Discrete random walk/risk process characterized by a step distribution with a heavy tail and a mean existing only as a Hadamard finite-part integral

Question

This question is an attempt to make more precise two previous questions of mine on another SE (this and this).

The random walk/risk process of interest

Consider the following discerete process. Let $n=1,\,2,\,3,\,\ldots$. Define $$\hspace{8em}R_{n}=u+c\,n-\sum_{j=1}^{n}V_{j}\,,\hspace{8em}(1)$$ where

1. The $V_{j}$'s are independent identically distributed random variables (i.i.d. RVs). Namely, for all $j=1,\,2,\,3,\,\ldots$, we have that $V_{j}=1/X_{j}^{2},$ where the $X_{j}$ are i.i.d. RVs whose common distribution is $\mathcal{N}(\mu,\,\sigma)$, the normal distribution with a mean $\mu$ and a standard deviation $\sigma>0$. We will require that $\boldsymbol{\mu\neq 0}$. Using standard methds, one can work out that probability density function of the distribution from which the $V_{j}$'s are drawn in
2. Note that the $V_{j}$'s are arriving at regularly spaced intervals. This is different than in the standard Cramér–Lundberg model (see e.g. here).
3. $u\geqslant 0$ is a fixed (i.e. independent of $n$) real number. (I suspect it won't matter much for what interests me.)
4. $c=(\eta+1)\mu_{V}$.
5. $\mu_{V}>0$ is a fixed real number.
6. $\eta>0$ is a fixed real number.

If the $V_{j}$'s had a finite mean, i.e. if the integral $\int_{-\infty}^{\infty}\frac{1}{x^{2}}\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x-\mu)^{2}}{2\sigma^{2}}}dx$ existed, then $\mu_{V}$ would be equal to it. But that integral doesn't exist, not even in the principal value sense (whereas the mean value of $1/x$ does exist in the principal value sense; more on that below).

My primary conjecture is that (for all nonzero $\mu$ and $\sigma$) there exists a number to which $\mu_{V}$ can be set such that the 'sharpened conjecture' (see below) is true. (If that is so, then it might make sense to call $\mu_{V}$ a 'pseudo-mean' or a 'quasi-mean' of the $V_{j}$', or something like that.)

My secondary conjecture is that if the primary conjecture is true, then $\mu_{V}$ is in fact the Hadamard finite part of $\int_{-\infty}^{\infty}\frac{1}{x^{2}}\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x-\mu)^{2}}{2\sigma^{2}}}dx$. It is given below in Eq. (9) and, in a different form, in Eqs. (4) and (10); I will denote it $\mu_{V}^{\text{H}}$. For more on the Hadamar finite part, see here. As I will discuss below, $\mu_{V}^{\text{H}}$ is analytically computable in terms of the Dawson function $F(x)=e^{-x^{2}}\int_{0}^{x}e^{t^{2}}dt$ as $$\hspace{5em}\mu_{V}^{\text{H}}=\frac{1}{\sigma^{2}}\left(\frac{\sqrt{2}\, \mu}{\sigma}F\left(\frac{\mu}{\sqrt{2}\,\sigma}\right)-1\right).\hspace{5em}(2)$$ (So my secondary conjecture is that $\mu_{V}=\mu_{V}^{\text{H}}$.)

Interpretation in the context of ruin theory

In the context of ruin theory,

1. $R_{n}$ is the reserve of the insurance company at time $n$ ('ruin' occurs if $R_{n}$ turns negative);
2. $u=R_{0}$ is the initial reserve;
3. $c>0$ is the rate of inflow of premium;
4. the $V_{j}$'s are claim sizes; and
5. $\eta$ is the safety loading (see e.g. here);

As usual, we will introduce the notion of ruin time $n_{\text{R}}$ (which in our case is a positive integer) as follows. For any given realization of the process $R_{n}$, $n_{\text{R}}$ is the lowest $n>0$ such that $R_{n}<0$.

