Study Guides (238,465)
STA261H1 (2)

# pracmidsol.pdf

5 Pages
189 Views

School
University of Toronto St. George
Department
Statistical Sciences
Course
STA261H1
Professor
Jonathan Lee
Semester
Winter

Description
1. a) The MLE isYR (n)(see solution for problem 8 in homework 4) ▯3y3 3▯ ▯ ^ 4Y▯ MME: E(Y ) = 0 ▯3dy = 4 = Y , so ▯ = 3 n▯1 y3 n▯1 3y2 3ny3n▯1 b) The pdf of Y (n)is fY(n)y) = nF(y) f(y) = n( ▯3) ▯3 = ▯3n , for 0 < y ▯ ▯ R ▯3ny3n▯1 3n R ▯ 3n 3n▯ E(Y ()) = 0 ▯3n ydy = ▯3n 0y dy = 3n+1 (3n+1)(n) Hence, the MLE is not unbiased but is an unbiased estimator of ▯ 3n n n▯1 2. a) The ▯rst factor should be n▯1 not n , as some of you noticed, to get an unbiased estimator. The estimator I wrote is not unbiased but it’s asymptotically unbiased. Below is the proof with the corrected factor in front. Obviously the other 2 estimators are unbiased 2 b) You can notice that ▯ i1 in fact S , the sample variance and we proved in 2 class that S is consistent for the variance if the 1st, 2nd and 4th moments are ▯nite which turns out to be true for Poisson(▯). The proof consists in using the LLN twice. (even the estimator with n▯1 in front is consistent) n Consistent estimators are about large sample (as n goes to in▯nity) but the second estimator is the sample mean of only two observations, so it doesn’t really make sense to talk about consistency for this estimator. ▯ is consis- n!1 3 tent since var(▯ 3 = ▯ ! 0 n ^ ^ MSE(▯ 3 var(▯3) ▯=n 2 c. eff(▯ 2▯ )3= MSE(▯ 2 = var(▯2)= ▯=2 = n since they are unbiased 1 3. a) Z E [X] = xfX(x;▯) dx ZX1 ▯ ▯ ▯ ▯ 1 x = x ▯ + 1 exp ▯ ▯ + 1 dx ▯0 ▯ ▯▯ 1 Z ▯ ▯ x 1 x = ▯xexp ▯ + exp ▯ dx ▯ ▯ + 1▯ 0 0▯▯ ▯ + 1 x 1 = 0 + ▯(▯ + 1)exp ▯ ▯ + 1 0 = ▯ + 1 " # ▯ ▯ 1 X E X = E X i n i=1 n 1 X = E [i ] n i=1 Xn = 1 (▯ + 1) n i=1 n(▯ + 1) = n = ▯ + 1 ▯ ▯ Bias X = 1 b) Consider the following estimator T (X) = X ▯ 1. ▯ ▯ E [T (X)] = E X ▯ 1 ▯ ▯ = E X ▯ 1 = ▯ + 1 ▯ 1 = ▯ 2 c) Z ▯ 2▯ 2 E X = x fX(x;▯) dx ZX1 ▯ ▯ ▯ ▯ 2 1 x = x exp ▯ dx ▯0 ▯▯+ 1 ▯▯ ▯ Z 1 ▯ ▯ x 1 1 x = ▯x exp ▯ ▯ ▯2xexp ▯ dx ▯ + 1 0 0 ▯ + 1 Z 1 ▯ ▯ = 2xexp ▯ x dx 0 ▯ + 1 ▯ ▯ ▯▯1 Z 1 ▯ ▯ x x = ▯2x(▯ + 1)exp ▯▯ + 1 ▯ ▯2(▯ + 1)exp ▯ ▯ + 1 dx Z ▯ ▯ 0 0 1 x = 2(▯ + 1)exp ▯ dx 0
More Less

Related notes for STA261H1

OR

Don't have an account?

Join OneClass

Access over 10 million pages of study
documents for 1.3 million courses.

Join to view

OR

By registering, I agree to the Terms and Privacy Policies
Already have an account?
Just a few more details

So we can recommend you notes for your school.