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?

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.