(a) Explain how k-fold cross-validation is implemented
“the k-fold CV approach”involves randomly dividing the set of observations into k groups, or folds, of approximately equal size. The first fold is treated as a validation set, and the method is fit on the remaining k-1 folds. The mean squared error, MSE, is then computed on the observations in the held-out fold. This procedure is repeated k times"
(b) What are the advantages and disadvantages of k-fold crossvalidation relative to:
i. The validation set approach?
less variance, more bias
ii. LOOCV
more variance, less bias
(a) Fit a logistic regression model that uses income and balance to predict default.
library(ISLR)## Warning: package 'ISLR' was built under R version 3.6.3glm.fit = glm(default ~ income + balance, data = Default, family = binomial)
summary(glm.fit)##
## Call:
## glm(formula = default ~ income + balance, family = binomial,
## data = Default)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -2.4725 -0.1444 -0.0574 -0.0211 3.7245
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -1.154e+01 4.348e-01 -26.545 < 2e-16 ***
## income 2.081e-05 4.985e-06 4.174 2.99e-05 ***
## balance 5.647e-03 2.274e-04 24.836 < 2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 2920.6 on 9999 degrees of freedom
## Residual deviance: 1579.0 on 9997 degrees of freedom
## AIC: 1585
##
## Number of Fisher Scoring iterations: 8(b) Using the validation set approach, estimate the test error of this model. In order to do this, you must perform the following steps:
i. Split the sample set into a training set and a validation set. 5.4 Exercises 199
ii. Fit a multiple logistic regression model using only the training observations.
iii. Obtain a prediction of default status for each individual in the validation set by computing the posterior probability of default for that individual, and classifying the individual to the default category if the posterior probability is greater than 0.5.
iv. Compute the validation set error, which is the fraction of the observations in the validation set that are misclassified.
fiveb = function()
{
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"
return(mean(glm.pred != Default[-train, ]$default))
}
fiveb()## [1] 0.0252(c) Repeat the process in (b) three times, using three different splits of the observations into a training set and a validation set. Comment on the results obtained.
fiveb()## [1] 0.0262fiveb()## [1] 0.027fiveb()## [1] 0.0264each run is usually between 24 and 28
(d) Now consider a logistic regression model that predicts the probability of default using income, balance, and a dummy variable for student. Estimate the test error for this model using the validation set approach. Comment on whether or not including a dummy variable for student leads to a reduction in the test error rate.
train = sample(dim(Default)[1], dim(Default)[1]/2)
glm.fit = glm(default ~ income + balance + student, data = Default, family = binomial,
subset = train)
glm.pred = rep("No", dim(Default)[1]/2)
glm.probs = predict(glm.fit, Default[-train, ], type = "response")
glm.pred[glm.probs > 0.5] = "Yes"
mean(glm.pred != Default[-train, ]$default)## [1] 0.0254The dummy variable for student doesnt seem to have an effect on the test error
##Problem 6 || We continue to consider the use of a logistic regression model to predict the probability of default using income and balance on the Default data set. In particular, we will now compute estimates for the standard errors of the income and balance logistic regression coefficients in two different ways: (1) using the bootstrap, and (2) using the standard formula for computing the standard errors in the glm() function. Do not forget to set a random seed before beginning your analysis.##
(a) Using the summary() and glm() functions, determine the estimated standard errors for the coefficients associated with income and balance in a multiple logistic regression model that uses both predictors.
(b) Write a function, boot.fn(), that takes as input the Default data set as well as an index of the observations, and that outputs the coefficient estimates for income and balance in the multiple logistic regression model.
(c) Use the boot() function together with your boot.fn() function to estimate the standard errors of the logistic regression coefficients for income and balance.
(d) Comment on the estimated standard errors obtained using the glm() function and using your bootstrap function
glm.fit1 = glm(default ~ income + balance, data = Default, family = binomial)
summary(glm.fit1)##
## Call:
## glm(formula = default ~ income + balance, family = binomial,
## data = Default)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -2.4725 -0.1444 -0.0574 -0.0211 3.7245
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -1.154e+01 4.348e-01 -26.545 < 2e-16 ***
## income 2.081e-05 4.985e-06 4.174 2.99e-05 ***
## balance 5.647e-03 2.274e-04 24.836 < 2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 2920.6 on 9999 degrees of freedom
## Residual deviance: 1579.0 on 9997 degrees of freedom
## AIC: 1585
##
## Number of Fisher Scoring iterations: 8boot.fn = function(data, index) return(coef(glm(default ~ income + balance, data = data, family = binomial, subset = index)))
library(boot)
boot(Default, boot.fn, 50)##
## ORDINARY NONPARAMETRIC BOOTSTRAP
##
##
## Call:
## boot(data = Default, statistic = boot.fn, R = 50)
##
##
## Bootstrap Statistics :
## original bias std. error
## t1* -1.154047e+01 -6.306052e-02 3.928227e-01
## t2* 2.080898e-05 5.308435e-08 4.755817e-06
## t3* 5.647103e-03 3.219385e-05 2.048734e-04Standard error estimates are pretty close using glm summary function versus bootstrap
##Problem 9 || We will now consider the Boston housing data set, from the MASS library.##
(a) Based on this data set, provide an estimate for the population mean of medv. Call this estimate ˆμ. (b) Provide an estimate of the standard error of ˆμ. Interpret this result. Hint: We can compute the standard error of the sample mean by dividing the sample standard deviation by the square root of the number of observations. (c) Now estimate the standard error of ˆμ using the bootstrap. How does this compare to your answer from (b)?
