Connection between Bootstrap and Maximum Likelihood Estimator I am trying to understand the connection between Bootstrap and Maximum Likelihood Estimator. Is the connection of the following format?
Let $ \mathcal{F}_{\theta} $ be a parametrized family of distributions.
Let us consider a distribution $ \mathcal{F}_{\theta_0} $ with unknown value of $ \theta_0 $.
Let $ \vec{X} $ be an iid sample of size N from distribution $ \mathcal{F}_{\theta_0} $.
And let $ \theta^* $ the MLE estimate of $ \theta $ under the above assumptions, that is
\begin{align*}
\theta^* = \arg\max_{\theta} \prod_{x \in \vec{X}} P_{\mathcal{F}_{\theta}}(x) \ .
\end{align*}
We are concerned about the variance of a statistics $ S(\vec{X}) $ over the sampled data $ \vec{X} $.
Let $ var_B S(\vec{X}) $ be a bootstrapped variance of statistics $ S $ computed over B rounds,
that is, we sample a sample $\vec{X}_i$ with replacement from $ \vec{X} $ and compute variance of
$ \{S(\vec{X}_i), i=1,\dots,B \} $.
Then
\begin{align*}
\lim _{B\to \infty} [var_B S(\vec{X})] =  var_{x \sim \mathcal{F}_{\theta^*}} S(x) \ .
\end{align*}
Is the limit correct? Does N have to go to infinity as well?
(I would appreciate answers beyond simple references).
Edit: Note that in the right side of the equation, $x$ is sampled from $\mathcal{F}_{\theta^*}$, not $\mathcal{F}_{\theta_0}$.

Ok, so after the edit it's clearer that the question is different from what I thought.  The answer is still 'No'.
Suppose $X$ has a Cauchy distribution and take $S(X)=\bar X$. $\mathrm{var}_B[\bar X]$ is always finite, and has a finite limit as $B\to\infty$ for fixed $N$, but $\mathrm{var}_{X\sim\textrm{Cauchy}(\theta^*)}[\bar X]$ is infinite.
Less extreme, take $X\sim N(\mu,1)$. Then $\mathrm{var}_{X\sim P_{\theta^*}}[\bar X]=1/N$, but
$$\lim_{B\to\infty}\mathrm{var}_{B}[\bar X]=\hat\sigma^2/N$$
where $\hat\sigma^2$ is the sample variance in that sample.
In the second case the equality straightforwardly holds for the limit as $N\to\infty$. In the first case it holds in the sense that the bootstrap variance increases a.s. without bound as $N\to\infty$.
