4
$\begingroup$

$\newcommand{\PV}{\operatorname{PV}}$ I have been working on evaluating the first negative moment of a random variable with a piecewise density function by means of the Cauchy principal value, i.e. \begin{equation} \tag{1} PV\!E(X^{-1})=\PV\int_{-\infty}^\infty\frac{f_X(x)}{x}\,\mathrm dx = \lim_{\epsilon\to0}\left(\int_{-\infty}^{-\epsilon}+\int_\epsilon^\infty \right)\frac{f_X(x)}{x}\,\mathrm dx, \end{equation} where $f_X$ is a probability density function. I was unable to get any traction using the definition in $(1)$ but was able to evaluate the moments of $X$ greater than one which yielded \begin{equation} \begin{aligned} E(X^{\epsilon-1}) &=\int_{-\infty}^0 x^{\epsilon-1}f_X(x)\,\mathrm dx% +\int_0^\infty x^{\epsilon-1}f_X(x)\,\mathrm dx,\\ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% &=\int_0^\infty x^{\epsilon-1}f_X(x)\,\mathrm{d}x% -e^{\mathrm i\pi\epsilon}\int_{-\infty}^0 |x|^{\epsilon-1}f_X(x)\,\mathrm dx,\\ &\propto\Gamma(\epsilon)\left(h_1(\epsilon)-e^{i\pi\epsilon}h_2(\epsilon)\right):=\Gamma(\epsilon)D(\epsilon). \end{aligned} \end{equation} I then tried taking the limit of the moments as $\epsilon\to0$. As $\epsilon\to0$, $D(\epsilon)\to0$ while the gamma term blows up. However, by writing the limit as \begin{equation} \lim_{\epsilon\to0}E(X^{\epsilon-1})\propto\Gamma(\epsilon+1)D(\epsilon)/\epsilon, \end{equation} we see that it is $0/0$ indeterminant. Using L'Hopitals rule and evaluating the limit I got something of the form \begin{equation} \lim_{\epsilon\to0}E(X^{\epsilon-1})= L-f_X(0)\pi\mathrm i. \end{equation} Numerical evaluation revealed that $L=PV\!E(X^{-1})$; thus, what I have found is \begin{equation} \tag{2} \lim_{\epsilon\to0}E(X^{\epsilon-1})= PV\!E(X^{-1})-f_X(0)\pi\mathrm i. \end{equation} I was able to come up with a proof of the result in $(2)$ by means of contour integration but was wondering if there other ways of getting to this result. So with that said, what is going on here? Why does adding the extra power to $t$ result in a residual imaginary term?

$\endgroup$
5
  • $\begingroup$ I may have missed a point, but why are you not saying the pdf of the difference is symmetric about $0$, i.e. $f_X(x)=f_X(-x)$ and so (1) evaluates to $0$? Do the independent gamma random variables have the same distribution? $\endgroup$
    – Henry
    Apr 17 '18 at 16:31
  • $\begingroup$ @Henry that is because the pdf is in general not symmetric since it is the pdf of the difference of two independent gamma RVs which can have different shape and rate parameters. It is only symmetric if the shape and rate parameters are equal in which case you are right that the PV would evaluate to $0$. $\endgroup$ Apr 17 '18 at 16:34
  • 1
    $\begingroup$ A specialization of this type of problem is Frullani's Theorem. A good series of proofs and expansions are in: math.stackexchange.com/questions/61828 $\endgroup$
    – rrogers
    Apr 21 '18 at 14:20
  • $\begingroup$ @rrogers nice find! This is helpful. $\endgroup$ Apr 21 '18 at 14:24
  • 1
    $\begingroup$ I found this to be more helpful (versus the AMS): jstor.org/stable/2306584 but it's a little hidden behind the jstor paywall; only freely readable online. It expands Frullani's without "Assume that there exists an inaccessible cardinal. Then there is a model of ZFC in which every set of reals definable from a countable sequence of ordinals is Lebesgue measurable." :) $\endgroup$
    – rrogers
    Apr 21 '18 at 14:40
2
$\begingroup$

What is going on here? Why does the time in which I add the extra power to $t$ matter?

