# Probability of an event with probability 0 happening at least once in infinite trials

This question here is confusing me a lot.
To summarize, let's say you have $$\text{i.i.d. }X_i \sim U(0, 1), i = 1,2,\ldots, n.$$

The question shows that $$Y_i = \max(X_1, \ldots, X_i) \rightarrow 1$$ in probability, and intuitively (for what it's worth) I can see that, but the math approached this way doesn't seem to agree with me.
$$\lim_{i\rightarrow \infty} P(|Y_i - 1| \geq\varepsilon ) =0 \text{ ?}$$

# Alternate Solution

\begin{align} & P(|Y_i - 1| \geq\varepsilon ) = P\left(\bigcup_i(X_i < 1)\right) \\[8pt] = {} & \prod_i [P(X_i \leq 1) - P(X_i = 1)] \\[8pt] = {} & \prod_i[F_{X_i}(1) - 0] = \prod_i 1 = 1 \neq 0. \end{align}

This shows that $$Y_n$$ does not converge in probability, which does not align with the answer to the question posted above.

# Possible flaws in my solution

However, I know while $$P(X_i = 1) = 0$$. If I can say that its limit is 0 but is in reality infinitesimal i.e. $$P(X_i = 1) = \varepsilon$$. I can show that $$\lim_{n\rightarrow\infty}(1-\varepsilon)^n = 0,$$ $$\varepsilon >0$$. However, that seems wrong to me.

What's confusing me is that as defined, $$X_i: i = 1, 2, \ldots, n$$ is a countably infinite number of "trials" whereas there is an uncountably infinite number of values a continuous RV could take. If I take a countably infinite number of "trials" when there are uncountably infinite possible outcomes. I know that it isn't almost sure that any single outcome will occur. It also seems to me that it won't occur in probability since the set of uncountably infinite $$\gg$$ countably infinite.

Thus I believe my solution is correct and $$Y_n$$ does not converge in p. However, the posted question's answer does not agree.

Would appreciate it if you could point out the flaw in my reasoning!

Thanks!

• you need to formulate the question clearly, separate it from the statements. what's the probability of drawing exactly 0 from standard normal if you keep trying every second to the end of world? It's zero. is this what you're looking for? Commented Sep 26, 2019 at 17:36
• Where you wrote "infinite trials", the correct term is "infinitely many trials." In standard mathematical terminology, "infinite trials" would mean trials each one of which is infinite. I don't know what an "infinite trial" would be, but if you had two of those, you'd have infinite trials, but you wouldn't have infinitely many trials. Commented Sep 26, 2019 at 17:48
• @Aksakal if it is 0, then wouldn't the above show that it $Y_n$ does not converge in p?
– dog
Commented Sep 26, 2019 at 17:53
• @MichaelHardy thanks for your edits and correction!
– dog
Commented Sep 26, 2019 at 17:55
• $P(X_i)\ne\varepsilon$, it's exactly zero: $P(X_i)=0$ Commented Sep 26, 2019 at 18:06

$$P(|Y_i - 1| \geq\varepsilon ) = P\left(\bigcup_i(X_i < 1)\right) \text{ ??}$$ Here you need $$P(|Y_i - 1| \geq\varepsilon ) = P\left(\bigcap_i(X_i \le 1 - \varepsilon)\right).$$ To say that the maximum observation is no more than $$1-\varepsilon$$ is to say that all of the observations are no more than $$1-\varepsilon,$$ so you need an intersection here, not a union. And how $$1-\varepsilon$$ got replaced by $$1$$ is not explained.

By independence, you have $$P\left( \bigcap_i \big[ X_i \le 1 - \varepsilon \big] \right) = \prod_i (1-\varepsilon) = 0.$$

• I see, thank you so much! I appreciate the time you took. I apologize for the careless mistake!
– dog
Commented Sep 26, 2019 at 18:23