# Finding the Cauchy Principal Value Mean of the pdf for $Z=XY$ where $\mathbb{E}[Y]$ does not exist

Let's first define the Cauchy Principal Value Mean (PVM). For a continuous random variable $V$ with pdf $f_{V}(v)$, the PVM of $f_{V}$ is $$\mathrm{PV}(\mathbb{E}[V])=\lim_{a\to\infty}\int_{-a}^{a}v f_{V}(v)\, \mathrm{d}v$$

The PVM is useful for assigning a 'location' to a pdf where the mean is undefined (ex. Cauchy).

Now, suppose I have two continuous, real-valued, independent random variables $X$ and $Y$ with pdf's $f_{X}(x)$ and $f_{Y}(y)$, respectively. For simplicity, let's assume $X$ is a positive random variable such that $f_{X}(x)=0$ for $x\in(-\infty, 0]$. Then the pdf of the product $$Z=XY,$$ can be found by $$f_{Z}(z)=\int _{0}^{\infty}\frac{1}{x}f_{X}(x)f_{Y}(z/x)\,\mathrm{d}x.$$

Furthermore, the mean of $f_{Z}$, i.e. $\mathbb{E}[Z]$ can be found by \begin{aligned} \mathbb{E}[Z] &=\int _{-\infty}^{\infty}\int _{0}^{\infty}\frac{z}{x}f_{X}(x)f_{Y}(z/x)\,\mathrm{d}x\mathrm{d}z\\ &=\int _{0}^{\infty}\frac{1}{x}f_{X}(x)\int _{-\infty}^{\infty}z\,f_{Y}(z/x)\,\mathrm{d}z\mathrm{d}x\\ &=\int _{0}^{\infty}\frac{1}{x}f_{X}(x)\int _{-\infty}^{\infty}x^{2}u\,f_{Y}(u)\,\mathrm{d}u\mathrm{d}x\\ &=\int _{0}^{\infty}x\,f_{X}(x)\int _{-\infty}^{\infty}u\,f_{Y}(u)\,\mathrm{d}u\mathrm{d}x\\ &=\mu_{X}\mu_{Y} \end{aligned}

My Question: Let's suppose that $\mathbb{E}[Y]$ does not exist. If this is true, one would think that $\mathbb{E}[Z]$ also does not exist. That said, would substituting the PVM of $f_{Y}$ into the above result in the PVM of $f_{Z}$?, i.e., \begin{aligned} \mathrm{PV}(\mathbb{E}[Z]) &=\lim_{a\to\infty}\int _{0}^{\infty}x\,f_{X}(x)\int _{-a}^{a}u\,f_{Y}(u)\,\mathrm{d}u\mathrm{d}x\\ &\overset{?}{=}\mu_{X}\times\mathrm{PV}(\mathbb{E}[Y]) \end{aligned}

Additional Information (7/11/17): I have created a product distribution in MATLAB where the mean of the positive random variable, $X$, exists, and the mean of the $Y$ pdf does not exist but its PVM does. I then computed the following: $$\mathrm{PV}(\mathbb{E}[Z])=\mu_{X}\times\mathrm{PV}(\mathbb{E}[Y])\,,$$ and $$\mathrm{PV}(\mathbb{E}[Z])=\lim_{a\to\infty}\int_{-a}^{a}z\, f_{Z}(z)\, \mathrm{d}z\,.$$

These values were computed for a variety of different parameters and are always in agreement. It seems like the answer to the question above is most likely "Yes" but numerical analysis certainly isn't sufficient to prove that hypothesis.

Attempt at Solution (7/11/17): Given the problem stated above, we want to determine if $$\mathrm{PV}(\mathbb{E}[Z])\overset{?}{=}\mu_{X}\times\mathrm{PV}(\mathbb{E}[Y])\,,$$ is true or false.

By definition, $\mathrm{PV}(\mathbb{E}[Z])$ is found by: $$\mathrm{PV}(\mathbb{E}[Z])=\lim_{a\to\infty}\int_{-a}^{a}z\, f_{Z}(z)\, \mathrm{d}z\,.$$

Substituting $f_{Z}$ yields $$\mathrm{PV}(\mathbb{E}[Z])=\lim_{a\to\infty}\int_{-a}^{a}z\, \int _{0}^{\infty}\frac{1}{x}f_{X}(x)f_{Y}(z/x)\,\mathrm{d}x\, \mathrm{d}z\,.$$

Switching the order of integration gives $$\mathrm{PV}(\mathbb{E}[Z])= \int _{0}^{\infty}\frac{1}{x}f_{X}(x) \left(\lim_{a\to\infty}\int_{-a}^{a}z\, f_{Y}(z/x)\,\mathrm{d}z\right) \mathrm{d}x\,.$$

Substitute $u=z/x$ for the inner integral (this seems sketchy...assumes $x\in(0,\ \infty)$, i.e. $x\neq0$ and $x\neq\infty$) gives: $$\mathrm{PV}(\mathbb{E}[Z])= \int _{0}^{\infty}x\,f_{X}(x) \left(\lim_{a\to\infty}\int_{-a/x}^{a/x}u\, f_{Y}(u)\,\mathrm{d}u\right) \mathrm{d}x\,.$$

If we take the limit now we get: \begin{aligned} \mathrm{PV}(\mathbb{E}[Z]) &=\int _{0}^{\infty}x\,f_{X}(x)\mathrm{PV}(\mathbb{E}[Y]) \mathrm{d}x\\ &=\mu_{X}\times\mathrm{PV}(\mathbb{E}[Y])\qquad\qquad\qquad\qquad\square \end{aligned}

The mathematics leading to this answer seems like it may be wrong (i.e. at the point of the substitution/limit). Thoughts?

• If you always write your improper integrals (the ones with $\infty$'s in the limits) with explicit limits, what do you need to be true to make each of your manipulations hold? If you're careful with your work you should see if you introduce any conditions at any step. – Glen_b Jul 11 '17 at 0:54
• You might wish to exclude the trivial counter-example of the distribution of $X$ being a delta on zero. – conjectures Jul 11 '17 at 12:34
• @Glen_b I tried working out "proof" (at end of question) which seems to indicate that the answer to my question is "Yes". That said, after working out the math I may have introduced a condition/assumption as you have eluded to. – Aaron Hendrickson Jul 11 '17 at 13:58
• At steps where you do things like swap order of integration or swap expectation and integration ... those are times to pay attention to when you can do that. – Glen_b Jul 12 '17 at 0:53