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 $q > 0$


(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.

  • $\begingroup$ @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 $C'$. $\endgroup$ – random_person Jan 15 '19 at 21:23
  • $\begingroup$ Yes, I noticed that just after pressing the "Add Comment" button... $\endgroup$ – Mateusz Kwaśnicki Jan 15 '19 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.

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