(a) Let moment generating function of random variable X exist in the neighbourhood of zero and if E(Xⁿ) = 1/5 + (-1)ⁿ 2/5 +…

(a)(i) Recovering the Distribution; P(|X − 0.75| ≤ 1.5)

Let M(t)=E(e^(tX)). Since the moment generating function exists in a neighbourhood of 0,

M(t)=1+Σ (n=1 to ∞) E(Xⁿ)tⁿ/n!.

Substituting the given moments,

M(t)=1+Σ (n=1 to ∞) [1/5+(−1)ⁿ 2/5+2^(n+1)/5]tⁿ/n!

=1+(1/5)(e^t−1)+(2/5)(e^(−t)−1)+(2/5)(e^(2t)−1)

=(1/5)e^t+(2/5)e^(−t)+(2/5)e^(2t).

Thus X has the distribution

P(X=−1)=2/5, P(X=1)=1/5, P(X=2)=2/5.

Now |X−0.75|≤1.5 means −0.75≤X≤2.25. Among the support values, X=1 and X=2 satisfy this, while X=−1 does not.

Therefore

P(|X−0.75|≤1.5)=P(X=1)+P(X=2)=1/5+2/5=3/5.

Final answer: 3/5.

(a)(ii) P(|X − μ| < σ)

From the distribution,

μ=E(X)=(−1)(2/5)+(1)(1/5)+(2)(2/5)=3/5=0.6.

Also,

E(X²)=(1)(2/5)+(1)(1/5)+(4)(2/5)=11/5.

Hence

σ²=Var(X)=E(X²)−μ²=11/5−(3/5)²=11/5−9/25=46/25=1.84.

So σ=√1.84=1.36, as given.

We need P(|X−μ|<σ)=P(|X−0.6|<1.36).

This is equivalent to

−1.36<X−0.6<1.36, i.e. −0.76<X<1.96.

Among X=−1,1,2, only X=1 lies in this interval.

Therefore

P(|X−μ|<σ)=P(X=1)=1/5.

Final answer: 1/5.

(b)(i) Importance of the Cramér–Rao Inequality and the Rao–Blackwell Theorem

The Cramer-Rao inequality states that, under regularity conditions, for an unbiased estimator T of φ(θ),

Var(T)≥[φ′(θ)]²/I(θ),

where I(θ) is the Fisher information. Its importance is that it gives a lower bound on variance, measures efficiency, helps identify UMVUEs, and guides sample-size determination. An unbiased estimator attaining the bound is the minimum variance bound (MVB) estimator; the bound is attained only when the score ∂ln L/∂θ is a linear function of T, which happens in the one-parameter exponential family.

The Rao-Blackwell theorem states that if T is a sufficient statistic and U is an unbiased estimator with finite variance, then E(U|T) is also unbiased and Var(E(U|T))≤Var(U). Its importance is that it improves an estimator by conditioning on a sufficient statistic; combined with completeness and the Lehmann-Scheffé theorem, it yields UMVUEs.

(b)(ii) UMVUE of θ(1 − θ)

As written, the question says X∼B(1,θ), i.e. a single Bernoulli observation. Any estimator is T(X) with T(0)=a and T(1)=b. Then

E[T(X)]=a(1−θ)+bθ=a+(b−a)θ,

which is linear in θ, whereas θ(1−θ)=θ−θ² is quadratic. No choice of a and b makes them equal for all θ. Hence no unbiased estimator of θ(1−θ) exists, and therefore no UMVUE exists.

If the intended statement is a random sample X₁,X₂,…,Xₙ∼B(1,θ) with n≥2, let T=ΣXᵢ. Then T is complete sufficient and T∼Bin(n,θ). The estimator

h(T)=T(n−T)/(n(n−1))

is unbiased because E[T(n−T)]=n(n−1)θ(1−θ). By the Lehmann-Scheffé theorem, it is the UMVUE.

Final answer: For one observation, no UMVUE exists. For a sample of size n≥2, the UMVUE is T(n−T)/(n(n−1)), where T=ΣXᵢ.

(c) MLEs of α and β

The density is

f(x;α,β)=C e^(−β(x−α)), α≤x<∞, β>0.

Normalization gives

1=∫ from α to ∞ C e^(−β(x−α)) dx=C∫ from 0 to ∞ e^(−βu) du=C/β,

so C=β. Hence

f(x;α,β)=β e^(−β(x−α)), α≤x<∞.

For a random sample X₁,…,Xₙ, the likelihood is

L(α,β)=βⁿ e^(−βΣ(Xᵢ−α)),

provided α≤min Xᵢ; otherwise L=0.

Let X₍₁₎=min(X₁,…,Xₙ). For fixed β,

L=βⁿ e^(−βΣXᵢ+nβα).

Since β>0, this increases as α increases. The constraint is α≤X₍₁₎. Therefore the maximum occurs at

α̂=X₍₁₎.

Substituting α̂=X₍₁₎,

ℓ=ln L=n ln β−βΣ(Xᵢ−X₍₁₎).

Differentiate:

dℓ/dβ=n/β−Σ(Xᵢ−X₍₁₎).

Set equal to zero:

β̂=n/Σ(Xᵢ−X₍₁₎).

Also,

d²ℓ/dβ²=−n/β²<0,

so this is a maximum. This requires that not all observations are equal; if all observations are equal, the likelihood is unbounded in β.

Final answer: α̂=min Xᵢ, and β̂=n/Σ(Xᵢ−min Xᵢ)=1/(X̄−min Xᵢ), assuming not all observations are equal.