where
Recall:
if $\epsilon\sim \mathcal{N}$ (normal), then the "best" estimator for $\theta$ is sample mean $\bar{X}$.
However, if $\epsilon$ is not normal, this may not be the case.
For example, if $f_\epsilon(x)=1/2 e^{-|x|}$ ("double xponential"), theb MLE is sample median.
This is important because if $F_\epsilon$ is not known, then it's hard to choose a good estimator.
The biggest problem with not knowing the $F_\epsilon$ is its tails might be very heavy.
(heavy tail refers to the speed at which the density goes to zero when its input goes to infinity)
This is a problem when the number of data points $n$ is not very large. Indeed, tjhen the CLT is NOT a good approximation, therefore the $\bar{X}$ is NOT approximately normal, and the MLE for $\theta$ is NOT $\approx \bar{X}$
Question: what to do when $F_\epsilon$ is not known and $n$ is not large? ($n=50$ is to low)
IDEA: trimmed mean $\bar{X}_\beta$ where $\beta\in(0,1)$:
throw away a proportion $\beta$ of data points from the left & right tail of your data
This certainly avoids outliers
IDEA: Let $\bar{\epsilon}_\beta = \bar{X}_\beta - \theta$
Consider the 0.025 & 0.975 quantiles $c_1$ and $c_2$ for $F_\epsilon$.
Then, we would be able to say $[\bar{X}_\beta-c_2, \bar{X}_\beta-c_1]$ is a $95\%$ confidence interval for $\theta$
Problem: Since we don't have $c_1$ and $c_2$, let us estimate them.
Solution: let us use a large number $B$ of bootstrap samples:
these are $X^*_{ij}$ where $i=1,..,n$ and $j=1,..,B$
then for each $j$ we compute the sample quantiles
then, we average these sample quantiles over $j=1,..,B$
the resulting estimates are $\hat{C}_1$ and $\hat{C}_2$; we cold call them bootstapped sample quantiles.
finally, the bootstrapped $95\%$ confidence interval for $\theta$ is
$$\left[\bar{X}_\beta - \hat{C}_2, \bar{X}_\beta - \hat{C}_1\right]$$
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