Regarding Asymptotic Behaviour of normal distibution

Asked Oct 18 '14 at 05:26

Active Oct 18 '14 at 05:33

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If $x_n -> \infty$ in such a way that $x_n^3/ \sqrt{n} -> 0$, then P{$S_n^*$ > $x_n$} ~ 1 - F($x_n$)

where $S_n^*$ is the standard normal variable and F is the normal distribution function.

(Note: This is a theorem and it has been discussed in Feller (An Introduction to Probability theory and its applications Chap. VII sec 6). I didn't get the proof which was given there)

edited Oct 18 '14 at 05:33

asked Oct 18 '14 at 05:26

Sriram Natarajan

what is your definition of $F$? – mookid Oct 18 '14 at 08:11
F is the normal distribution function – Sriram Natarajan Oct 18 '14 at 08:44
isn't this a direct consequence of the definition of the distribution function? – mookid Oct 18 '14 at 08:51
I also thought the same. But the proof in Feller is more involved. I also don't understand the need for $x_n^3$/$\sqrt{n}$ -> 0 – Sriram Natarajan Oct 18 '14 at 08:57
are you sure of your definition of $S_n^*$? it sounds more like a Cesaro mean according to the notation. – mookid Oct 18 '14 at 09:01
I am fairly confident. – Sriram Natarajan Oct 18 '14 at 09:08

Regarding Asymptotic Behaviour of normal distibution

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