# Difference between $P(f(x*,w)>0)→1,P(f(x,w)>0)→1$ and $P(min(f(x*,w),f(x,w))>0)→1$ when dimension grows

Let $$f(x_1,\cdots,x_n,w)$$ be a function from $$R^{n+1}\rightarrow R$$, where $$x_1,\cdots,x_n$$ are deterministic variables, and $$w$$ be random variable. As a simple example, $$f$$ can be $$(x_1+\cdots+x_n)w$$. For simplicity, I would just write $$f(\vec x,w)$$ in the following context.

Is there any difference between the following two?

(1) $$P(f(\vec x^*,w)>0)\rightarrow 1$$ and $$P(f(\vec x,w)>0)\rightarrow 1$$ when $$n$$ goes to infinity.

(2) $$P\big(\min(f(\vec x^*,w),f(\vec x,w))>0\big)\rightarrow 1$$ when $$n$$ goes to infinity.

I encountered this question while attempting to prove a random function $$f(\vec x,w)$$ is positive with high probability (probability of $$f>0$$ goes to 1), where $$\vec x\in A\subset \mathbb{R}$$ is a variable (deterministic), and $$w$$ is a random variable.

Initially, I began by proving $$P(\inf_{\vec x\in A} f(\vec x,w))\rightarrow 1$$ using an $$\epsilon$$-net argument, as the region $$A$$ is not discrete. However, I recently realized that, for any $$x\in A$$, the probability $$P(f(\vec x,w))>0\rightarrow 1$$. Therefore, it seems unnecessary to go through the effort of proving $$P(\inf_{\vec x\in A} f(\vec x,w))\rightarrow 1$$ since I already know that the function is positive at all points with high probability.

I am uncertain about which statement I should focus on proving. I am also confused on when should we use epsilon net argument. Consequently, I am curious to determine the precise distinction between (1) and (2).

Thanks for any suggestion!

• What is P(x_1>0)? Commented May 25, 2023 at 16:15
• @AndrewZhang sorry for the confusion. I've re-edited Commented May 25, 2023 at 16:17

You have sequences of random variables $$(X_n)$$ and $$(Y_n)$$ and you are wondering whether

(1) $$P(X_n > 0) \to 1$$ and $$P(Y_n > 0) \to 1$$

and

(2) $$P(\min(X_n,Y_n)>0) \to 1$$

are equivalent. The answer is yes.

Proof. Note that $$\min(X_n, Y_n) \leq X_n$$ and $$\min(X_n, Y_n) \leq Y_n,$$ so $$P(\min(X_n, Y_n) > 0) \leq P(X_n > 0)$$ and $$P(\min(X_n, Y_n) > 0) \leq P(Y_n > 0).$$ This shows (2) implies (1).

Suppose now (1) and let $$\varepsilon \in (0, 1).$$ Then, with probability at least $$1-\frac{\varepsilon}{2},$$ there exists an $$N$$ such that for every $$n > N,$$ the events $$\{X_n > 0\}$$ and $$\{Y_n > 0\}$$ have probabilities at least $$1 - \frac{\varepsilon}{2}.$$ If $$A$$ and $$B$$ are events, then the inclusion-exclusion formula shows that $$P(A \cap B) = P(A) + P(B) - P(A \cup B) \geq P(A) + P(B) - 1.$$ Then, $$P(\min(X_n, Y_n)>0) = P(\{X_n > 0\} \cap \{Y_n > 0\}) \geq 1 - \varepsilon,$$ for $$n \geq N.$$ Thus, (2) follows. QED

The above proof can easily be extended (by induction say), to the case of a finite number of sequences but it will fail for infinitely many sequences.

• thank you very much for your answer. In my case since $A$ is a connected subset in $R^{n+1}$, we have infinite many sequences. Thus this proof won't imply my case. But I was wondering whether we need the $w$ in $f(x^*,w)$ and the $w$ in $f(x,w)$ to be independent? Commented May 25, 2023 at 18:37
• $(X_n)$ and $(Y_n)$ are assumed any sequences. Commented May 25, 2023 at 18:41
• And are (1) (2) equivalent in infinite case? Commented May 25, 2023 at 19:01
• @happyle they shouldn't be. $\{ \min_{t\in [a,b]} X_t > 0\} \subset \bigcap_{t \in [a,b]} \{X_t > 0\}$ but the two sets might differ. However, you are using nets, so likely you have finitely many points in your net. Commented May 25, 2023 at 22:33
• why $\{\min_{t\in[a,b]}X_t>0\}\subset\cap_{t\in[a,b]}\{X_t>0\}$? sorry it might be very basic. Shouldn't they be equivalent? If $X_t>0$ on every $t$ then the minimum of them must $>0$ as well. Commented May 26, 2023 at 14:34