STAT310: Introduction to Probability and Mathematical Statistics II 1
Homework 5: Fisher Information and more Maximum Likelihood
Solutions
Problem 1 (The Pareto distribution). The Pareto distribution is often a good model for
the distribution of wealth or other resources in a society. It has a density
(
θxθ0 x−(θ+1) if x ≥ x0
f (x; x0 , θ) =
0 otherwise,
where x0 > 0 and θ > 1.
(a) Show that f (x; x0 , θ) indeed defines a probability density. That is, show that it is
non-negative and integrates to 1.
(b) Derive a method of moments estimate for θ under the assumption that x0 is known.
(c) Derive the MLE for θ under the assumption that x0 is known.
(d) Find the Fisher information for θ under the assumption that x0 is known. You may
assume that the density of the Pareto distribution obeys the “suitable regularity
conditions” discussed in lecture.
(e) Use part (d) to construct an approximate 100(1 − α)% confidence interval for θ.
Solution:
(a) It is simple to check that f (x; x0 , θ) ≥ 0, so we just need to check that it integrates
to 1. Z ∞ Z ∞
θ −(θ+1)
θx0 x dx = θx0θ
x−(θ+1) dx
x0 x
0 −θ ∞
−x
= θxθ0
θ x0
x−θ
0
= θxθ0 = 1.
θ
(b) We have that EX1 = x0 θ/(θ − 1). Rearranging, we find Letting M = EX1 /x0 and
rearranging, we find
µ1 /x0 µ1
θ= = ,
µ1 /x0 − 1 µ 1 − x0
from which we have
X̄
θ̂MoM = .
X̄ − x0
Homework 5: Fisher Information and more Maximum Likelihood
Solutions
Problem 1 (The Pareto distribution). The Pareto distribution is often a good model for
the distribution of wealth or other resources in a society. It has a density
(
θxθ0 x−(θ+1) if x ≥ x0
f (x; x0 , θ) =
0 otherwise,
where x0 > 0 and θ > 1.
(a) Show that f (x; x0 , θ) indeed defines a probability density. That is, show that it is
non-negative and integrates to 1.
(b) Derive a method of moments estimate for θ under the assumption that x0 is known.
(c) Derive the MLE for θ under the assumption that x0 is known.
(d) Find the Fisher information for θ under the assumption that x0 is known. You may
assume that the density of the Pareto distribution obeys the “suitable regularity
conditions” discussed in lecture.
(e) Use part (d) to construct an approximate 100(1 − α)% confidence interval for θ.
Solution:
(a) It is simple to check that f (x; x0 , θ) ≥ 0, so we just need to check that it integrates
to 1. Z ∞ Z ∞
θ −(θ+1)
θx0 x dx = θx0θ
x−(θ+1) dx
x0 x
0 −θ ∞
−x
= θxθ0
θ x0
x−θ
0
= θxθ0 = 1.
θ
(b) We have that EX1 = x0 θ/(θ − 1). Rearranging, we find Letting M = EX1 /x0 and
rearranging, we find
µ1 /x0 µ1
θ= = ,
µ1 /x0 − 1 µ 1 − x0
from which we have
X̄
θ̂MoM = .
X̄ − x0
