We want to draw inference abouta parameter \(\theta\), which can be a vector.
NB: \(T(X_1,...,X_n)\) can equal \(\theta(\hat{F})\), in which case it is a plug-in estimator
Alternative estimator: \(\hat{\theta_2} = T(\hat{F})\) = sample median
\(\hat{F} \rightarrow\) empirical distribution based on sample \(x_1,...,x_n\), a discrete probability distribution with mass \(\frac{1}{n}\) for each observation
A method to estimate a parameter from a sample
\[\theta=\theta(F)\]
\[\hat{\theta_B}=\theta(\hat{F})\]
Bootstrap makes use of the principle of plugging in this estimator.
\(\hat{F}\) is an empirical distribution function, which serves as an estimate of \(F\)
We treat \(\hat{F}\) as if it is the population and draw rebeated samples from it
Assume that the behaviour of \(T(X_1^{*i},...,T_n^{*i})\) relative to \(\hat{\theta_B}\) is the same as that of \(T(X_1,...,X_n)\) relative to \(\theta\)
\(T(X_1^{*i},...,T_n^{*i})\) is the estimate from the \(i^{th}\) Bootstrap Sample
\(B\) is the number of bootstrap samples
\[\bar{T^*}=\frac{1}{B} \sum_{i=1}^B T(X_1^{*i},...,X_n^{*i})\]
Bias:
\[E(\hat{\theta})-\theta\]
Bootstrap estimate of bias:
\[\bar{T^*} - \hat{\theta_B}\]
Plug in an estimate, which may differ from \(T(X_1,...,X_n)\)
Standard error:
\[\widehat{SE}(T(X_1,...,X_n))=\Big[\frac{1}{B} \sum_{i=1}^{B} (T(X_1^{*i},...,X_n^{*i})-\hat{T^*})^2\Big]^{\frac{1}{2}}\]
\(c_1=B(\frac{\alpha}{2})^{th} \space order \space statistic\) (from bootstrap distribution}
\(c_2=B(\frac{1- \alpha}{2})^{th} \space order \space statistic\)
\[L = 2 \hat{\theta_B}-c_2\]
\[U = 2 \hat{\theta_B}-c_1\]
\[L = \hat{\theta_B}+T(X_1,...,X_n)\]
\[U = \hat{\theta_B}+T(X_1,...,X_n)\]
If the bootstrap is distributed symmetrically around \(T(X_1,...,X^n)\), then:
\[T(X_1,...,X_n)-c_2 = -(T(X_1,...,X_n)-c_1)\]
and the confidence interval simplifies to \((c_1,c_2)\)
No assumptions about model, distribution, or sample observations
Regression is done on sample pairs: \((x_i,y_i)\)
Assume model is correct, only parameter values unknown, sample from:
\[F_Y(y;\hat{\theta})\]
Regression is done by sampling residuals from \(N(0,\hat{\sigma}^2)\), calculate:
\[y_i^*=\hat{\alpha}+\hat{\beta}x_i+e_i^*\]
Assume residuals are iid, but do not assume a distribution
Regression is done by resampling from observed residuals, bootstrapping data as before,
\[y_i^*=\hat{\alpha}+\hat{\beta}x_i+e_i^*\]