuming the counterexample holds, I believe it shows there's no hope of that. (There's a fixed sequence of moments that is associated with an uncountable class of absolute first moments). But, perhaps (a) I'm missing something obvious, (b) I'm misunderstanding your aim, or (c) your Q1 was intended only for distributions determined by their moments. $\endgroup$ – cardinal Oct 13 '11 at 16:25
  • $\begingroup$ I was mostly interested in Q1 for distributions determined by their moments, but wondered if it might be true more widely. Pietro's answer is the sort of thing I was looking for. $\endgroup$ – Brendan McKay Oct 14 '11 at 11:03

As to the question of relating the absolute moments of a measure $m$ to its even order moments, a possibility is the following, at least for a measure on a bounded interval. Assume w.l.o.g. that the measure is concentrated on the interval $I:=[-1,1]$.

From the binomial series with exponent $1/2$ we get a series absolutely convergent on $I$ $$|x|^k=\left(1+(x^{2k}-1)\right)^{1/2} =\sum_{h=0}^\infty\Big({\frac{1}{2} \atop h}\Big)(x^{2k}-1)^h $$ So we can integrate by series and get $$\int_I|x|^k dm=\sum_{h=0}^\infty \Big({\frac{1}{2} \atop h}\Big)a_{h,k}$$ where the coefficients $a_{h,k}$ are $$a_{h,k}:=\int_I (x^{2k}-1)^h dm =\sum_{j=0}^h (-1)^{h-j}\Big({h \atop j}\Big)\mu_{2kj}\, . $$

Notice that no symmetry assumptions on on the distribution is needed (which is not surprising, after all). On these lines, the case of a measure on $\mathbb{R}$ may be treated taking the limit as $T\to \infty$ of the moments on $[-T,T]$, extending the validity of the above formulas to the moments on $\mathbb{R}$ (of course, some assumption on the measure, or on the sequence of its moments, is needed, in order to allow to pass to the limit in the double sum).

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  • $\begingroup$ As a double sum, it is conditionally convergent in a strong way: the coefficient of $\mu_{2kj}$ is 0 for $j=0$ and a divergent summation for $j\ge 1$. $\endgroup$ – Brendan McKay Oct 14 '11 at 11:05

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