Stat 431 Solution to Assignment 1 Winter 2020
Question 1 [10 marks]
Suppose that Y is a random variable from the exponential distribtuion with rate
... [Show More] parameter λ > 0 and
probability density function:
f(y; λ) = λe−λy
(a) [2 marks] Show that the distribution of Y is a member of the exponential family by identifying the
canonical parameter, the dispersion parameter, and the functions a(φ), b(θ), c(y; φ).
Recall a distribtuion is a member of the exponential family if its pdf/pmf can be written in the
form:
f(y; θ, φ) = exp
yθ − b(θ)
a(φ)
+ c(y; φ)
For the exponential distribtion we can write
f(y; λ) = λe−λy = exp {−(yλ − log λ)}
Therefore the exponential distribtuion is a member of the expoential family with
θ = λ b(θ) = log λ = log θ
φ = 1 a(φ) = −1 c(y; φ) = 0
Note: The above is not unique. Students could also set θ = −λ and proceed from there.
(b) [2 marks] Obtain an expression for the mean and variance of Y and identify the canonical link.
The mean and variance of Y are given by:
E[Y ] = b
0
(θ) = 1/θ = λ
−1
V ar[Y ] = b“(θ)a(φ) = −1/θ2
(−1) = λ
−2
To find the canonical link we set g(µ) = θ = η = x
T β. Here:
µ = 1/θ therefore g(µ) = 1/µ
This is the reciprocal or inverse link.
(c) [3 marks] Suppose Yi
, i = 1, . . . , n are iid and for each Yi there is vector of explanatory variables
xi = (1, xi1, . . . , xi,p−1)
0
. Consider the linear predictor ηi = x
0
iβ and the canonical link found in (b).
Find the specific form of the score vector and information matrix for β.
To find the Score and Information we can either substitute into the Likelihood using the parameter
relationship defined by the canonical link: x
0β = η = θ = 1/µ = λ i.e. λ = x
0β or we can use the
general result discussed in class (course notes pages 6-8). Since we have a random sample we can
omit the subscript i and find the contributions from a single observation.
Using direct subsitution λ = x
0β
`(λ) = −yλ + log λ
`(β) = −yx
0β + log x
0β
Sj (β) = ∂`
∂βj
= −yxj +
xj
x0β
Ijk(β) = −
∂
2
`
∂βj∂βk
=
xjxk
(x0β)
2 [Show Less]