# Asymptotic rate for the expected value of the square root of sample average

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?

• Taylor expansion of square root at 1 yields $\sqrt{S_n}=1+(S_n-1)/2+O((S_n-1))$. – user35593 Mar 21 '19 at 7:07
• 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. – Florian Tramèr Mar 21 '19 at 7:18
• The second order term in the Taylor expansion should be $O((S_n - 1)^2)$ of course. – Florian Tramèr Mar 21 '19 at 7:26
• Yes, I could not correct it. expectation of second order term gives $O(1/n)$. – user35593 Mar 21 '19 at 7:33
• 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? – Florian Tramèr Mar 21 '19 at 7:37

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.