Do the principal values of higher moments exist for the Cauchy distribution?

Question

It is known that the Cauchy distribution has undefined moments, and that the expectation has a principal Cauchy value $\operatorname{PV}\left( \mathbb{E} [X] \right)$ of zero.

I wonder if $\operatorname{PV}\left( \mathbb{E} \left[ (X - \operatorname{PV}\left( \mathbb{E} [X] \right))^n \right] \right)$ exists for all $n$, or if there is a highest principal Cauchy moment for the Cauchy distribution.

Do all the higher moment expectations have defined principal Cauchy values for the Cauchy distribution?

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The first simplification of the above is to note that $\operatorname{PV} \left(\mathbb{E}[X]\right) = 0 \implies \operatorname{PV}\left( \mathbb{E} \left[ (X - \operatorname{PV}\left( \mathbb{E} [X] \right))^n \right] \right) = \operatorname{PV}\left( \mathbb{E} \left[ X^n \right] \right)$.

From the definition of the Cauchy distribution and the Cauchy principal value we have

$\lim_{\epsilon \rightarrow 0^+} \left[ \int_{-\infty}^{0-\epsilon} x^n \frac{1}{\pi \gamma \left[ 1 + \left(\frac{x-x_0}{\gamma} \right)^2 \right]}dx + \int_{0+\epsilon}^{\infty} x^n \frac{1}{\pi \gamma \left[ 1 + \left(\frac{x-x_0}{\gamma} \right)^2 \right]}dx \right].$

So my question reduces to whether the above expression is defined.

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Wolfram Alpha tells me that

$\int x^n \frac{1}{\pi \gamma \left[ 1 + \left(\frac{x-x_0}{\gamma} \right)^2 \right]}dx = \frac{\gamma x^{n+1}}{2 \pi \sqrt{- \gamma^2} (n+1)} \left( \frac{_2F_1 \left(1, n+1;n+2;\frac{x}{m - \sqrt{- \gamma^2}} \right)}{m-\sqrt{- \gamma^2}} - \frac{_2F_1 \left(1, n+1;n+2;\frac{x}{m + \sqrt{- \gamma^2}} \right)}{m + \sqrt{- \gamma^2}} \right) + C$

where $_2F_1 (\cdot, \cdot ; \cdot ; \cdot )$ is a hypergeometric function.

Answer 1

The odd integral moments have principal values of zero while the even integral moments have infinite principal values.

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By definition, the Cauchy Principal Value [w.r.t. infinity] of an integral over the real line is

$$\operatorname{pv}\int_\mathbb{R} g(x)\,\mathrm{d}x = \lim_{R\to\infty} \int_{-R}^R g(x)\,\mathrm{dx}.$$

When $g$ is antisymmetric -- that is, $g(x) = -g(x)$ for all $x,$ obviously the right hand side is always zero, whence its limit is zero.

The Cauchy density is proportional to $1/(1+x^2),$ which is symmetric about $0.$ Consequently, for any antisymmetric integrable function $h,$

$$g(x) = \frac{h(x)}{1+x^2}$$

is also integrable and antisymmetric, whence

$$\operatorname{pv} \int_\mathbb R \frac{h(x)}{1+x^2}\,\mathrm{d}x = 0.$$

This is the case for $h(x) = x^{2k+1}$ when $k$ is a natural number. It is not the case when $h(x) = x^{2k}.$ With $k=0,$ the integral converges, but for $k\ge 1$ use the simple fact that when $|x|\ge 1,$

$$\frac{x^{2k}}{1+x^2} \ge \frac{x^{2k}}{x^2+x^2} = \frac{1}{2}x^{2k-2} \ge \frac{1}{2}$$ to estimate

$$\begin{aligned} \lim_{R\to\infty}\int_{-R}^R \frac{x^{2k}}{1+x^2}\,\mathrm{d}x &\ge \lim_{R\to\infty}\left(\int_{-R}^{-1} + \int_1^R\right) \frac{x^{2k}}{1+x^2}\,\mathrm{d}x\\ &\ge \lim_{R\to\infty}\left(\int_{-R}^{-1} + \int_1^R\right) \frac{1}{2}\,\mathrm{d}x\\ &= \lim_{R\to\infty} R - 1 =\infty. \end{aligned}$$

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A similar argument by symmetry handles the case of $negative$ integral $n,$ but now the principal value is computed by excising a small neighborhood $(-\epsilon,\epsilon)$ around zero and taking the limit (from above) as $\epsilon\to 0.$ Again there's cancellation for odd $n$ but divergence for even $n.$ Alternatively, change variables from $x$ to $1/x,$ which reduces the integral to the form explicitly considered here.