Let $\frac{1}{n}\sum_{i=1}^n X_i^j \overset{\text{p}}{\to} \mu^j$ for each $j$ (as $n \to \infty$ ). Under what conditions can we guarantee that $$ \frac{1}{nm}\sum_{j=1}^m\sum_{i=1}^nX_i^j \overset{\text{p}}{\to} E[\mu]? $$

Is there a way to guarantee it that doesn't involve assuming uniform convergence in probability? The sequences are uniformly convergent in probability if $$ \sup_{j} \left| \frac{1}{n}\sum_{i=1}^n X_i^j - \mu^j \right| \overset{\text{p}}{\to} 0. $$

  • $\begingroup$ $E[\mu]$ is undefined. Is $m$ fixed or does $m \rightarrow \infty$? $\endgroup$ – Michael May 17 at 18:31
  • $\begingroup$ @Michael I'm flexible on this, but what I had in mind was $m \to \infty$ and $\mu$ was a continuous random variable. I was thinking along the lines that $m$ was a multiple of $n$ or something $\endgroup$ – Taylor May 17 at 19:55

Assuming $E[\mu] = \frac{1}{m} \sum_{j=1}^m \mu^j$ and $m \rightarrow \infty$. If you're ok with passing to a subsequence $n_m$ so that $$ P( | \frac{1}{n_m}\sum_{i=1}^{n_m} X_{i}^j - \mu^j | < \frac{1}{m^2} ) > 1 - \frac{1}{m^2}, $$ then the result would hold for the subsequence (more precisely, sub-array).

In general, it is not true that $$ \epsilon^j_n \stackrel{p}{\rightarrow} 0 \; \mbox{ as } n \rightarrow \infty, \;\; \forall j $$
implies $$ \frac{1}{m} \sum_{j = 1}^m \epsilon^j_n \stackrel{p}{\rightarrow} 0 \; \mbox{ as } n, \; m \rightarrow \infty. $$

Indeed, it is not even true for deterministic sequences, which is a special case. Let $$ \epsilon^j_n = j \cdot (\log (n+1) - \log(n)), $$ then $\epsilon^j_n \rightarrow 0$ as $n \rightarrow \infty$ for all $j$. Averaging across $j$ gives $$ \frac{1}{m} \sum_{j = 1}^m \epsilon^j_n = O\left(m \cdot ( \log (n+1) - \log(n) ) \right) $$ which does not converge to $0$ as $n, m \rightarrow 0$.

(You can easily cook up a sequence $X^j_n$ so that $\frac{1}{n}\sum_{i=1}^nX_n^j = \epsilon^j_n$, making it a counter-example to the claim.)

| cite | improve this answer | |

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.