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I have iid random variables $X_1, \dots, X_n$ with $X_i \geq 0$, $E[X_i]=1$ and $V[X_i] = \sigma^2$. Let $S_n = \frac{\sum_{i=1}^n X_i}{n}$.

I'd like to say that $E[\sqrt{S_n}] = 1-O(1/n)$.

My first approach was to write $E[\sqrt{S_n}] = \sqrt{E[S_n] - V[\sqrt{S_n}]} = \sqrt{1-V[\sqrt{S_n}]}$.

I'm then left with showing that $V[\sqrt{S_n}] = O(1/n)$.

I'm unsure how to go about this. First, can I hope to prove such an asymptotic bound in general? If not, are there extra assumptions that can be made on the $X_i$ so that this holds true?

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  • $\begingroup$ Taylor expansion of square root at 1 yields $\sqrt{S_n}=1+(S_n-1)/2+O((S_n-1))$. $\endgroup$
    – user35593
    Commented Mar 21, 2019 at 7:07
  • $\begingroup$ Right, this seems to yield something similar to the expression in terms of the variance I have above. Taking expectations on the Taylor expansion, I'd get $E[\sqrt{Sn}] = 1 + E[O(S_n - 1)]$. I'm not sure what to make of that second term. $\endgroup$
    – Florian Tramèr
    Commented Mar 21, 2019 at 7:18
  • $\begingroup$ The second order term in the Taylor expansion should be $O((S_n - 1)^2)$ of course. $\endgroup$
    – Florian Tramèr
    Commented Mar 21, 2019 at 7:26
  • $\begingroup$ Yes, I could not correct it. expectation of second order term gives $O(1/n)$. $\endgroup$
    – user35593
    Commented Mar 21, 2019 at 7:33
  • $\begingroup$ How does the asymptotic growth of the second term follow? E.g., why wouldn't this be $O(1/\sqrt{n})$ or $O(1/\log{n})$ or anything else? $\endgroup$
    – Florian Tramèr
    Commented Mar 21, 2019 at 7:37

1 Answer 1

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Substituting $S_n$ for $u$ in the inequalities $$\frac{1+u-(u-1)^2}2\le\sqrt u\le\frac{1+u}2$$ for $u\ge0$, taking the expectations, and using that $ES_n=1$ and $E(S_n-1)^2=V(S_n)=\sigma^2/n$, we have $$1-\frac{\sigma^2}{2n}\le E\sqrt{S_n}\le1,$$ so that $E\sqrt{S_n}=1-O(1/n)$, as desired.

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