1. Consider a two sample problem with (X1,X2, . . . ,Xm) i.i.d. N(p,02) and (Y1, Y2, . . . ,Yn) i.i.d. N(p,, T2). The sample of Xi's is also taken to be inde pendent of the sample of lg's. For now, we assume that 0' and 'r are known and p is the only unknown parameter of interest. Let Mi, Z, pa) denote the joint density of the sample (i, K) under parame ter value a. Letting fats, n) denote the N (a, 0'2) density and fm, p.) denote the N ((1., 7'2) density, we have: p(3:1#) = H31f0(X1-3 I\") . H?=lfT(}/.}ip) ' Set (,Z, p) := log pIX'XLM)' and denote its rst and second derivatives with respect to ,LL by E and I? respectively. The information about u in the sample is given by: no Im,n(#) = [laLll - (i) Show that m n Immm) = g + 13 (ii) Consider the estimator am? + n l7 where am = m/(n + m) and n = n/(m+n). Show that this is unbiased but is not the best unbiased estimator of p in general. Under what relation between 0' and 1' is it the best unbiased estimator? (iii) Consider all estimators of the form Tm := w? | (1 m)? where 0 g 11) g 1. Find the best unbiased estimator in this class and show that this is also the best unbiased estimator by showing that its variance equals the information bound for M. (iv) The best unbiased estimator obtained above should have the form: T2 Y | J2/m T2+J2/m T2+02/m Tbest 5: 3; - Now suppose that 0'2 and 7'2 are unknown. We will construct a new estimate of Tbest by imputing estimates of T2 and 02 into the above estimator. So, consider the estimates: .2 ,2 23%- Y)2 J -2 2:09 ?)2 ml and T := \"1 Denote by The\" the estimator obtained by plugging in the above estimates of the variances into the expression for Tbsst. Show.r that E [Tbgst] Hint: From the developments in 426 notes8. pdf it follows that the random vector X Y,o \"2 $2 are all mutually independent Their distributions can also be determined in terms of normal and X2 distributions. Use this. (v) Show that Varest) can be expressed explicity in terms of m,n,cr,'r and the moments of a random variable menyg assuming values between 0 and 1 which can be expressed as a 11 transformation of a random variable following the me distribution