统计学习导论：基于R应用——第五章习题

时间：2023-12-15 11:48:38

第五章习题

1.

我们主要用到下面三个公式：

统计学习导论：基于R应用——第五章习题

统计学习导论：基于R应用——第五章习题

统计学习导论：基于R应用——第五章习题

根据上述公式，我们将式子统计学习导论：基于R应用——第五章习题化简为统计学习导论：基于R应用——第五章习题

对统计学习导论：基于R应用——第五章习题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)

　　

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