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MATH5905 Assignment One Solutions Statistical Inference
MATH5905 - Statistical Inference
Assignment 1 Solutions
Problem One
A single die is tossed; then n coins are tossed where n is the number shown on the die. Show that the
probability of exactly two heads is close to 0.2578.
Solution: Let U1, U2, . . . , U6 denote the events “uppermost 1, 2, . . . , 6” respectively and A denote the event
in question. We have P (U1) = · · · = P (U6) = 16 . The formula of total probability gives
P (A) = P (A|U1) ∗ P (U1) + P (A|U2)P (U2) + . . . P (A|U6)P (U6) = 1
6
(P (A|U1) + . . . P (A|U6)).
Obviously P (A|U1) = 0 holds. For the remaining conditional probabilities, considering ‘Head’ as a success
and ’Tail’ as a failure, we need to get the probability of a number of two successes out of i independent
Bernoulli trials by calculating the ratio favourable outcomes
total number of outcomes
. The total number of outcomes is 2i and
the favourable ones are
(
i
2
)
, that is, P (A|Ui) = (
i
2)
2i , i = 2, 3, 4, 5, 6. Hence we get
P (A) =
1
6
(0 + 1/4 + 3/8 + 6/16 + 20/64 + 30/128) = 33/128 ≈ 0.2578.
Problem Two
A certain river floods every year. Suppose that the low-water mark is set at 1 and the high-water mark
X has a distribution function
FX(x) = P (X ≤ x) = 1− 1
x2
, 1 ≤ x <∞
1. Verify that FX(x) is a cumulative distribution function
2. Find the density fX(x) (specify it on the whole real axis)
3. If the (same) low-water mark is reset at 0 and we use a unit of measurement that is 110 of that used
previously, express the random variable Z for the new measurement as a function of X. Then find
precisely the cumulative distribution function and the density of Z.
Solution: a) Obviously limx→−∞ FX(x) = 0 (as FX(x) is a constant 0 for x ≤ 1). Also,
lim
x→∞FX(x) = limx→∞ 1−
1
x2
= 1.
For x > 1, ddxFX(x) = 2/x
3 > 0 which implies that FX(x) is increasing. So-for all x on the real axis, FX(x)
is non-decreasing.
b) The density as a derivative was calculated in a) already for x > 1 and it is zero for x ≤ 1.
c) FZ(z) = P (Z ≤ z) = P (10(X − 1) ≤ z) = P (X ≤ (z/10) + 1) = FX((z/10) + 1). Hence
FZ(z) = 1− ( 1
[z/10 + 1]2
)
when z > 0 (and zero else).
The density is obtained via differentiation of the cdf leading to
fZ(z) =
1
5(z/10 + 1)3
, z > 0
(and zero else).
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MATH5905, T1 2023 Assignment One Solutions Statistical Inference
Problem 3
Suppose X = (X1, . . . , Xn) are i.i.d. Poisson(θ) with a probability mass function
f(x, θ) =
e−θθx
x!
, x ∈ {0, 1, 2, . . . }, θ > 0
The prior on the unknown parameter θ is assumed to be Gamma(α, β) distribution with density
τ(θ) =
1
Γ(α)βα
θα−1e−θ/β , α, β > 0, θ > 0.
a) Find the posterior distribution for θ.
b) Hence or otherwise determine the Bayes estimator of θ with respect to the quadratic loss function
L(a, θ) = (a− θ)2.
c) Suppose the following ten observations were observed:
4, 0, 1, 0, 0, 2, 3, 1, 1, 0.
Using a zero-one loss with the parameters α = 2 and β = 1 for the prior, what is your decision
when testing H0 : θ ≤ 1.2 versus H1 : θ > 1.2. (You may use the integrate function in R or
another numerical integration routine from your favourite programming package to answer the
question.)
Solution:
a) We use the fact that the posterior is proportional to the prior and the likelihood:
p(θ|X) ∝ L(X|θ)τ(θ)
=
e−nθθ
∑n
i=1Xi∏n
i=1 xi!
1
Γ(α)βα
θα−1e−θ/β
∝ θα+
∑n
i=1Xi−1e−θ(n+
1
β ).
Hence we recognise this as a gamma density with parameters
α˜ = α+
n∑
i=1
Xi and β˜ =
1
n+ 1β
=
β
nβ + 1
b) From lectures, the Bayes estimator with respect to quadratic loss is the posterior mean given the
sample.
