Suppose that X1, X2, ..., Xn are i.i.d. N(mu, sigma2), where mu and sigma2 are e
ID: 3177477 • Letter: S
Question
Suppose that X1, X2, ..., Xn are i.i.d. N(mu, sigma2), where mu and sigma2 are estimated by the method of maximum likelihood, with resulting estimates mu and sigma2. Suppose the bootstrap is used to estimate the sampling distribution of mu. b. Explain why the bootstrap estimate of the distribution of mu - mu_0 is N (0, siagm^2/n). c. According to the result of the previous part, what is the form of the bootstrap confidence interval for mu, and how does it compare to the exact confidence interval based on the t distribution?Explanation / Answer
Given that X1,X2,...Xn are idd normaldistribution, aand 2 are estimated bythe method of maximum likelihood, with resulting estimates and2 . suppose the bootstrap is used toestimate the sampling dstribution of .
a)
According to the Central limit theorem , the sampling distributionof the sample mean is approximately normal with mean and variance 2 /n.
Given that bootstrap is used toestimate the sampling dstribution of .
In the bootstrap techinque, we take a large number of randomsamples of size 'm' using with replacement techinque from the random sample of size 'n' , where m <= n.
a)
For the large number of random samples of size 'm' , we calculatethe sample mean for each and every sample.
Therefore according to the Central limit theorem, thebootstrap estimate of the distribution of is N(,2 /n).
b)
For the large number ofrandom samples of size 'm' , we calculate the sample mean for eachand every sample.
Therefore according to the Central limit theorem, thebootstrap estimate of the distributionof 0 is N(,2 /n).
The bootstrap estimate of thedistribution of -0 is
E( -0) = - = 0 and
Var(-0 ) = Var() + Var(0) - 2Cov( , 0)
= 0 + 2 /n-2*0
= 2 /n
Therefore the bootstrap estimateof the distribution of -0 isN(0,2 /n).
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