rebuild bagging

schedule/slides/17-nonlinear-classifiers.qmd

@@ -346,14 +346,46 @@ err <- map_dbl(1:kmax, ~ mean(knn.cv(dat1[, -1], dat1$y, k = .x) != dat1$y))
 ```{r}
 #| echo: false
 ggplot(data.frame(k = 1:kmax, error = err), aes(k, error)) +
-  geom_point(color = red) +
-  geom_line(color = red)
+  geom_point(color = orange) +
+  geom_line(color = orange)
 ```
 
 I would use the _largest_ (odd) `k` that is close to the minimum.  
 This produces simpler, smoother, decision boundaries.
 
 
+## Alternative (using deviance loss, I think this is right)
+
+```{r}
+#| code-fold: true
+dev <- function(y, prob, prob_min = 1e-5) {
+  y <- as.numeric(as.factor(y)) - 1 # 0/1 valued
+  m <- mean(y)
+  prob_max <- 1 - prob_min
+  prob <- pmin(pmax(prob, prob_min), prob_max)
+  lp <- (1 - y) * log(1 - prob) + y * log(prob)
+  ly <- (1 - y) * log(1 - m) + y * log(m)
+  2 * (ly - lp)
+}
+knn.cv_probs <- function(train, cl, k = 1) {
+  o <- knn.cv(train, cl, k = k, prob = TRUE)
+  p <- attr(o, "prob")
+  o <- as.numeric(as.factor(o)) - 1
+  p[o == 0] <- 1 - p[o == 0]
+  p
+}
+dev_err <- map_dbl(1:kmax, ~ mean(dev(dat1$y, knn.cv_probs(dat1[, -1], dat1$y, k = .x))))
+```
+
+```{r}
+#| echo: false
+ggplot(data.frame(k = 1:kmax, error = dev_err), aes(k, error)) +
+  geom_point(color = orange) +
+  geom_line(color = orange)
+```
+
+
+
 
 ## Final version
 

schedule/slides/18-the-bootstrap.qmd

@@ -92,10 +92,26 @@ Can I get a 95% confidence interval for $Pr(y_0=1 \given x_0)$?
 $\hat{F}$ is the "empirical" distribution of the bootstraps. 
 
 
+## Empirical distribution
+
+```{r}
+#| code-fold: true
+r <- rexp(50, 1 / 5)
+ggplot(tibble(r = r), aes(r)) + 
+  stat_ecdf(colour = orange) +
+  geom_vline(xintercept = quantile(r, probs = c(.05, .95))) +
+  geom_hline(yintercept = c(.05, .95), linetype = "dashed") +
+  annotate(
+    "label", x = c(5, 12), y = c(.25, .75), 
+    label = c("hat(F)[boot](.05)", "hat(F)[boot](.95)"), 
+    parse = TRUE
+  )
+```
+
 
 ## Very basic example
 
-* Let $X_i\sim Exponential(1/5)$. The pdf is $f(x) = \frac{1}{5}e^{-x/5}$
+* Let $X_i\sim \textrm{Exponential}(1/5)$. The pdf is $f(x) = \frac{1}{5}e^{-x/5}$
 
 
 * I know if I estimate the mean with $\bar{X}$, then by the CLT (if $n$ is big), 
@@ -368,7 +384,7 @@ where $\theta^*_q$ is the $q$ quantile of $\hat{\Theta}$.
 Let $\hat{\theta}$ be our sample statistic, $\hat{\Theta}$ be the resamples
 
 $$
-[\hat{\theta} - t_{\alpha/2}\hat{s},\ \hat{\theta} - t_{\alpha/2}\hat{s}]
+[\hat{\theta} - t_{\alpha/2}s,\ \hat{\theta} - t_{\alpha/2}s]
 $$
 
 where $\hat{s} = \sqrt{\Var{\hat{\Theta}}}$