6. [R] Estimation of the parameter for exponential distribution 1 (Jan12,14,17)

6. [R] Estimation of the parameter for exponential distribution 1 (Jan12,14,17)

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# Estimation of the parameter for exponential distribution # E(X) = 1/lam lam = .5 n = 40 x = rexp(n, lam) # 2 of many possible estimators lam.hat.1 = 1/mean(x) lam.hat.2 = 1/sqrt(var(x)) lam.hat.1 lam.hat.2 # Which is better? What is their sampling distribution? # sampling distribution # numerical aproximation to obtain this based on Monte Carlo simulation experiment M = 1000 lh.1.rep = rep(0, M) lh.2.rep = rep(0, M) lambda = lam for(i.rep in 1:M){ x.sim = rexp(n, lambda) lh.1.rep[i.rep] = 1/mean(x.sim) lh.2.rep[i.rep] =1/sqrt(var(x.sim)) } # the simulated pairs (lam.hat.1 , lam.hat.2) allows us to obtain approximations to various properties # of the sampling distribution of (lam.hat.1 , lam.hat.2) # Below we explore some of these in order to have some understanding of which simple summary measures # are useful in order to compare these two estimators. # scatter plot of lam.hat.1 versus lam.hat.2 # sometimes lam.hat.1 < lam and lam.hat.2 > lam plot( lh.1.rep, lh.2.rep) # based on this simulated values of lam.hat.1 and lam.hat.2 we can approximate many properties of # the r.v. (lam.hat.1 , lam.hat.2)
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