Chapter 5: Exercise 2
-a
-\( 1 - 1/n \)
-b
+\( 1 - 1/n \)
-c
-In bootstrap, we sample with replacement so each observation in the bootstrap -sample has the same 1/n (independent) chance of equaling the jth observation. -Applying the product rule for a total of n observations gives us \( (1 - 1/n)^n \).
-d
+\( Pr(in) = 1 - Pr(out) = 1 - (1 - 1/5)^5 = 1 - (4/5)^5 = 67.2\% \)
+Chapter 5: Exercise 2
+a
+\(1 - 1/n\)
+b
+\(1 - 1/n\)
+c
+In bootstrap, we sample with replacement so each observation in the bootstrap sample has the same 1/n (independent) chance of equaling the jth observation. Applying the product rule for a total of n observations gives us \((1 - 1/n)^n\).
+d
+\(Pr(in) = 1 - Pr(out) = 1 - (1 - 1/5)^5 = 1 - (4/5)^5 = 67.2\%\)
+e
+\(Pr(in) = 1 - Pr(out) = 1 - (1 - 1/100)^{10} = 1 - (99/100)^{100} = 63.4\%\)
+f
+\(1 - (1 - 1/10000)^{10000} = 63.2\%\)
+g
+pr = function(n) return(1 - (1 - 1/n)^n)
+x = 1:100000
+plot(x, pr(x))
The plot quickly reaches an asymptote of about 63.2%. This is what we should expect, because a standard result from calculus tells us that \(\lim_{n \to \infty} (1 + \frac{x}{n})^n = e^{x}\) for any \(x \in \mathbb{R}\), which means that
\[ +1 - \lim_{n \to \infty} \left( 1 - \frac{1}{n}\right)^n = 1 - e^{-1} \approx 0.632 +\]
+h
+set.seed(1)
+store = rep(NA, 1e4)
+for (i in 1:1e4) {
+ store[i] = sum(sample(1:100, rep=T) == 4) > 0
+}
+mean(store)## [1] 0.6408The numerical results show an approximate mean probability of 64.1%, close to our theoretically derived result.
+\( Pr(in) = 1 - Pr(out) = 1 - (1 - 1/100)^{10} = 1 - (99/100)^{100} = 63.4\% \)
-f
-\( 1 - (1 - 1/10000)^{10000} = 63.2\% \)
-