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STA261H1
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University of Toronto St. George

Statistical Sciences

STA261H1

Jonathan Lee

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

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