STAT 4101 Quiz 5, Spring 2021
First name: Last name:
Hawkid: @uiowa.edu
Please read the following instructions before you start the exam.
• There are three pages including the cover page.
• The quiz is available from Apr 23rd, 2021 noon and due on Apr 26th, 2021 noon.
• There are 2 problems. The total is 10 ∗ 2 = 20 points.
• Show all work for each problem below. If you only present a correct answer
without stating how you get it, no credit will be given.
• Calculator ready solutions are not sufficient to obtain full credits. A numer- ical answer is needed if a problem asks for it.
• You must stop writing immediately after the invigilator signals the end of the exam.
• Any distribution of this material is prohibited.
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1. (10pts) Let X have the following pdf,
f(x) = 1e−2|x|/θ, −∞ < x < ∞.
θ
24 ∞ 1 −2|x|/θ 0 1 2x/θ
(a) Find the Fisher information I(θ).
Solution: We first note that E|X| = θ and Var|X| = θ2 :
E|X|= 2
∞1 −2x/θ dx+ θxe
θ dx= 2.
θ2 dx= 2,
θ|x|e dx= θ(−x)e
−∞ −∞ 0
E(|X|) =
which gives Var(|X|) = θ2 . We then see
∞ 1 2 −2|x|/θ 0 1 2 2x/θ ∞ 1 2 −2x/θ θxe dx= θxe dx+ θxe
and therefore
giving I(θ) = 1 . θ2
−∞ −∞ 0 4
λ(x; θ) = − log θ − 2|X|/θ and ∂λ(x; θ) = −1 + 2|X|, ∂θ θ θ2
∂2λ(x;θ) = 1 − 4|X|, ∂θ2 θ2 θ3
(b) Let X1,...,Xn be a random sample from this distribution. Show that the maximum likelihood esti- mator is an efficient estimator of θ.
Solution:
It is easy to check that θˆn = 2 ni=1 |Xi|/n is the mle.
ˆ θ2 1
Therefore the mle is an efficient estimator of θ.
(c) Find the asymptotic distribution of θˆ.
Var(θn) = n = nI(θ).
Solution:
√ˆ 1 2 n(θ−θ)∼N 0,I(θ) ,whichisN0,θ .
(d) Find an approximate 95% CI for θ. Solution:
gives
θˆ − θ D ˆ P θ/√n → N(0, 1) and θ → θ,
θˆ−θ →D N(0,1), θˆ/√n
by Slutsky’s theorem. Thus the approximate 95% CI is ˆ θˆ
θ ± z0.025 √n . 2
2. (10pts) Let X1, . . . , Xn be the lifetime of n electronic devices of the same kind. Suppose the sample follows an exponential distribution with mean θ > 0. Suppose X ̄ = 1.2 and n = 20.
Findp-valuesofthetestH0 :θ=2againstH1 :θ̸=2,using (a) −2logΛ.
Solution: We have
and the likelihood function is
We then find the mle:
θ
f(x;θ)= 1e−x, θ>0, x>0,
θ
1 ni=1xi L(θ;x1,…,xn) = θn exp − θ
ni=1 xi l(θ;x1,…,xn) = −nlogθ − θ
′ n ni=1xi l (θ;x1,…,xn) = −θ + θ2 .
ˆ ni=1xi θ=n.
n
i=1
.
,
We have
Thus
ni=1 xi 2
l(2) = −n log 2 −
l(θˆ) = n log n − n log xi − n.
,
which gives a p-value of 0.035. (b) A Wald-type statistic.
Solution: Since
i=1 i=1
∂2logf 1 I(θ)=E − ∂θ2 =θ2,
nn −2logΛ=−2l(2)+2l(θˆ)=2n(log2n−1)+xi −2nlogxi =4.433,
we have
which gives a p-value of 0.003. (c) Rao’s score statistic.
Solution: We have
2 χR =
which gives a p-value of 0.074.
2 2 n 2 2n 2 Wn
χ = nI(θ) θ − 2 = (θ − 2) = n 1 −
= 8.89,
l′ (2) nI (2)
2
−n/2 + n i=1
xi/42
θ2
i=1 xi
= 3.2,
= n/4
End of the exam!
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