# The Expectation of a little 0 probability term

My question is that, say, if $Y_n=o_p(\frac 1 n)$ or $nY_n\rightarrow 0$ pointwise, do we have $E(Y_n)=o(1/n)$ in general?

This arises from the solution to a particular problem: $X_1,X_2,...$ are i.i.d., each having exponential density with parameter $\lambda$. We define $g(\lambda)=P_{\lambda}\{X_i>1\}$. Now we can apply Taylor's theorem and get $g(\bar{X}_n)=g(\lambda)+g'(\lambda)(\bar{X}_n-\lambda)+\frac{g''(\lambda)}{2} (\bar{X}_n-\lambda)^2+R(\bar{X}_n-\lambda)$, where $\bar{X}_n$ is the sample mean and $R$ is the remainder term with $R(x)=o(x^2)$. By Law of Large Numbers, we have $R(\bar{X}_n-\lambda)=o_p(\frac 1 n)$ or $nR(\bar{X}_n-\lambda)\rightarrow 0$ pointwise (because $Var(\bar{X}_n)=\frac{\lambda^2}{n}$). Then do we have $E(R(\bar{X}_n-\lambda))=o(\frac 1 n)$?

The solution directly says "we have $E(R(\bar{X}_n-\lambda))=o(\frac 1 n)$", and I am just confused about this step.

Thank you in advance!

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So basically your question is: is it true that convergence in probability to $0$ of a sequence $(Z_n)$ implies convergence of the sequence $(E(Z_n))$ ? This is not true in general (at least the expectations must exist). I do not remember the practical conditions under which this is true. –  Stéphane Laurent Oct 18 '12 at 7:46
Thanks. I do know that this usually requires Dominated Convergence Theorem. But it doesn't look applicable here. –  Zariski Oct 18 '12 at 20:24
Maybe "uniform integrability" is a good criterion (I vaguely remember) –  Stéphane Laurent Oct 18 '12 at 20:28