Existence of fractional moment of cauchy distribution

Question

For the standard Cauchy distribution, do fractional moments exist? If $Y \sim C(1,0)$, is it possible to evaluate $E(Y^{1/3})$?

Answer 1

Let's apply the law of the unconscious statistician, so we want to be able to evaluate

$\int_{-a}^b \frac{x^k}{1+x^2} dx$

in the limit as $a,b$ each $\to\infty$.

As mentioned in comments, for most fractional $k$ there's problems for negative $x$.

Let's step aside from that issue, and just talk about convergence of the right side.

Consider then, the convergence of

$I(b)=\int_0^b \frac{x^k}{1+x^2} dx\,$, for $0<k<1$

First consider splitting into $\int_0^1$ and $\int_1^b$ for $b>1$

for $\int_0^1 \frac{x^k}{1+x^2} dx$ note that when $x$ and $k$ are between 0 and 1 and that $x^k\leq 1$, so that integral is bounded.

Now note that for $x>0$, we have $\frac{1}{(1+x^2)}<x^{-2}$, so

$\int_1^b \frac{x^k}{1+x^2} dx<\int_1^b x^{k-2} dx = \left. \frac{x^{k-1}}{k-1}\right|_1^b = \frac{b^{k-1}-1}{k-1}=\frac{1-b^{k-1}}{1-k}$

which doesn't get any bigger than $\frac{1}{1-k}$ (when $b>1$ and $0<k<1$, as here).

So in the limit as $b\to\infty$, the integral converges.

As a result, $\int_{-\infty}^\infty \frac{|x|^k}{1+x^2} dx$ converges, and so when $x^k$ is real, $\int_{-\infty}^\infty \frac{x^k}{1+x^2} dx$ will converge to a real answer.

Specifically, the $\frac{1}{3}$ moment of a standard Cauchy is therefore 0 (as are all reciprocal-of-odd-positive-integer moments), by symmetry.

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Note that according to Wolfram alpha (I didn't try to integrate it myself) $\int_{0}^\infty \frac{x^{1/3}}{1+x^2} dx=\frac{\pi}{\sqrt{3}}$.

Answer 2

To complement @Glen_b's answer, I asked maple for help, and it gives the following answer, which is a complex number since we ask for fractional powers of negative numbers! If such a complex moment is of any use I do not know, probably one would have to take the moments of the positive and negative part separately, as is done in definition of The Mellin transform https://en.wikipedia.org/wiki/Mellin_transform As that page shows, these fractional moments is essentially the Mellin transform. The Mellin transform is used in studying products of random variables.

The result is (I did not check on maple ...) $$ \int_{-\infty}^\infty x^\alpha \frac1{\pi (1+x^2)}\; dx = \frac{(-1)^\alpha + 1}{2 \cos (\pi\alpha/2)}, \quad \alpha < 1 $$ That's mostly curiosa, more interesting is the fractional moment of the absolute value $$ 2\int_0^\infty |x|^\alpha \frac1{\pi (1+x^2)}\; dx = \sec(\pi\alpha/2), \quad \alpha<1 $$ Some papers which seems to have more details: On Distribution of Product of Stable Laws and On the Use of Fractional Calculus for the Probabilistic Characterization of Random Variables and MELLIN-BARNES INTEGRALS FOR STABLE DISTRIBUTIONS AND THEIR CONVOLUTIONS (I did'nt look much at them yet, but they seem to be relevant)

Answer 3

The fractional moments $E[x^k]$ do exist for $k<1$. You have to be careful with fractional moments of distributions with negative values, of course. The value $x^k$ may end up being a complex number.

The absolute moments $E[|x|^k]$ were studied in Goria, M. N. "Fractional absolute moments of the Cauchy distribution." Quaderni di Statistica e Matematica applicata alle scienze Economico-Sociali 1 (1992): 3-9.

Answer 4

Let $\mu \in \mathbb{R}$ and $\sigma > 0$ be the location and scale parameters of a Cauchy distribution respectively. Let $r > 0$ and $\theta \in (0, \pi)$ such that $r\exp(i \theta) = \mu + \sigma i$, where $i$ denotes the imaginary unit. Let $-1 < \text{Re}(a) < 1$. Assume that $x^a = |x|^a \text{sign}(x), \ x \ne 0$ and $0^a = 0$. Then, by (3.0.3) in V. M. Zolotarev, One-dimensional stable distributions, AMS, 1986 https://bookstore.ams.org/mmono-65,

$$ E[|X|^a] = \frac{\cos\left(a (\pi /2 - \theta)\right)}{\cos(a\pi /2)} $$ and $$ E[X^a] = \frac{\sin\left(a (\pi/2 - \theta) \right)}{\sin(a \pi/2)}, \ a \ne 0, $$ and $$ E[X^a] = 1 - \frac{2\theta}{\pi}, \ a = 0. $$