The Hadamard finite part

The main complication in my case is that the claims come from a distribution whose tail is so heavy that the mean value doesn't exist even in a principal value sense, but only in the Hadamard finite part sense. Using standard methods, one can show that he probability density function for the $V_{j}$'s is $$\hspace{4em}p(z)=\frac{1}{z^{3/2}}\frac{1}{2\, \sqrt{2 \pi }\, \sigma}\Big(e^{-\frac{1}{2 \sigma ^2}(\mu +1/\sqrt{z})^{2}}+e^{-\frac{1}{2 \sigma ^2}(\mu -1/\sqrt{z})^{2}}\Big)\,.\hspace{4em}(3)$$ We see that the meain value $\int_{0}^{\infty}z\,p(z)\,dz$ doesn't exit because $z\,p(z)$ goes to zero only as $1/\sqrt{z}$ for large $z$. Note that this same (nonexistent) mean value may also be written as $\int_{-\infty}^{\infty}\frac{1}{x^{2}}\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x-\mu)^{2}}{2\sigma^{2}}}dx$, a form in which it is more obvious how we might try to use the principal value; for example, the mean value of $1/X$, $\int_{-\infty}^{\infty}\frac{1}{x}\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x-\mu)^{2}}{2\sigma^{2}}}dx$, does exist in the principal value sense. Indeed, I will use this latter fact in the Appendix, where I show that the Hadamard finite part of the mean value integral $\int_{0}^{\infty}z\,p(z)\,dz$ is $$\hspace{2em}\mu_{V}^{\text{H}}= \int_{0}^{\infty}z\,\left(p(z)-{C}/{z^{3/2}}\right)\,dz=\frac{1}{\sigma^{2}}\left(\frac{\sqrt{2}\, \mu}{\sigma}F\left(\frac{\mu}{\sqrt{2}\,\sigma}\right)-1\right)\,,\hspace{2em}(4) $$ where $C={\displaystyle\lim_{s\to+\infty}} s^{3/2}p(s)$.

As a comparison, in the case of e.g. the Cauchy distribution, a prototypical example of a heavy-tailed distribution, the mean value does exist in the principal value sense; see e.g. the footnote here.

Motivation: a vague conjecture

My hope is that this random walk/risk process will make possible a precise formulation of the following vague conjecture:

Vague conjecture: As $n$ grows larger and larger, the average $\frac{1}{n}\sum_{j=1}^{n}V_{j}$ gets closer and closer to $\mu_{V}^{\text{H}}$ until the latest $V_{j}$ suffers a single catastrophic jump.

Numerical evidence for the vague conjecture

This is indeed what I see in numerical simulations with $\mu=0.60$ and $\sigma=0.10$, which results in $\mu_{V}^{\text{H}}=3.05003$. Here is my procedure.

(1) Draw a sample $\left\{x_{j}\right\}_{j=1}^{N}$ of $N=10^{8}$ points from $\mathcal{N}(\mu,\,\sigma)$ (that's close to the maximal $N$ that my laptop can handle).
(2) Define the sample $\left\{v_{j}\right\}_{j=1}^{N}$ by setting $v_{j}=1/x_{j}^{2}$.
(3) Check whether $\frac{1}{N}\sum_{j=1}^{N}v_{j}$ differs from $\mu_{V}^{\text{H}}$ by more than $\eta\, \mu_{V}^{\text{H}}$, where $\eta=10^{-3}$.
(4) If it doesn't, go back to (1). If it does, stop, and produce a plot of $\left(\frac{1}{n}\sum_{j=1}^{n}v_{j}\right)$ vs. $n$ for $n=1,\,2,\,\ldots,\,N$. Here is a typical plot that comes out (only every millionth point is plotted, and I start from $n/10^{6}=3$, omitting an initial transient):

Here the sample maximum was about $v_{j_{\text{max}}}\approx 2\times10^{6}$, while the second-largest value was about 800. The maximum occurred at $j_{\text{max}}\approx 65\times10^{6}$ in the sample and caused the catastrophic jump away from $\mu_{V}^{\text{H}}$ that is visible in the plot.

The discrepancy criterion of $\eta=10^{-3}$ was chosen by experimentation. For larger discrepancies (e.g. $10^{-2}$), it takes too long to produce a plot—too many cycles from (3) to (1). For smaller discrepancies (e.g. $10^{-4}$), there isn't a manifest single jump away from $\mu_{V}^{\text{H}}$.

How a sharpened conjecture (a more precise formulation of the vague conjecture) might look like, in terms of the random walk/risk process

Intuitively, we need $n$ to be large enough so that the 'non-catastrophic' fluctuations in $V_{j}$'s get averaged out. But we also need to condition this on the assumption that the first catastrophic jump hasn't happened yet.

We will use the notion of ruin time $n_{\text{R}}$ (which we introduced above) to characterize 'catastrophic jumps'. Both the numerics and general experience with heavy-tailed distributions suggest that in the presence of heavy tails, ruin occurs as a result of a single big step (see e.g. here).

