Product of independent random variables and tail deconvolution

Suppose $$X, Y$$ are two independent non-negative random variables. The conditions

(i) $$\mathbb{P}(X > t) = \frac{C}{t^p} + o(t^{-p})$$

(ii) $$\mathbb{P}(Y > t) = o(t^{-q})$$ for any $$q > 0$$

imply

(iii) $$\mathbb{P}(XY > t) = \frac{C \mathbb{E}[Y^p]}{t^p} + o(t^{-p})$$.

(Of course here I am talking about the asymptotic behaviour as $$t \to \infty$$ and $$p > 0$$.)

My question concerns a converse of this statement: if I know (ii) and (iii), does that imply (i)?

(While I would very much love to see that this is true, I have the impression that this claim is false but just haven't come up with a counter-example.)

I am aware that the converse holds at the exponential level, i.e. $$\lim_{t \to \infty} \frac{\log \mathbb{P}(X > t)}{\log t} = -p.$$

One may consider random variable $$Y$$ with a density (which is sufficient for my purpose) if that helps. In case a counter-example for the "full" converse can be found, I would like to know if the "full" converse can still hold when (a) $$Y$$ is a lognormal random variable or slightly more generally (b) $$Y$$ has a tail upper bounded by that of some lognormal.

Update: to clarify, $$Y$$ is a given random variable, the distribution of which is hence given and cannot be chosen freely. In particular $$Y$$ is not a constant (otherwise the converse is trivially true unless $$Y = 0$$ a.s., in which case the converse is trivially false).

Update 2: if a counter-example is found (in the case where $$Y$$ has density), I will be interested to see if there is some way to save the claim by strengthening the assumption slightly.

• @MateuszKwaśnicki I may not have made it clear enough but what I mean is that $Y$ is some given random variable that you don't get to choose. (Of course condition (ii) implies $\mathbb{E}[Y^p] < \infty$ so you can always rewrite a constant $C$ as the product of $\mathbb{E}[Y^p]$ and a new constant $C'$. – random_person Jan 15 at 21:23
• Yes, I noticed that just after pressing the "Add Comment" button... – Mateusz Kwaśnicki Jan 15 at 21:41

For every $$a > 0$$ we have $$\mathbb{P}(XY>at) \geqslant \mathbb{P}(X>t)\mathbb{P}(Y>a) ,$$ and therefore (with $$D = C \, \mathbb{E}(Y^p)$$) $$\mathbb{P}(X > t) \leqslant \frac{\mathbb{P}(XY>a t)}{\mathbb{P}(Y>a)} = \frac{D}{t^p} + o(t^{-p}) .$$ That is, $$\mathbb{P}(X>t) = O(t^{-p})$$. However, $$\mathbb{P}(X>t)$$ need not be asymptotic with $$C t^{-p}$$.

Consider the following example. Let $$X$$ be a discrete random variable such that $$\mathbb{P}(X = 2^{n/p}) = 2^{-n}$$ for $$n = 1, 2, \ldots\,$$ Then $$\mathbb{P}(X > 2^{n/p}) = 2^{-n} = (2^{n/p})^{-p} ,$$ but $$\mathbb{P}(X \geqslant 2^{n/p}) = 2^{1-n} = 2 \times (2^{n/p})^{-p} .$$ It follows that $$\mathbb{P}(X \geqslant t)$$ oscillates between $$t^{-p}$$ and $$2 t^{-p}$$.

Now let $$Y$$ have an absolutely continuous distribution on $$[2^{-1/p}, 1]$$, with density $$p t^{-1-p}$$. Then it is quite easy to see that $$X Y$$ is absolutely continuous on $$[1, \infty)$$ with density function $$p t^{-1-p}$$. Thus, $$P(X Y > t) = t^{-p} = t^{-p} + o(t^-p),$$ as desired.

This example shows that the answer to your additional question (b) is negative, too: $$Y$$ has compact support, and so the tail of $$Y$$ is bounded by the tail of log-normal distribution.

It is easier to understand the above example when one writes $$A = \log X$$, $$B = \log Y$$, so that $$\log(X Y) = A + B$$. We assume that $$\mathbb{P}(A + B > t) = D e^{-p t} + o(e^{-p t})$$ and $$\mathbb{P}(B > t) = o(e^{-q t})$$ for every $$q > 0$$, and ask whether $$\mathbb{P}(A > t) = C e^{-p t} + o(e^{-p t})$$. In the counterexample, we take $$A$$ geometrically distributed on positive integers, and $$B$$ with exponential density function on $$[-1, 0]$$. The sum $$A + B$$ therefore has the usual exponential distribution.

This sheds some light on your additional question (a), where $$Y$$ has a log-normal distribution, that is, $$B$$ is normally distributed. A positive answer would therefore follow from a Karamata-type theory for rapidly decaying functions. There's a whole section on these in the Bingham–Goldie–Teugels book on regular variation, but I have never read it. Time permits, I will do that later and update the answer.