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

This is the answer. But again, as an extra example, I can suggest a MME based on the
second moment. Indeed, Eλ(X 2 ) = V arλ(X) + (EλX) 2 = λ + λ 2 and this yields that
˜λMME + ˜ λ 2 n
MME = n−1 X
i=1
2 i .
Then you need to solve this equation to get the MME. Obviously it is a more complicated
estimator, but it is yet another MME.
3. Problem 10.56. Let X1, . . .,Xn be iid according to the pdf
gθ(x) = θ −1 e −(x−δ)/θ I(x > δ).
Please note that this is a location-exponential family because
X = δ + Z,
where Z is a classical exponential RV with f Z θ (z) = θ −1 e −z/θ I(z > 0). I can go either further
by saying that we are dealing with a location-scale family because
X = δ + θZ0,
where f Z0 (z) = e −z I(z > 0).
So now we know the meaning of parameters δ and θ: the former is the location (shift)
and the latter is the scale (multiplier).
Note that this understanding simplifies all calculations because you can easily figure out
(otherwise do calculations) that
Eδ,θ(X) = δ + θ, V arδ,θ(X) = θ 2 .
These two familiar results yield Eδ,θ(X 2 ) = θ 2 + (δ + θ) 2 , and we get the following system of
two equations to find the pair of MMEs:
ˆδ + ˆ θ = ¯ X,
2ˆ θ 2 + 2ˆ δˆ θ + ˆ δ 2 = n −1
n
Xi.
i=1
To solve this system, we square the both sides of the first equality and then subtract the
obtained equality from the second equality. We get a new system
ˆδ + ˆ θ = ¯ X,
ˆθ 2 = n −1
n
X
i=1
2 i − ¯ X 2 .
This, together with a simple algebra, yields the answer
ˆδMME = ¯ X − [n −1 n
i=1
X 2 i − ¯ X 2 ] 1/2 , θMME
ˆ = [n −1 n
X 2 i − ¯ X 2 ] 1/2 .
2
i=1