Clarification on $m$ and $n$ I'n the $m$-out-of-$n$ bootstrap I've asked questions on the $m$-out-of-$n$ bootstrap here before. Responses have been quite valuable, but one key aspect still sparks some confusion from non-statisticians.
Chernick (2007, 2011) states that both $m$ and $n$ approach infinity (with $m$ < $n$) (i.e., $(m, n) \rightarrow \infty$) while $\frac{m}{n} \rightarrow 0$.
My question is: when is the above condition statisfied and how might i go about explaining it to non experts?
 A: I think it's helpful to consider an example such as the maximum.  The obvious problem with an n of n bootstrap in the maximum is that the maximum of the bootstrap sample is the maximum of the original sample with probability 0.632, which is a very discrete distribution.  To make the m of n bootstrap work you need m small enough that the distribution of the subsample maximum isn't very discrete -- so you need m much smaller than n.
Asymptotically, you need the probability of the sample maximum being in the subsample to go to zero.
A: This condition just means that the "speed" with which both quantities go to infinity is different: $n$ goes faster than $m$. For example, if you take $m =\sqrt{n}$, then the condition is satisfied: $\frac{m}{n} = \frac{\sqrt{n}}{n} = \frac{1}{\sqrt{n}} \to 0$ as $n \to \infty$.
One way of explaining it would be through a graph. If you have a plot where $n$ is on the $x$ axis and $m$ is on the $y$ axis, then the curve should not only lie under the line $m=n$ (in fact, in the beginning it can even lie above it), but should go "quite fast" away from it. "quite fast" is difficult to quantify without going into too much detail, but you can draw $m = \sqrt{n}$ or $m = \log n$ and say that this is definitely more than enough.
Hope this helps!
