I need to figure out whether the following swapping of limit operators is allowed:
$\lim\limits_{n \longrightarrow \infty} \lim\limits_{k \longrightarrow \infty} P(Z_n > \frac{1}{k}) = \lim\limits_{k \longrightarrow \infty} \lim\limits_{n \longrightarrow \infty} P(Z_n > \frac{1}{k})$,
where, $Z_n, n\ge 0$ is a branching process if that helps (I do not think inclusion of this fact is crucial). Clearly the sequence $P(Z_n > \frac{1}{k})$ is non-decreasing in $k \in \mathbb{N}$. Also, clearly $\lim\limits_{k \longrightarrow \infty} P(Z_n > \frac{1}{k}) = P(Z_n > 0)$ exists and also using Markov inequality I am able to show that for fixed $k$ $\lim\limits_{n \longrightarrow \infty} P(Z_n > \frac{1}{k}) = \lim\limits_{n \longrightarrow \infty} P(Z_n > 0) = 0$. I am trying to prove that swapping the limits is allowed here using the above observations. I would appreciate any pointers and help. Thanks!
This is unlikely to be valid. By continuity of measure, the left hand side is just $\lim_{n\to\infty}P(Z_n>0)$, whereas the right hand side need not behave so simply. Suppose for instance that $\{X_i\}$ are iid $\mathcal N(0,1)$ random variables and $Z_n=\frac1n\sum_{i=1}^nX_i$. Then the left hand side is $\frac12$ whereas the right hand side is $0$.