Limit probability for product of RVs smaller than constant tending toward lower bound of support

Question

Suppose we have a list $(X_1, \dots, X_n)$ of continuous random variables with support $[0,1]$ (not necessarily independent) .

I am wondering whether, as $t \rightarrow 0$, the probability of

$$\left[\prod_{i=1}^n X_i \leq t\right] \qquad (a)$$

tends towards the probability of

$$\left[X_i \leq t^{1/n} \text{ for all } i \in \{1,\dots, n\}\right]. \qquad (b)$$

So just to be clear, my question is whether

$$\lim_{t\rightarrow 0} \frac{P\left(\prod\limits_{i=1}^n X_i \leq t\right)}{P\left(X_i \leq t^{1/n} \text{ for all } i \in \{1,\dots, n\}\right)} = 1. \qquad (q)$$

What makes me think the probabilities of (a) and (b) may be identical in the limit (in the sense of (q)) is that if we have (a) but not (b), then we must have

$$[X_i \leq t^{1/(n-1)} \text{ for at least } (n-1) \text{ of the } i \in \{1,\dots, n\}]. \qquad (c)$$

Intuitively, it seems that, as $t$ approaches $0$, the probability that (c) is true goes to $0$ much faster than the probability that (b) is true, which would give (b) a dominating first-order effect in the limit. But maybe my intuition is misguided and even if it isn't, I am struggling to spell this out and prove it formally.

(If (q) only held under "minor" regularity condition on the joint distribution of the $X_i$'s I wouldn't mind learning about this either :))

Answer 1

The case $U_1, U_2$ are i.i.d. uniform over $[0,1]$ should be given best answer. This just summarizes my earlier comments, with a minor note on large deviation theory:

Some counterexamples for $n=2$

1. Let $X_1 \sim Unif([0,1])$ and $X_2 = 1-X_1$ (dependent). Then both $X_1$ and $X_2$ are uniform but $X_1$ and $X_2$ cannot both be small and so $P[X_1\leq t^{1/2}, X_2\leq t^{1/2}]=0$ for $0<t<1/4$.

2. Let $X_1 \sim Unif([0,1])$ and $X_2 \sim Unif([1/2,1])$. Then $P[X_1\leq t^{1/2}, X_2 \leq t^{1/2}]=0$ if $0<t< 1/4$.

Some possible relation to large deviation theory

A minor comment: Let $X, Y$ be independent. Fix $t \in \mathbb{R}$. Then for all $x \in \mathbb{R}$ we have $$ \{X > x, Y>t-x\} \subseteq \{X+Y>t\}$$ so $$P[X>x]P[Y>t-x] \leq P[X+Y>t] $$ This holds for all $x \in \mathbb{R}$ so $$ \sup_{x \in \mathbb{R}} P[X>x]P[Y>t-x] \leq P[X+Y>t]$$ On the other hand, assuming moment generating functions exist, we have by the Chernov bound (for all $r>0)$: $$ P[X+Y>t]\leq E[e^{rX}]E[e^{rY}]e^{-rt}$$ So $$ \sup_{x \in \mathbb{R}} P[X>x] P[Y>t-x]\leq P[X+Y>t]\leq \inf_{r>0}E[e^{rX}]E[e^{rY}]e^{-rt}$$ and I think the upper and lower bounds can be close for large $t$.

Answer 2

In the comments, @Michael has a very simple counter-example to (q) that, I believe, will soon be turned into an answer. That example relies on dependent RVs. As it turns out, the conjecture is not true for independent RVs either. In fact, the conjecture is not even true in the simple case where $X_1$ and $X_2$ are independent $U[0,1]$, something I really should have checked before posting my question.

Suppose $X_1$ and $X_2$ are independent $U[0,1]$. Then,

• $P(X_1*X_2 \leq t) = t - t\log(t)$ (see product distribution of two uniform distribution, what about 3 or more)
• Also, $P(X_1 \text{ and } X_2 \leq t) = P(X_1 \leq t)*P(X_2 \leq t) = t^2$.
• Therefore, $P(X_1 \text{ and } X_2 \leq t^{1/2}) = (t^{1/2})^2 = t$.

Overall, we have

$$ \frac{P(X_1*X_2 \leq t)}{P(X_1 \text{ and } X_2 \leq t^{1/2})} = \frac{t - t\log(t)}{t} = 1-\log(t),$$

which converges to $\infty$ (rather than $1$) as $t \rightarrow 0$ (i.e., in this simple case, $P(X_1*X_2 \leq t)$ converges to zero "significantly" faster than $P(X_1,X_2 \leq t^{1/2})$).