第五章习题
1.
我们主要用到下面三个公式：
根据上述公式，我们将式子化简为
对求导即可得到得到公式5-6。
2.
(a)
1 - 1/n
(b)
自助法是有有放回的，所以第二个的概率还是1 - 1/n
(c)
由于自助法是有放回的，且每次抽样都是独立事件，所以概率是(1 - 1/n)^n
(d)
答案是1-(1-1/5)^5 = 67.2%
(e)
63.4%
(f)
63.2%
(g)
pr = function(n) return(1 - (1 - 1/n)^n)
x = 1:1e+05
plot(x, pr(x))
3题和4题略
5.
(a)
library(ISLR)
summary(Default) attach(Default) set.seed(1)
glm.fit = glm(default ~ income + balance, data = Default, family = binomial)
(b)
train = sample(dim(Default)[1], dim(Default)[1]/2)
glm.fit = glm(default ~ income + balance, data = Default, family = binomial, subset = train)
glm.pred = rep("No", dim(Default)[1]/2)
glm.probs = predict(glm.fit, Default[-train, ], type = "response")
glm.pred[glm.probs > 0.5] = "Yes"
mean(glm.pred != Default[-train, ]$default)
(c)
把(b)跑三遍。。。
(d)
上面代码在拟合逻辑回归的时候加个变量即可
6.
(a)
library(ISLR)
summary(Default)
attach(Default) set.seed(1)
glm.fit = glm(default ~ income + balance, data = Default, family = binomial)
summary(glm.fit)
(b)
boot.fn = function(data, index) return(coef(glm(default ~ income + balance, data = data, family = binomial, subset = index)))
(c)
library(boot)
boot(Default, boot.fn, 50)
7.
(a)
library(ISLR)
summary(Weekly)
set.seed(1)
attach(Weekly) glm.fit = glm(Direction ~ Lag1 + Lag2, data = Weekly, family = binomial)
summary(glm.fit)
(b)
glm.fit = glm(Direction ~ Lag1 + Lag2, data = Weekly[-1, ], family = binomial)
summary(glm.fit)
(c)
predict.glm(glm.fit, Weekly[1, ], type = "response") > 0.5
(d)
count = rep(0, dim(Weekly)[1])
for (i in 1:(dim(Weekly)[1])) {
glm.fit = glm(Direction ~ Lag1 + Lag2, data = Weekly[-i, ], family = binomial)
is_up = predict.glm(glm.fit, Weekly[i, ], type = "response") > 0.5
is_true_up = Weekly[i, ]$Direction == "Up"
if (is_up != is_true_up)
count[i] = 1
}
sum(count)
(e)
mean(count)
8.
(a)
n为100，p为2
(b)
set.seed(1)
y = rnorm(100)
x = rnorm(100)
y = x - 2 * x^2 + rnorm(100)
plot(x, y)
(c)
library(boot)
Data = data.frame(x, y)
set.seed(1) glm.fit = glm(y ~ x)
cv.glm(Data, glm.fit)$delta glm.fit = glm(y ~ poly(x, 2))
cv.glm(Data, glm.fit)$delta glm.fit = glm(y ~ poly(x, 3))
cv.glm(Data, glm.fit)$delta glm.fit = glm(y ~ poly(x, 4))
cv.glm(Data, glm.fit)$delta
(d)
set.seed(10)
glm.fit = glm(y ~ x)
cv.glm(Data, glm.fit)$delta glm.fit = glm(y ~ poly(x, 2))
cv.glm(Data, glm.fit)$delta glm.fit = glm(y ~ poly(x, 3))
cv.glm(Data, glm.fit)$delta glm.fit = glm(y ~ poly(x, 4))
cv.glm(Data, glm.fit)$delta
结果一样。。。
(e)
二次的最小
9.
(a)
library(MASS)
summary(Boston) set.seed(1)
attach(Boston) medv.mean = mean(medv)
medv.mean
(b)
medv.err = sd(medv)/sqrt(length(medv))
medv.err
(c)
boot.fn = function(data, index) return(mean(data[index]))
library(boot)
bstrap = boot(medv, boot.fn, 1000)
bstrap
(d)
t.test(medv)
c(bstrap$t0 - 2 * 0.4119, bstrap$t0 + 2 * 0.4119)
(e)
medv.med = median(medv)
medv.med
(f)
boot.fn = function(data, index) return(median(data[index]))
boot(medv, boot.fn, 1000)
(g)
medv.tenth = quantile(medv, c(0.1))
medv.tenth
(h)
boot.fn = function(data, index) return(quantile(data[index], c(0.1)))
boot(medv, boot.fn, 1000)