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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 $$ \sup_{j} \left| \frac{1}{n}\sum_{i=1}^n X_i^j - \mu^j \right| \overset{\text{p}}{\to} 0. $$

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  • $\begingroup$ $E[\mu]$ 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
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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.)

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By the triangle-inequality

\begin{align*} 0 &\le \left|\frac{1}{nm}\sum_{j=1}^m\sum_{i=1}^nX_i^j - E[\mu]\right| \\ &\le \left|\frac{1}{nm}\sum_{j=1}^m\sum_{i=1}^nX_i^j - m^{-1}\sum_{j=1}^m\mu^j \right| + \left|m^{-1}\sum_{j=1}^m\mu^j - E[\mu]\right| \\ &\le \sup_{\mu \in \Theta} \left|\frac{1}{n}\sum_{i=1}^n \bar{X}_\mu - m^{-1}\sum_{j=1}^m\mu^j \right| + \left|m^{-1}\sum_{j=1}^m\mu^j - E[\mu]\right|. \end{align*} With some nonstandard assumptions, you can invoke Theorem 2.1 of this to guarantee the first term converges to $0$. The second term converges to $0$ by the standard law of large numbers.

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