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probability

suppose I tell you that INT(f(x)^2) dx is finite, where f(x) is the dist. function of X.

show that n* P(|X|>e*sqrt(n)) -> 0 for all e>0.

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I know that P(|X|>e*sqrt(n)) = O(1/(n*e^2))

so, this should be O(1). So has no reason to tend to 0.

????
Reply 1
Avatar for nota bene
nota bene
15
Hmm, maybe I'm missing something, but aren't you looking at lim⁡n→∞O(1ne2)=0\displaystyle\lim_{n\to\infty} O(\frac{1}{ne^2})=0?
Reply 2
Avatar for Chewwy
Chewwy
OP
17
nota bene
Hmm, maybe I'm missing something, but aren't you looking at lim⁡n→∞O(1ne2)=0\displaystyle\lim_{n\to\infty} O(\frac{1}{ne^2})=0?

n times this.
Reply 3
Avatar for nota bene
nota bene
15
Chewwy
n times this.

woops, so you wrote...! In which case I'm inclined to agree - I can't see a reason why it'd tend to zero, clearly the limit is some non-zero constant...

Is there any chance there is some other fact about P(|X|>e*sqrt(n)) known?

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