I didn't quite follow the way CI of mean was calculated in the last lab session for bootstrap sampling distribution.
[ (m_hat - np.quantile(result - m_hat, 0.95)), (m_hat - np.quantile(result - m_hat, 0.05)) ]
But I wanted to ask if the following is correct:
Since the bootstrap sampling distribution of the statistic in question would be normal in most cases (CLT), assuming it to be symmetric, can't we find the CI as [lb, ub] where,
lb = np.quantile(T, .025)
ub = np.quantile(T, .975)
you can use the CI [np.quantile(T, .025),np.quantile(T, 0.975)], it is a valid confidence interval (and this does not use the CLT, I don't know why you would need symmetry to have this, the proof is more complicated than simply symmetry) but we didn't give you this interval because the pivotal interval introduced during the course and later used in practical session is a better interval. Both are approximate confidence interval, the approximation comes from the bootstrap part, but the pivotal confidence interval is a better approximation.
There are a number of different methods of using the
bootstrap to find approximate confidence intervals. The "pivotal" CI presented in the course has the methodological avantage that its construction is based on the approximation of \hat\theta-\theta by the bootstrap distribution of \hat\theta^*-\hat\theta.
Another popular method, the bootstrap percentile method, uses the quantiles
of the bootstrap sampling distribution of \hat\theta^* directly.
As written in Rice’s book : "although this CI may seem
initially appealing, its rationale is somewhat obscure. If the bootstrap distribution is
symmetric, the two methods are equivalent".
And actually, simulation experiments in various models show that the CI of the course is better (its simulated confidence level is closer to the theoretical confidence level) than the percentile CI.