Are min$(X_1,\ldots,X_n)$ and min$(X_1Y_1,\ldots,X_nY_n)$ independent for $n$ to infinity?

Question

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?

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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?

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The answer can be found here:

https://math.stackexchange.com/questions/3403441/are-minx-1-ldots-x-n-and-minx-1y-1-ldots-x-ny-n-independent-for-n-to/3405002#3405002

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

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.