\[\newcommand{\E}{\mathrm{E}} \newcommand{\Var}{\mathrm{Var}} \newcommand{\Cov}{\mathrm{Cov}} \newcommand{\se}{\text{se}}\]
Parametric model
\[\xi \equiv f(x; \mu, \sigma) = \frac{1}{\sigma \sqrt{2 \pi}} \exp \left[ -\frac{1}{2\sigma^2} (x - \mu)^2 \right], \quad \mu \in \Re, \sigma > 0\]General form
\[\xi \equiv f(x; \theta) : \theta \in \Theta\]\(\xi_{\text{DENS}}\) is the set of all PDFs and
\[\xi_{\text{SOB}} \equiv f: \int (f''(x))^2 dx < \infty\]
General form
\[g(x) = \frac{1}{nh} \sum_{i = 1}^n f(x)\]
\[f(x) = \frac{1}{\sqrt{2 \pi}}\exp\left[-\frac{1}{2} x^2 \right]\]
\[f(x) = \frac{1}{2} \mathbf{1}_{\{ |x| \leq 1 \} }\]
\[\mu = E(\text{Income}|\text{Education}) = f(\text{Education})\]
\[\mu = E(Y|x) = f(x)\]
\[X_1 \in \{1, 2, \dots ,10 \}\] \[X_2 \in \{1, 2, \dots ,10 \}\] \[X_3 \in \{1, 2, \dots ,10 \}\]
\(10^3 = 1000\) possible combinations of the explanatory variables and \(1000\) conditional expectations of \(Y\) given \(X\):
\[\mu = E(Y|x_1, x_2, x_3) = f(x_1, x_2, x_3)\]
Let \(X_1, \ldots, X_n\) be \(n\) IID data points from some distribution \(F\). A point estimator \(\hat{\theta}_n\) of a paramater \(\theta\) is some function of \(X_1, \ldots, X_n\):
\[\hat{\theta}_n = g(X_1, \ldots, X_n)\]
Standard error
\[\se = \se(\hat{\theta}_n) = \sqrt{\Var (\hat{\theta}_n)}\]Mean squared error
\[ \begin{align} \text{MSE} &= \E_\theta (\hat{\theta}_n - \theta)^2 \\ &= \text{bias}^2(\hat{\theta}_n) + \Var_\theta (\hat{\theta}_n) \end{align} \]
Estimators are approximately Normally distributed
\[\frac{\hat{\theta}_n - \theta}{\se} \leadsto N(0,1)\]
Let \(\hat{\pi}_n = \frac{1}{n} \sum_{i=1}^n X_i\). Then
\[\E(\hat{\pi}_n) = \frac{1}{n} \sum_{i=1}^n \E(X_i) = \pi\]
so \(\hat{\pi}_n\) is unbiasedStandard error is
\[\se = \sqrt{\Var (\hat{\pi}_n)} = \sqrt{\frac{\pi (1 - \pi)}{n}}\]
\[\widehat{\se} = \sqrt{\frac{\hat{\pi} (1 - \hat{\pi})}{n}}\]
\(\E_\pi (\hat{\pi}_n) = \pi\) so \(\text{bias} = \pi - \pi = 0\)
\[ \begin{align} \text{bias}(\hat{\pi}_n) &= \E_\pi (\hat{\pi}) - \pi \\ &= \pi - \pi \\ &= 0 \end{align} \]Consistency \[\se = \sqrt{\frac{\pi (1 - \pi)}{n}} \rightarrow 0\]
\(b = b(X_1, \ldots, X_n)\)
\[\Pr_{\theta} (\theta \in C_n) \geq 1 - \alpha, \quad \forall \theta \in \Theta\]On day 1, you collect data and construct a 95% confidence interval for a parameter \(\theta_1\). On day 2, you collect new data and construct a 95% confidence interval for a parameter \(\theta_2\). You continue this way constructing confidence intervals for a sequence of unrelated parameters \(\theta_1, \theta_2, \ldots\). Then 95% of your intervals will trap the true parameter value.
Let \(\Phi\) be the CDF of a standard Normal distribution
\[z_{\frac{\alpha}{2}} = \Phi^{-1} \left(1 - \frac{\alpha}{2} \right)\]
\[\Pr (Z > \frac{\alpha}{2}) = \frac{\alpha}{2}\]
\[\Pr (-z_{\frac{\alpha}{2}} \leq Z \leq z_{\frac{\alpha}{2}}) = 1 - \alpha\]
where \(Z \sim N(0,1)\)Let
\[C_n = (\hat{\theta}_n - z_{\frac{\alpha}{2}} \widehat{\se}, \hat{\theta}_n + z_{\frac{\alpha}{2}} \widehat{\se})\]Then
\[ \begin{align} \Pr_\theta (\theta \in C_n) &= \Pr_\theta (\hat{\theta}_n - z_{\frac{\alpha}{2}} \widehat{\se} < \theta < \hat{\theta}_n + z_{\frac{\alpha}{2}} \widehat{\se}) \\ &= \Pr_\theta (- z_{\frac{\alpha}{2}} < \frac{\hat{\theta}_n - \theta}{\widehat{\se}} < z_{\frac{\alpha}{2}}) \\ &\rightarrow \Pr ( - z_{\frac{\alpha}{2}} < Z < z_{\frac{\alpha}{2}}) \\ &= 1 - \alpha \end{align} \]
\(\alpha = 0.05\) and \(z_{\frac{\alpha}{2}} = 1.96 \approx 2\)
From this,
\[\Pr (\pi \in C_n \geq 1 - \alpha)\]Approximate confidence interval
\[ \begin{align} \Var (\hat{\pi}_n) &= \frac{1}{n^2} \sum_{i=1}^n \Var(X_i) \\ &= \frac{1}{n^2} \sum_{i=1}^n \pi(1 - \pi) \\ &= \frac{1}{n^2} n\pi(1 - \pi) \\ &= \frac{\pi(1 - \pi)}{n} \\ \se &= \sqrt{\frac{\pi(1 - \pi)}{n}} \\ \widehat{\se} &= \sqrt{\frac{\hat{\pi}(1 - \hat{\pi})}{n}} \end{align} \]
\[\hat{\pi}_n \pm z_{\frac{\alpha}{2}} \widehat{\se} = \hat{\pi}_n \pm z_{\frac{\alpha}{2}} \sqrt{\frac{\hat{\pi}(1 - \hat{\pi})}{n}}\]
Let
\[X_1, \ldots, X_n \sim \text{Bernoulli}(\pi)\]Reasonable to reject \(H_0\) if
\[T = | \hat{\pi}_n - 0.5|\] is large