Then I suspect that the following is (a) true and (b) a good candidate for a rigorous version of the vague conjecture above:

Sharpened conjecture:

Primary conjecture: Set $u=0$. For every $\mu\neq 0$ and every $\sigma>0$, there exists a $\mu_{V}>0$ such that the following holds.

For every $\eta>0$, no matter how small, and every $\epsilon>0$, no matter how small, there exists a sufficiently large $n^{*}>0$ such that if $n_{\text{R}}> n^{*}$, then for every $n$ such that $n^{*}\leqslant n < n_{\text{R}}$, the probability that $\left|\frac{1}{n}\sum_{j=1}^{n}V_{j}-\mu_{V}\right|<\eta\,\mu_{V}$ is at least $1-\epsilon$.

Secondary conjecture: If the primary conjecture is true, then $\mu_{V}=\mu_{V}^{\text{H}}$.

The primary conjecture is kind of inspired by Theorem 10.7.2 here (and, if true, would correspond to some sort of a baby version of that theorem).

My questions

1. Is it the case that the 'sharpened conjecture' is both true and a good candidate for a rigorous version of the 'vague conjecture'? What would it take to prove it?
2. If not, is there a statement that is both true and a good candidate for a rigorous version of the 'vague conjecture'? (It doesn't have to necessarily be stated in the language of random walks or risk processes.)

Prior research

It is hard to believe that what I'm asking isn't well-known.

However, extensive searches on google and google books for keywords 'random walk', 'Hadamard finite part', and 'heavy tail' have so far turned up nothing.

Of course processes similar to my $R_{n}$ are studied in risk/ruin theory, queuing theory, and related fields. In these fields, the processes under study are in one sense typically more complicated than my $R_{n}$, because the $V_{j}$'s are not equally spaced but arrive as a Poisson process. But in another sense their processes are usually simpler, because normally $V_{j}$'s are assumed to have a well-defined mean value even if they have a heavy tail (e.g. here). Nevertheless, models with infinite means have been studied. One example is D. Denisov, S. Foss, and D. Korshunov, 'Tail Asymptotics for the Supremum of a Random Walk when the Mean Is not Finite', Queueing Syst. 46, 15–33 (2004) (published version here, arXiv version here).

Nevertheless, so far I haven't found anything that would help me answer the questions I'm posing here.

Appendix: How the Hadamard finite part comes in

Let $Y=1/X,$ where $X$ is a normally distributed random variable with a mean $\mu$ and a standard deviation $\sigma$. Note that the integral representing $\mathbb{E}[Y]$ (the mean value of $Y$) exists in the principal value sense. It may be related to the Hilbert transform of a Gaussian, which, in turn, is expressible in terms of the Dawson function $F(x)=e^{-x^{2}}\int_{0}^{x}e^{t^{2}}dt$. For references, see e.g. here, here, this paper and this question on Mathematics SE. In the end, one obtains $$\hspace{4em}\mathbb{E}[Y]=\text{P.V.}\;\int_{-\infty}^{\infty}\frac{1}{x}\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x-\mu)^{2}}{2\sigma^{2}}}dx=\frac{\sqrt{2}}{\sigma}F\left(\frac{\mu}{\sqrt{2}\,\sigma}\right)\,.\hspace{4em}(5)$$

Let us change variables, like so: $$\text{P.V.}\;\int_{-\infty}^{\infty}\frac{1}{x}\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x-\mu)^{2}}{2\sigma^{2}}}dx = \text{P.V.}\;\int_{-\infty}^{\infty}\frac{1}{u+\mu}\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{u^{2}}{2\sigma^{2}}}du\,.\hspace{4em}(6)$$

