Student Notes To Accompany MS4214: STATISTICAL INFERENCE

Outline Solutions
1. The cdf is F (t) = P (T ≤ t) = � t
0 λe−λυ dυ = λ � − 1
λ e−λυ� t
0 = 1 − e−λυ . Next
P (a ≤ T ≤ b) = F (b) − F (a) = e −λa − e −λb . Assume all settlements are inde-
pendent. Then P (50 in first week) = {F (1)} 50 = (1 − e −λ ) 50 , because T ≤ 1
for these 50 settlements. Likewise, 1 < T ≤ 2, for the 35 in the second week,
so we have P (35 in second week) = {F (2) − F (1)} 35 = (e −λ − e −2λ ) 35 . The re-
maining 15 have T > 2, which has probability 1 − P (T ≤ 2) = e −2λ , and thus
P (15 after week two) = (e −2λ ) 15 . The likelihood function is therefore the product
L(λ) = (1 − e −λ ) 50 (e −λ − e −2λ ) 35 (e −2λ ) 15 . Taking logarithms (always based e),
∴ d
dλ
ln L(λ) = 50 ln(1 − e −λ ) + 35 ln � e −λ (1 − e −λ ) � + 15 ln(e −2λ )
= 85 ln(1 − e −λ ) − (35 + 30)λ = 85 ln(1 − e −λ ) − 65λ.
85e−λ
85
ln L(λ) = − 65 =
1 − e−λ e−λ − 65.
− 1
Equating to zero, 85 = 65(e −λ − 1) or e λ = 150/65, so that ˆ λ = ln(150/65) =
0.836. This is indeed a maximum; e.g. d2
dλ 2 ln L(λ) = −85/(e λ − 1) 2 < 0 ∀ λ. Next
1 − e −0.836 = 0.5666; e −0.836 − e −1.672 = 0.43344 − 0.18787 = 0.2456. Hence out of
100 invoices, 56.66, 24.56 and 18.78 would be expected to be paid, on this model,
in weeks 1, 2 and later. The actual numbers were 50, 35 and 15. The prediction
for the second week is a long way from what happened, balanced by smaller
discrepancies in the other two periods. This does not seem very satisfactory.
2. nA + nB = n ⇒ nB = n − nA. Suppose we observe the sequence A, B, A, B, A,
then L1(nA) = nA
n
× n−nA
n−1
× nA−1
n−2
× n−nA−1
n−3
sequence A, A, B, B, A, then L2(nA) = nA
n
nA−2
× . Next, suppose we observe the
n−4
× nA−1
n−1
× n−nA
n−2
× n−nA−1
n−3
nA−2
× n−4
. If it
is known that 3As and 2Bs are drawn but the exact sequence is unknown then
L3(nA) = P (Y = y|nA) = � �� � � � nA n−nA n
/ , where y = 3. This third likelihood
y 5−y 5
function expands to give L3(nA) = 10 × nA(nA−1)(nA−2)(n−nA)(n−nA−1)
. Clearly
n(n−1)(n−2)(n−3)(n−4)
L1(nA) = L2(nA) = L3(nA) ÷ 10. The first two likelihood functions are identical.
The third likelihood function is a constant times the other two, and as only the
ratio of likelihood functions are meaningful, L3(nA) carries the same information
about our preferences for the parameter nA as the other functions.
3. L(θ) = 4 −n (2 + θ) a (1 − θ) b+c (θ) d so, ℓ(θ) = −n ln(4)+a ln(2+θ)+(b+c) ln(1−
θ) + d ln(θ). Differentiating we get S(θ) = a b+c d − + and setting S(θ) = 0
2+θ 1−θ θ
leads to the quadratic equation nθ2 − {a − 2b − 2c − d}θ − 2d = 0, of which
the positive root, ˆ θ, satisfies the condition of maximum likelihood. If S(θ) is
differentiated again with respect to θ, and expected values substituted for a, b, c,
and d, we obtain Var( ˆ θ) ≈ {E[I(θ)]} −1 = {I(θ)} −1 = 2θ(1−θ)(2+θ)
(1+2θ)n
45
.