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I have a question from Durrett which I don't quite get the solution.

The question is

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and the solution is

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I think I understand up to when $|S_n - n| \leq n^{2/3}$ is an event w.p.1, this is because $P(|S_n - n| \leq n^{2/3}) = P(|S_n - n|/\sigma n^{1/2} \leq n^{1/6}/\sigma)$. The former converges to a standard normal and the latter increases as n increases so the prob. converges to 1.

However, I do not understand the first inequality below (where the integral is less than $n^{2/3}$ times the big bracket). I don't quite see why it holds. Please help, thanks.

jderzol
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  • Since this was reactivated 26 months later, one can note that the desired result is direct from the identity $$2\left(\sqrt{S_n}-\sqrt{n}\right)=2\frac{\frac{S_n-n}{\sqrt{n}}}{1+\sqrt{{}{}{}{}\frac{S_n}{n}}}.$$ – Did May 03 '16 at 12:29

1 Answers1

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It follows from the fact that: $$\int^b_a f(x)dx\leqslant |b-a| \max_{x\in [a;b]} |f(x)| $$ and the fact that $|\int_a^b f|=|\int_b^a f| $ Therefore you get :

\begin{align} \left |\int_n^{S_n}x^{-1/2}-n^{-1/2}dx\right|&\leqslant |S_n-n|\times \max _{x\in [n;S_n]}|f(x)| \\ &\leqslant n^{2/3} \times \max_{[n-n^{2/3};n+n^{2/3}] } |f(x)| \end{align} Where $f(x)=x^{-1/2}-n^{-1/2}$ and the abuse of notation when $S_n<n$ then $[n;S_n]$ is actually $[S_n;n]$. The second inequality is implied by the hypothesis meaning that $|S_n-n|\leqslant n^{2/3}$, implies $[n;S_n]\subset [n-n^{2/3};n+n^{2/3}] $

Finally, the function $f$ is decreasing, so the maximal value of $|f|$ over $[a;b]$ is either at $a$, either at $b$. An easy calculation gives that the maximum of $f$ at $n-n^{2/3}$

Sefi
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