This is similar to 9.b
(d) Based on your bootstrap estimate from (c), provide a 95 % confidence interval for the mean of medv. Compare it to the results obtained using t.test(Boston$medv). Hint: You can approximate a 95 % confidence interval using the formula [ˆμ − 2SE(ˆμ), μˆ + 2SE(ˆμ)].
bootstrap is .02 from t test
(e) Based on this data set, provide an estimate, ˆμmed, for the median value of medv in the population. (f) We now would like to estimate the standard error of ˆμmed. Unfortunately, there is no simple formula for computing the standard error of the median. Instead, estimate the standard error of the median using the bootstrap. Comment on your findings.
Median of 21.2 with a standard error of 0.380. Small standard error relative to median value.
(g) Based on this data set, provide an estimate for the tenth percentile of medv in Boston suburbs. Call this quantity ˆμ0.1. (You can use the quantile() function.) (h) Use the bootstrap to estimate the standard error of ˆμ0.1. Comment on your findings.
library(MASS)
summary(Boston)## crim zn indus chas
## Min. : 0.00632 Min. : 0.00 Min. : 0.46 Min. :0.00000
## 1st Qu.: 0.08204 1st Qu.: 0.00 1st Qu.: 5.19 1st Qu.:0.00000
## Median : 0.25651 Median : 0.00 Median : 9.69 Median :0.00000
## Mean : 3.61352 Mean : 11.36 Mean :11.14 Mean :0.06917
## 3rd Qu.: 3.67708 3rd Qu.: 12.50 3rd Qu.:18.10 3rd Qu.:0.00000
## Max. :88.97620 Max. :100.00 Max. :27.74 Max. :1.00000
## nox rm age dis
## Min. :0.3850 Min. :3.561 Min. : 2.90 Min. : 1.130
## 1st Qu.:0.4490 1st Qu.:5.886 1st Qu.: 45.02 1st Qu.: 2.100
## Median :0.5380 Median :6.208 Median : 77.50 Median : 3.207
## Mean :0.5547 Mean :6.285 Mean : 68.57 Mean : 3.795
## 3rd Qu.:0.6240 3rd Qu.:6.623 3rd Qu.: 94.08 3rd Qu.: 5.188
## Max. :0.8710 Max. :8.780 Max. :100.00 Max. :12.127
## rad tax ptratio black
## Min. : 1.000 Min. :187.0 Min. :12.60 Min. : 0.32
## 1st Qu.: 4.000 1st Qu.:279.0 1st Qu.:17.40 1st Qu.:375.38
## Median : 5.000 Median :330.0 Median :19.05 Median :391.44
## Mean : 9.549 Mean :408.2 Mean :18.46 Mean :356.67
## 3rd Qu.:24.000 3rd Qu.:666.0 3rd Qu.:20.20 3rd Qu.:396.23
## Max. :24.000 Max. :711.0 Max. :22.00 Max. :396.90
## lstat medv
## Min. : 1.73 Min. : 5.00
## 1st Qu.: 6.95 1st Qu.:17.02
## Median :11.36 Median :21.20
## Mean :12.65 Mean :22.53
## 3rd Qu.:16.95 3rd Qu.:25.00
## Max. :37.97 Max. :50.00attach(Boston)
medv.mean = mean(medv)
medv.mean## [1] 22.53281medv.err = sd(medv)/sqrt(length(medv))
medv.err## [1] 0.4088611boot.fn = function(data, index) return(mean(data[index]))
library(boot)
bstrap = boot(medv, boot.fn, 1000)
bstrap##
## ORDINARY NONPARAMETRIC BOOTSTRAP
##
##
## Call:
## boot(data = medv, statistic = boot.fn, R = 1000)
##
##
## Bootstrap Statistics :
## original bias std. error
## t1* 22.53281 0.02950257 0.4023026t.test(medv)##
## One Sample t-test
##
## data: medv
## t = 55.111, df = 505, p-value < 2.2e-16
## alternative hypothesis: true mean is not equal to 0
## 95 percent confidence interval:
## 21.72953 23.33608
## sample estimates:
## mean of x
## 22.53281c(bstrap$t0 - 2 * 0.4119, bstrap$t0 + 2 * 0.4119)## [1] 21.70901 23.35661medv.med = median(medv)
medv.med## [1] 21.2boot.fn = function(data, index) return(median(data[index]))
boot(medv, boot.fn, 1000)##
## ORDINARY NONPARAMETRIC BOOTSTRAP
##
##
## Call:
## boot(data = medv, statistic = boot.fn, R = 1000)
##
##
## Bootstrap Statistics :
## original bias std. error
## t1* 21.2 -0.01405 0.3709416medv.tenth = quantile(medv, c(0.1))
medv.tenth## 10%
## 12.75boot.fn = function(data, index) return(quantile(data[index], c(0.1)))
boot(medv, boot.fn, 1000)##
## ORDINARY NONPARAMETRIC BOOTSTRAP
##
##
## Call:
## boot(data = medv, statistic = boot.fn, R = 1000)
##
##
## Bootstrap Statistics :
## original bias std. error
## t1* 12.75 0.0049 0.5255637