6
$\begingroup$

Let $X_n$ be a uniformly integrable (UI) sequence of random variables. If we have $$ X_n = \mu + O_p(n^{-1}), $$ then for $0 \le \delta < 1$ this implies $$ X_n = \mu + o_p(n^{-\delta}) \quad \quad \implies \quad \quad n^\delta(X_n - \mu) = o_p(1). $$ Since $X_n$ is UI and $n^\delta(X_n - \mu)$ converges in probability to zero it converges to zero in expecation and we get $$ E[n^\delta(X_n - \mu)] = o(1) \quad \quad \implies \quad \quad E[(X_n - \mu)] = o(n^{-\delta}). $$ So we have converted the convergence in probability to convergence in expectation: $$ X_n = \mu + O_p(n^{-1}) \quad \quad \text{to} \quad \quad E[X_n] = \mu + o(n^{-\delta}). $$ But we have lost some precision as we have moved from a Big $O_p$ with a specific $n^{-1}$ to a little $o$ with an $n^{- \delta}$ that is less than $1$ but can be arbitrarily close to it. I want to know is it possible to go from a Big $O_p$ to a Big $O$ and say: $$ X_n = \mu + O_p(n^{-1}) \quad \quad \text{to} \quad \quad E[X_n] = \mu + O(n^{-1}). $$

$\endgroup$

1 Answer 1

6
$\begingroup$

Here is a counterexample:

$P(X_n = 1) = \frac{1}{\sqrt{n}}$

$P(X_n = 0) = 1 - \frac{1}{\sqrt{n}}$

To show that $X_n = O_p(\frac{1}{n})$: given $\epsilon > 0$, let $M = N > \frac{1}{\epsilon^2}$.

Then for $n > N$, $P(n|X_n| > M) = P(|X_n| > \frac{M}{n}) = P(X_n = 1) = \frac{1}{\sqrt{n}} < \epsilon$ as required.

But $E(X_n) = \frac{1}{\sqrt{n}}$, which is not $O(\frac{1}{n})$.

$\endgroup$
8
  • $\begingroup$ That is not a uniformly integrable sequence. We can go from convergence in probability to convergence in expecation once we have uniform integrability. My question is can we go from Big Op in probability to Big O in expectation, because at the moment all I know is we can go from Big Op in probability to little o in expectation. $\endgroup$
    – sonicboom
    Dec 16, 2020 at 14:22
  • 3
    $\begingroup$ @sonicboom sorry, I didn't notice the uniform integrability condition. I've updated the answer with a uniformly integrable example. $\endgroup$
    – fblundun
    Dec 16, 2020 at 17:14
  • 1
    $\begingroup$ @SextusEmpiricus I believe it's that although $X_n$ is UI, $n^\delta (X_n - \mu)$ might not be UI (and in my example is not for $\delta = 0.5$), so might not converge in expectation to 0. $\endgroup$
    – fblundun
    Dec 16, 2020 at 18:24
  • 2
    $\begingroup$ @SextusEmpiricus "$nX_n$ is UI" isn't a necessary condition for OP's conclusion (though it might be sufficient) - for example if $P(Y_n = 1) = n^{-1}$ and $P(Y_n = 0) = 1 - n^{-1}$ then $E(Y_n) = n^{-1}$, but $nY_n$ is not UI. Maybe the necessary condition is "whenever $f(n) = o(n)$, $f(n)X_n$ is UI"? $\endgroup$
    – fblundun
    Dec 17, 2020 at 12:09
  • 1
    $\begingroup$ @fblundun Your counterexample shows the statement doesn't hold in the general case. What about in the case where $X_n = \sum_{i=1}^n x_i/\sum_{i=1}^n x_i^2$ where $\{x_i\}_{i=1}^n$ are iid with $E[x_i^k]= \mu_k < \infty$? Here $$ \begin{align} X_n &= (n\mu_1 + O_p(n^{1/2}))/(n\mu_2 + O_p(n^{1/2})) \\ &= (O_p(n) + O_p(n^{1/2}))/(O_p(n) + O_p(n^{1/2})) \\ &= O_p(n)/O_p(n) \\ & = O_p(1). \end{align} $$ Since the $\{x_i\}_{i=1}^n$ are iid the numerator and denominator are very well-behaved as $n$ increases, I think in this case $X_n = O_p(1) \implies E[X_n] = O(1)$. But I'm not sure. $\endgroup$
    – sonicboom
    Dec 17, 2020 at 18:29

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.