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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!

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  • $\begingroup$ 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? $\endgroup$ – Aksakal Sep 26 '19 at 17:36
  • $\begingroup$ 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. $\endgroup$ – Michael Hardy Sep 26 '19 at 17:48
  • $\begingroup$ @Aksakal if it is 0, then wouldn't the above show that it $Y_n$ does not converge in p? $\endgroup$ – dog Sep 26 '19 at 17:53
  • $\begingroup$ @MichaelHardy thanks for your edits and correction! $\endgroup$ – dog Sep 26 '19 at 17:55
  • $\begingroup$ $P(X_i)\ne\varepsilon$, it's exactly zero: $P(X_i)=0$ $\endgroup$ – Aksakal Sep 26 '19 at 18:06
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$$ P(|Y_i - 1| \geq\varepsilon ) = P\left(\bigcup_i(X_i < 1)\right) \text{ ??} $$ 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 $$ P\left( \bigcap_i \big[ X_i \le 1 - \varepsilon \big] \right) = \prod_i (1-\varepsilon) = 0. $$

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  • $\begingroup$ I see, thank you so much! I appreciate the time you took. I apologize for the careless mistake! $\endgroup$ – dog Sep 26 '19 at 18:23

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