Assume that we have given two continuous iid random variables $X$ and $Y$ with support $[1,c)$, where $c$ is some constant greater than one. Now assume I have a given iid sample $X_1, \ldots,X_n$ and $Y_1, \ldots,Y_n$ (so absolutely no dependence here).

Imagine that I know that:

$$(1): \mathbb P \left(\frac{\min(X_1,\ldots,X_n)-a_n}{b_n}\leq x_1\right) \sim F(x_1), \text{ for }n \to \infty$$

$$(2):\mathbb P \left(\frac{\min(X_1Y_1,\ldots,X_nY_n)-\bar a_n}{\bar b_n}\leq x_2\right) \sim G(x_2), \text{ for }n \to \infty,$$

where $F$ and $G$ are just the limit distributions. Is it true that then it also follows that

$$(3):\mathbb P \left(\frac{\min(X_1,\ldots,X_n)-a_n}{b_n}\leq x_1,\frac{\min(X_1Y_1,\ldots,X_nY_n)-\bar a_n}{\bar b_n}\leq x_2\right) \sim F(x_1) G(x_2),$$

for $n$ to infinity?

At first I thought this cannot work since they are obviously dependent - but then I thought that the probability that $\min(X_1,\ldots,X_n)$ and $\min(X_1Y_1,\ldots, X_nY_n)$ will be obtained in the same realization will converge to zero and since the sample is iid, it is actually true? So am I right?

edit: Okay, since noone is answering so far I just explain a little bit my idea why I think it could be true; We have for some $i$:

\begin{align*} &\mathbb P\big(X_i=\min(X_1,\ldots,X_n), X_iY_i=\min(X_1Y_1,\ldots,X_nY_n)\big) \\ \leq &\mathbb P\big(X_i=\min(X_1,\ldots,X_n), Y_i \leq \min(X_1Y_1,\ldots,X_nY_n)\big) \\ = &\mathbb P\big(X_i=\min(X_1,\ldots,X_n)\big) \mathbb P\big( Y_i \leq \min(X_1Y_1,\ldots,X_nY_n) \vert X_i=\min(X_1,\ldots,X_n) \big) \\ = &1/n \mathbb P\big( Y_i \leq \min(X_1Y_1,\ldots,X_nY_n) \vert X_i=\min(X_1,\ldots,X_n) \big) \end{align*}

where the latter probability converges to zero, since $\min(X_1Y_1,\ldots,X_nY_n)$ gets arbitrarily close to 1 for $n \to \infty$. Therefore, the probability, that the minimum is realized in the same observation is something like $n \cdot 1/n \cdot o(1)=o(1)$, so converges to zero...

Now this is obviously not a rigorous proof but are my thoughts correct?

The answer can be found here:


Sangchul Lee gave me a good idea, doing it with the easist case, the uniform distribution and I could extend it the general case of atom-free random variables

  • $\begingroup$ What are $a_n$ and $b_n$? Is it enough to focus on the case $a_n=0$, $b_n=1$? $\endgroup$
    – Matt F.
    Commented Oct 21, 2019 at 15:24
  • 1
    $\begingroup$ No, it is not; You can see it as the sequences in Extreme Value Theory; In Extreme Value Theory you have something like $\mathbb P (\frac{\max(X_1,\ldots,X_n)-a_n}{b_n}\leq x)$ which converges to some Extreme Value Distribution, given some weak conditions; With Minima you can do the same; I think in this case, $a_n=1$ and $b_n$ goes to zero, as $n$ goes to infinity $\endgroup$
    – Mark
    Commented Oct 21, 2019 at 15:35

1 Answer 1


Just a quick thought of mine:

The condition that the product $X_i Y_i$ is smaller than $X_2$ removes all possibilities of $X_i$ being larger than $(X2_/b_n)+a_n$.

EDIT: Yes you understood me correctly! Imagine the conditional probability of both (the product and Xi alone) being small enough: P(Xi|XiYi)! Given the product XiYi is small enough, since Yi must be larger than 1, there is no possibility of of Xi being larger than (x2/bn)+an. This means the conditional probability P(Xi|Xi*Yi) is always 1, if x1 is big enough. Meaning your combined probability equals G(x2), although F(x1) can still be smaller than 1.

  • 1
    $\begingroup$ So I think I got your idea; (I am still not sure if this is really what you want to say); If $\frac{\land XY-\bar a_n}{\bar b_n}\leq x_2$, then it holds that $\land XY\leq x_2 \bar b_n+\bar a_n$ and therefore in particular it holds that: $\land X\leq x_2 \bar b_n+\bar a_n$; So if $x_2 \bar b_n+\bar a_n \leq x_1 b_n+a_n$, then $\frac{\land XY-\bar a_n}{\bar b_n}\leq x_2$ already implies $\frac{\land X- a_n}{ b_n}\leq x_1$; $\endgroup$
    – Mark
    Commented Oct 21, 2019 at 16:24
  • $\begingroup$ This, however, would not work if $x_2$ and $x_1$ are fixed, as $a_n=1$ and $b_n$ goes to zero and the same holds for $\bar a_n$ and $\bar b_n$; However, even if $x_1$ is infinity I don't quite think your arguments are working; Because then the probability $\mathbb P(\frac{\land X- a_n}{ b_n}\leq x_1)=1$ and then you still have equality; And even if $a_n$ and $b_n$ are defined differently, it would not work because then $\frac{\land XY-\bar a_n}{\bar b_n}$ respectively $\frac{\land X- a_n}{ b_n}$ just converge a.s. to some number; But I still gave you a thumbs up, maybe I dont get your point.. $\endgroup$
    – Mark
    Commented Oct 21, 2019 at 16:31
  • $\begingroup$ I made an edit. I hope I understood you correctly! $\endgroup$
    – KaPy3141
    Commented Oct 21, 2019 at 18:09
  • $\begingroup$ I still do not understand you; In particular, I have no idea what $P(X_i | vert X_i Y_i)$ supposed to mean; But let's go a step back; I was wondering if (1) and (2) holds, that then also (3) must hold; What I did later was just an intuition why it should be correct; So right now only the equations (1), (2) and (3) matter; As I already said: $\frac{\land XY-\bar a_n}{\bar b_n}\leq x_2$ implies $\frac{\land X- a_n}{ b_n}\leq x_1$ if $x_2 \bar b_n+ \bar a_n \leq x_1 b_n + a_n$; Do we agree on this? But this is equivalent to $x_1\geq (x_2 \bar b_n + \bar a_n- a_n)/b_n$ $\endgroup$
    – Mark
    Commented Oct 21, 2019 at 19:01
  • 1
    $\begingroup$ Also check the same question on math.stackexchange: math.stackexchange.com/questions/3403441/… I added a bunch of explanations :-) $\endgroup$
    – Mark
    Commented Oct 22, 2019 at 10:10

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