In the first case, for $t < 0$ you multiply the integrand with $e^{i\pi\varepsilon}\cdot \lvert t\rvert^{\varepsilon}$, and in the second case you multiply it just with $\lvert t\rvert^{\varepsilon}$. When we're looking only at the part of the integral for $t < 0$, it is clear by the properties of $f_X$ that

$$I(\varepsilon) := \int_{-\infty}^0 \lvert t\rvert^{\varepsilon}\cdot \frac{f_X(t)}{t}\,dt$$

tends to $-\infty$ as $\varepsilon \searrow 0$. What is not immediately obvious is that the decrease of $I(\varepsilon)$ is essentially proportional to $\varepsilon^{-1}$, and that causes the appearance of the purely imaginary term, since

$$e^{i\pi\varepsilon} = \cos (\pi\varepsilon) + i\sin (\pi\varepsilon) = 1 + i\pi\varepsilon + O(\varepsilon^2)\,.$$

The real part, $\cos (\pi \varepsilon) I(\varepsilon) = I(\varepsilon) + O\bigl(\varepsilon^2 I(\varepsilon)\bigr) = I(\varepsilon) + O(\varepsilon)$, combines with the corresponding integral over $t > 0$ to yield the Cauchy principal value of the integral, and the imaginary part $\sin (\pi\varepsilon)I(\varepsilon)$ produces the $-f_X(0)\pi$.

To see this, split the integral some more. Since $f_X$ is integrable we have

$$\lim_{\varepsilon \to 0} \int_{-\infty}^{-1} \lvert t\rvert^{\varepsilon} \frac{f_X(t)}{t}\,dt = \int_{-\infty}^{-1} \frac{f_X(t)}{t}\,dt$$

by the dominated convergence theorem. We can write the integral from $-1$ to $0$ as

$$\int_{-1}^0 \lvert t\rvert^{\varepsilon} \frac{f_X(t)}{t}\,dt = \int_{-1}^0 \lvert t\rvert^{\varepsilon} \frac{f_X(t) - f_X(0)}{t}\,dt + f_X(0)\int_{-1}^0 \frac{\lvert t\rvert^{\varepsilon}}{t}\,dt\,.$$

If $f_X$ is well-behaved at $0$, the first integral on the right is harmless. Well-behaved means that

$$\int_{-1}^0 \biggl\lvert \frac{f_X(t) - f_X(0)}{t}\biggr\rvert\,dt < +\infty$$

here. For that it is sufficient that $f_X$ is $\alpha$-Hölder continuous for some $\alpha > 0$ (for then the integrand is dominated by a multiple of $\lvert t\rvert^{\alpha - 1}$). This is the case for your $f_X$. Then, again by the dominated convergence theorem, we have

$$\lim_{\varepsilon \to 0} \int_{-1}^0 \lvert t\rvert^{\varepsilon} \frac{f_X(t) - f_X(0)}{t}\,dt = \int_{-1}^0 \frac{f_X(t) - f_X(0)}{t}\,dt\,.$$

And finally we have

$$\int_{-1}^0 \frac{\lvert t\rvert^{\varepsilon}}{t}\,dt = -\int_0^1 u^{\varepsilon - 1}\,du = -\frac{1}{\varepsilon}\,.$$

The analogous splitting of the integral for positive $t$ also produces two parts that tend to a finite limit as $\varepsilon \searrow 0$, plus

$$f_X(0) \int_0^1 t^{\varepsilon - 1}\,dt = \frac{f_X(0)}{\varepsilon}\,.$$

We therefore have

$$\lim_{\varepsilon \searrow 0} \int_{-\infty}^{+\infty} t^{\varepsilon}\frac{f_X(t)}{t}\,dt = \operatorname{PV} \int_{-\infty}^{\infty} \frac{f_X(t)}{t}\,dt + \lim_{\varepsilon \searrow 0}\: f_X(0) \underbrace{\frac{1 - e^{i\pi\varepsilon}}{\varepsilon}}_{\to -\pi i}$$

with

$$\operatorname{PV} \int_{-\infty}^{\infty} \frac{f_X(t)}{t}\,dt = \int_{-1}^1 \frac{f_X(t) - f_X(0)}{t}\,dt + \int_{\lvert t\rvert > 1} \frac{f_X(t)}{t}\,dt\,.$$

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.