θˆbayes = E(θ|X) = α˜β˜ = (α+
n∑
i=1
Xi)(
β
nβ + 1
).
c) The structure of the test is as follows:
φ =
{
1 if P (θ < 1.2|X) < 0.5
0 if P (θ < 1.2|X) ≥ 0.5
We have n = 10,
∑10
i=1 xi = 12 so the posterior distribution given the sample is:
θ|X ∼ Gamma
(
2 + 12,
1
10× 1 + 1
)
= Gamma(14, 1/11)
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MATH5905, T1 2023 Assignment One Solutions Statistical Inference
Then using R we can compute the posterior probability under these conditions as
P (θ < 1.2|X) =
∫ 1.2
0
1
Γ(14)(1/11)14
θ13e−11θdθ ≈ 0.44893
Hence, since the posterior probability is smaller than 0.50 we reject H0.
Problem Four
The Premier of NSW has to take an important decision whether or not to include an additional line
(extension) to Sydney’s metro system. The decision is based on the financial viability of the extension.
He is willing to apply decision theory in making his decision. He uses the independent opinion of two
consulting experts. The data he uses is the number X of viability recommendations of the two experts
(so X = 0, 1 or 2).
If the Premier decides not to go ahead and the extension turns out not to be financially viable or
if he decides to go ahead and the project is financially viable, nothing is lost. If the project is not
financially viable and he has decided to go ahead, his subjective judgement is that his loss would equal
three times the loss of not going ahead but the project is financially viable.
The Premier has investigated the history of viability predictions of the two consulting experts and it is
as follows. When a project is financially viable, both experts have correctly predicted its viability with
probability 4/5 (and wrongly with a probability 1/5). When a project has not been financially viable,
both experts had a prediction of 3/5 for it to be viable. The Premier listens to the recommendations
of the two experts and makes his decision based on the value of X.
a) There are two possible actions in the action space A = {a0, a1} where action a0 is to go ahead
and action a1 is not to go ahead with the extension. There are two states of nature Θ = {θ0, θ1}
where θ0 = 0 represents “Extension is financially viable” and θ1 = 1 represents “Extension is not
financially viable”. Define the appropriate loss function L(θ, a) for this problem.
b) Compute the probability mass function (pmf) for X under both states of nature.
c) The complete list of all the non-randomized decisions rules D based on x is given by:
d1 d2 d3 d4 d5 d6 d7 d8
x = 0 a0 a1 a0 a1 a0 a1 a0 a1
x = 1 a0 a0 a1 a1 a0 a0 a1 a1
x = 2 a0 a0 a0 a0 a1 a1 a1 a1
For the set of non-randomized decision rules D compute the corresponding risk points.
d) Find the minimax rule(s) among the non-randomized rules in D.
e) Sketch the risk set of all randomized rules D generated by the set of rules in D. You might want
to use R (or your favorite programming language) to make this sketch more precise.
f) Suppose there are two decisions rules d and d′. The decision d strictly dominates d′ ifR(θ, d) ≤ R(θ, d′)
for all values of θ and R(θ, d) < (θ, d′) for at least one value θ. Hence, given a choice between d
and d′ we would always prefer to use d. Any decision rules which is strictly dominated by another
decisions rule (as d′ is in the above) is said to be inadmissible. Correspondingly, if a decision rule
d is not strictly dominated by any other decision rule then it is admissible. Show on the risk plot
the set of randomized decisions rules that correspond to the admissible decision rules.
g) Find the risk point of the minimax rule in the set of randomized decision rules D and determine
its minimax risk. Compare the two minimax risks of the minimax decision rule in D and in D.
Comment.
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MATH5905, T1 2023 Assignment One Solutions Statistical Inference
h) Define the minimax rule in the set D in terms of rules in D.
i) For which prior on {θ1, θ2} is the minimax rule in the set D also a Bayes rule?
j) Prior to listening to the two experts, the Premier’s belief in the viability is 50%. Find the Bayes
rule and the Bayes risk with respect to his prior.
k) For a small positive ϵ = 0.1, illustrate on the risk set the risk points of all rules which are ϵ-
minimax.
Solution: a) There are two actions: a0 : decide to go ahead and a1 : do not go ahead. Two states of
nature, θ0 = 0, ”Extension is financially viable”, and θ1 = 1 ”Extension is not financially viable”. Let x be
the number of experts predicting financial viability. The loss function is given by
L(θ1, a0) = 3, L(θ1, a1) = 0, L(θ0, a0) = 0, L(θ0, a1) = 1.
