Absolute moments of symmetrical distributions

Question

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?

==========ADDITION==========

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 $m=\lfloor \nu/2\rfloor$.

Answer 1

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 $$ \mathbb E |Y_a| = e^{1/32} + a e^{\frac{1}{32}-2\pi^2} = e^{1/32}(1 + a e^{-2 \pi^2}). $$

Answer 2

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).