统计学习导论：基于R应用——第五章习题

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)