SOLUTION FOR HOMEWORK 3, STAT 4352 Welcome to your third ...

Then
6. Problem 10.66. Let X1, . . .,Xn be iid according to the pdf
gθ(x) = θ −1 e −(x−δ)/θ I(x > δ).
LX n(δ, θ) = θ−n e − n
l=1 (Xl−δ)/θ I(X(1) > δ).
Recall that X(1) = min(X1, . . .,Xn) is the minimal observation [the first ordered observation].
This is the case that I wrote you about earlier: it is absolutely crucial to take into account
the indicator function (the support) because here the parameter δ defines the support.
By its definition,
Note that
( ˆ δMLE, ˆ θMLE) := arg max ln(LXn(δ, θ)).
δ∈(−∞,∞),θ∈(0,∞)
L(δ, θ) := ln(LXn(δ, θ)) = −n ln(θ) − θ−1 n
(Xl − δ) + ln I(X(1) ≥ δ).
Now the crucial step: you should graph the loglikelihood L as a function in δ and visualize
that it takes on maximum when δ = X(1). So we get ˆ δMLE = X(1). Then by taking a
derivative we get that ˆ θMLE = n −1 n l=1 (Xl − X(1)).
Answer: ( ˆ δMLE ˆ θMLE) = (X(1), n −1 n l=1 (Xl − X(1)). Please note that ˆ δMLE is a biased
estimator; this is a rather typical outcome.
7. Problem 10.73. Consider iid uniform observations X1, . . .,Xn with the parametric pdf
l=1
fθ(x) = I(θ − 1/2 < x < θ + 1/2).
As soon as the parameter is in the indicator function you should be very cautious: typically
a graphic will help you to find a MLE estimator, and not a differentiation. Also, it is very
helpful to figure out the nature of the parameter. Here it is obviously a location parameter,
and you can write
X = θ + Z, Z ∼ Uniform(−1/2, 1/2).
The latter helps you to guess about a correct estimator and check a suggested one and, if
necessary, simplify calculations of descriptive characteristics (mean, variance, etc.)
Well, now we need to write down the likelihood function (recall that this is just a joint
density only considered as a function in the parameter given a vector of observations):
LXn(θ) =
n
I(θ − 1/2 < Xl < θ + 1/2) = I(θ − 1/2 < X(1) ≤ X(n) < θ + 1/2).
l=1
Note that the latter expression implies that (X(1), X(n)) is a sufficient statistic (due to the
Factorization Theorem). As a result, any good estimator, and the MLE in particular, must
be a function of only these two statistics. Another remark is: it is possible to show (there
exists a technique how to do this which is beyond this class objectives) that this pair of
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