only question 3 and 4
STAT 346/446: Theoretical Statistics II Extra Problems – last updated January 14, 2019 1. Let X1, X2, . . . , Xn be a random sample from N(µ1, σ 2) and Y1, Y2, . . . , Ym be a ran- dom sample from N(µ2, σ 2). Also assume that the random vectors (X1, X2, . . . , Xn) and (Y1, Y2, . . . , Ym) are mutually independent. Notice that the two populations have the same variance, but different means. (a) Let S2 = ∑n i=1(Xi −X)2 + ∑m i=1(Yi − Y )2 n+m− 2 Show that (n+m− 2)S2 σ2 ∼ χ2n+m−2 and E(S2) = σ2 (b) Let T = (X − Y )− (µ1 − µ2) S √ 1/n+ 1/m where S = √ S2 from part (a). Show that T ∼ tn+m−2 2. Let Y ∼ Binomial(n, p) where p does not depend on n. Without using the WLLN (but you can use e.g. Chebychev’s inequality) show that (a) Y/n P−→ p as n→∞ (b) (1− Y/n) P−→ (1− p) as n→∞ 3. Coding required. Use Monte Carlo methods to calculate the following probabilities (a) P (X > 320) where X ∼ Poisson(300) (b) P (1/X > 5) where X ∼ Gamma(2, 4) 4. Coding required. The Central Limit Theorem holds for most distributions but the approximation may be faster to kick in for some distributions than others. Here you will investigate how well CLT holds for various population distributions and the samples sizes n = 2, 5, 10, 30, 100. • For each case produce a Figure with 6 graphs in a 2 by 3 matrix: – one of the pdf for original population distribution – For each value of n show a histogram of the sample means and the pdf of the corresponding approximate normal distribution Example code for Expo(10) is shown on the next page. Cases: 1 (a) Uniform(0, 1) (b) Beta(0.5, 0.5) (c) Beta(10, 2) (d) Expo(2) (e) Gamma(2, 4) Do you notice what features of the population distribution tend to require larger n for the CLT to be a reasonable approximation? Example R code for population density Expo(10) set.seed(7443) nval <- c(2,5,10,30,100)="" n.sim=""><- 100000="" b=""><- 10="" xn=""><- matrix(na,="" nrow="N.sim," ncol="length(nval))" for(k="" in="" 1:length(nval)){="" for(i="" in="" 1:n.sim){="" xn[i,k]=""><- mean(rexp(n = nval[k], rate = 1/b)) } } # takes a couple of seconds par(mfrow=c(2,3), mar=c(3.5,3.5,2,1), mgp=c(2,0.8,0)) curve(dexp(x, rate = 1/b), from = 0, to = 60, main="population density") for(k in 1:length(nval)){ hist(xn[,k], freq = f, main = paste("n =", nval[k])) curve(dnorm(x, mean = b, sd = b/sqrt(nval[k])), add = t, from = min(xn[,k]), to = max(xn[,k]), col=’blue’) } 2 mean(rexp(n="nval[k]," rate="1/b))" }="" }="" #="" takes="" a="" couple="" of="" seconds="" par(mfrow="c(2,3)," mar="c(3.5,3.5,2,1)," mgp="c(2,0.8,0))" curve(dexp(x,="" rate="1/b)," from="0," to="60," main="Population density" )="" for(k="" in="" 1:length(nval)){="" hist(xn[,k],="" freq="F," main="paste("n" =",="" nval[k]))="" curve(dnorm(x,="" mean="b," sd="b/sqrt(nval[k]))," add="T," from="min(xn[,k])," to="max(xn[,k])," col="’blue’)" }="">