Next, we differentiate with respect to $\mu$. I will assume without proof that the differentiation commutes with the limit used to define the principal value (I don't have a rigorous proof, but I do believe it is true). On the left-hand side, we get $$\frac{d}{d\mu}\text{P.V.}\;\int_{-\infty}^{\infty}\frac{1}{x}\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x-\mu)^{2}}{2\sigma^{2}}}dx=\frac{1}{\sigma^{2}}\left[1-\mu\left(\text{P.V.}\;\int_{-\infty}^{\infty}\frac{1}{x}\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x-\mu)^{2}}{2\sigma^{2}}}dx\right)\right]$$ $$\hspace{10em}=\frac{1}{\sigma^{2}}\left(1-\frac{\sqrt{2}\mu}{\sigma}F\left(\frac{\mu}{\sqrt{2}\,\sigma}\right)\right)\,.\hspace{4em}(7)$$ When we differentiate on the right-hand side, we must take into account the fact that, unlike on the left-hand side, here $\mu$ appears in the limits of integration used to define the principal value: $$\frac{d}{d\mu} \lim_{\epsilon\to 0^{+}} \left(\int_{-\infty}^{-\mu-\epsilon}\frac{1}{u+\mu}\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{u^{2}}{2\sigma^{2}}}du+\int_{-\mu+\epsilon}^{\infty}\frac{1}{u+\mu}\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{u^{2}}{2\sigma^{2}}}du\right).$$ We get $$\lim_{\epsilon\to 0^{+}} \left[\frac{1}{\epsilon}\frac{1}{\sqrt{2\pi}\sigma}\left(e^{-\frac{(\mu+\epsilon)^{2}}{2\sigma^{2}}}+e^{-\frac{(\mu-\epsilon)^{2}}{2\sigma^{2}}}\right)\right.$$ $$\left.-\int_{-\infty}^{-\mu-\epsilon}\frac{1}{(u+\mu)^{2}}\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{u^{2}}{2\sigma^{2}}}du-\int_{-\mu+\epsilon}^{\infty}\frac{1}{(u+\mu)^{2}}\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{u^{2}}{2\sigma^{2}}}du\right]\,.\hspace{2em}(8)$$ The first term, which comes from the differentiation of the limits of integration, diverges (when $\epsilon\to 0$) as $\frac{1}{\epsilon}\frac{2}{\sqrt{2\pi}\sigma}e^{-\frac{\mu^{2}}{2\sigma^{2}}}$, and presumably exactly cancels the divergences from the two integrals that follow. Putting it all together, after changing variables once again and mupliplying both sides by $-1$, we get $$\lim_{\epsilon\to 0^{+}} \left[\int_{-\infty}^{-\epsilon}\frac{1}{x^{2}}\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x-\mu)^{2}}{2\sigma^{2}}}dx+\int_{\epsilon}^{\infty}\frac{1}{x^{2}}\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x-\mu)^{2}}{2\sigma^{2}}}dx-\frac{1}{\epsilon}\frac{2}{\sqrt{2\pi}\sigma}e^{-\frac{\mu^{2}}{2\sigma^{2}}}\right]$$ $$\hspace{10em}=\mu_{V}^{\text{H}}=\frac{1}{\sigma^{2}}\left(\frac{\sqrt{2}\, \mu}{\sigma}F\left(\frac{\mu}{\sqrt{2}\,\sigma}\right)-1\right)\,.\hspace{4em}(9)$$ The left-hand side of Eq. (9) is precisely the Hadamard finite part of the integral $\int_{-\infty}^{\infty}\frac{1}{x^{2}}\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x-\mu)^{2}}{2\sigma^{2}}}dx$.

It may be useful to present this in yet another way. Using standard methods, we can show that the probability density function for $V=1/X^{2}$ is (this is Eq. (3) above) $$p(z)=\frac{1}{z^{3/2}}\frac{1}{2\, \sqrt{2 \pi }\, \sigma}\Big(e^{-\frac{1}{2 \sigma ^2}(\mu +1/\sqrt{z})^{2}}+e^{-\frac{1}{2 \sigma ^2}(\mu -1/\sqrt{z})^{2}}\Big).$$ It is easy to see that $p(z)$ (defined for $z>0$) is heavy-tailed and that its mean value $\int_{0}^{\infty}z\,p(z)\,dz$ diverges, since $z\,p(z)$ goes to zero only as $1/\sqrt{z}$ for large $z$. But Mathematica, for example, can work out the following integral, which we obtain from the left-hand side of Eq. (9) after a change of variables and after bringing the third term on the left-hand side of Eq. (9) inside the integral: $$\hspace{4em} \mu_{V}^{\text{H}}=\int_{0}^{\infty}z\,\left(p(z)-\frac{1}{z^{3/2}}\frac{2}{\sqrt{2\pi}\sigma}e^{-\frac{\mu^{2}}{2\sigma^{2}}}\right)\,dz\,.\hspace{4em}(10) $$ The result, of course, is the same as the right-hand side of Eq. (9), $=\frac{1}{\sigma^{2}}\left(\frac{\sqrt{2}\, \mu}{\sigma}F\left(\frac{\mu}{\sqrt{2}\,\sigma}\right)-1\right)$.