91Ó°ÊÓ

There are no assumptions for the distribution F, meaning the nonparametric bootstrap shall be used. The estimator of interest is the variance of the sample median.

To initialize the process, the sample median of the initial sample shall be computed. To find the bootstrap variance estimate, find the sample variances of the generated samples\({X^{*(i)}},i = 1,2, \ldots ,n\) from the distribution\(\hat F\). The bootstrap variance estimate is then the average of these values. The following code gives it in RStudio. Note that the variable of interest is Corn.

The sample median from the initial sample is equal to 24. The approximation of the variance, or the estimated variance, is 19.446. Note that this number will change every time you run the code. The number of samples used was \(n = 2000 = \nu .\)

For the bias to be between -0.005 and 0.005 from the actual variance, one should look into the following probability

\(P ( - 0.005 < \theta < 0.005) = 0.95\)as in the exercise, the probability shall be at least 0.95. Then, one would need the estimator's expected value and standard deviation. Instead, take that the sample size is large and use the simulated sample standard deviation as an estimate (note here one should use the sample standard deviation of the n generated sample medians) and set the mean to be equal to zero. Then,

\(\begin{aligned}{l}P ( - 0.005 < \theta < 0.005) = P\left( {\frac{{ - 0.005 - 0}}{{\sqrt {4.1/n} }} < \frac{{\theta - 0}}{{\sqrt {4.1/n} }} < \frac{{0.005 - 0}}{{\sqrt {4.1/n} }}} \right)\\ \\ = P\left( {\frac{{ - 0.005 - 0}}{{\sqrt {4.1/n} }} < Z < \frac{{0.005 - 0}}{{\sqrt {4.1/n} }}} \right) \\\\ = 0.95,\quad \end{aligned}\)

or equally, because P(Z<1.96) =0.975 (standard normal is symmetric), it is true that

\(\frac{{0.005}}{{\sqrt {4.1/n} }} = 1.96\)

which leads to

\(n = {\left( {\frac{{4.1}}{{0.005}} \times 1.96} \right)^2} = 2,988,027\)

Instead of using the initial n=2000, use \(n \ge 2,988,027\) it to obtain the desired result. By rerunning the code with a changed value of n, the standard error of the simulation will be close to 0.0035<0.005. The estimate of the variance is close to 20.53.

#The library for built-in bootstrap functions

library(boot)

#Read data

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Exercise6.txt",

# a column instead of a list

\(Corndata = matrix(unlist(data(1)))\)

# a way to find the sample median

\(OriginalSamplemedian = median\left( {CornData} \right)\)

# The number of bootstrap samples

n=2988027

xstarimedian = numeric (n)

Sample Bias = numeric (n)

\#generate sample \({x^ \wedge }i\) and the biases

for \((iin(1:n))\{ \)

#Generate the bootstrap sample with replacement

\(Xstar i = sample(CornData,Length(CornData)2 + 5,replace = T)\)

#Median of the bootstrap sample

\(Xstarimedian (i) = median(Xstari)\)}

#Average the differences

\(BootstrapEstimateVariance = var(XstariMedian)\)

#The second part

#The estimate of the standard deviation of the n sample medians

\(BootstrapEstimatesD = sd\left( {Xstarimedian} \right)\)

\(n = {( qnorm (0.975,0,1)* Bootstrap Estimate SD /0.005)^ \wedge }2\)

\(Bootstrap Estimate SD Median = sd (SampleBias)/sqrt (n)\)