(a) Let Y₁, Y₂, Y₃, ... be independent and identical Poisson random variables with parameter 1. Use central limit theorem to establish
n! ≃ √(2π n)(n/e)ⁿ
for large value of positive integer n. (20 marks)

(b) Let X₁, X₂, ..., Xₙ be a random sample su

Question

(a) Let Y₁, Y₂, Y₃, ... be independent and identical Poisson random variables with parameter 1. Use central limit theorem to establish
n! ≃ √(2π n)(n/e)ⁿ
for large value of positive integer n. (20 marks)

(b) Let X₁, X₂, ..., Xₙ be a random sample such that log Xᵢ ~ N(θ, θ) distribution with θ > 0 unknown. Show that one of the solutions of the likelihood equation is the unique MLE of θ. Obtain asymptotic distribution of MLE of θ. (15 marks)

(c) (i) State the sufficient conditions for a function φ(t) to be a characteristic function.
(ii) Investigate if the following functions are characteristic functions :
1. e⁻ᵗ⁴
2. [1 + |t|]⁻¹
Justify your answer. (5+10 marks)

UPSC Answer Check · Accepted Answer

(a) Let Sₙ = Y₁ + Y₂ + ⋯ + Yₙ. Since Yᵢ are independent Poisson(1), Sₙ has Poisson(n) distribution. Hence
P(Sₙ = n) = e⁻ⁿ nⁿ/n!.

Set Zₙ = (Sₙ − n)/√n. By the central limit theorem, Zₙ → N(0, 1). For lattice random variables, the local central limit theorem gives, for k near n,
P(Sₙ = k) ≈ [1/√(2π n)] exp[−(k−n)²/(2n)].
Putting k = n,
P(Sₙ = n) ≈ 1/√(2π n).
Therefore
e⁻ⁿ nⁿ/n! ≈ 1/√(2π n).
Rearranging,
n! ≈ √(2π n)(n/e)ⁿ.
This is valid for large n; the relative error tends to 0.

(b) Let Zᵢ = log Xᵢ. Then Zᵢ ~ N(θ, θ). Its density is
f_Z(z; θ) = [1/√(2πθ)] exp[−(z−θ)²/(2θ)].
Since x = eᶻ, the density of Xᵢ is
f_X(x; θ) = [1/(x√(2πθ))] exp[−(log x−θ)²/(2θ)], x > 0.
The likelihood is
L(θ) = (2πθ)^(−n/2) ∏xᵢ⁻¹ exp[−(1/(2θ))Σ(log xᵢ−θ)²].
Thus the log-likelihood is
l(θ) = constant − (n/2)log θ − (1/(2θ))Σ(zᵢ−θ)².
Expanding the square,
Σ(zᵢ−θ)² = Σzᵢ² − 2θΣzᵢ + nθ².
So, absorbing constants,
l(θ) = C − (n/2)log θ − S/(2θ) − nθ/2, where S = Σzᵢ².
Differentiate:
l′(θ) = −n/(2θ) + S/(2θ²) − n/2
= n/(2θ²)(S/n − θ − θ²).
The likelihood equation l′(θ) = 0 gives
θ² + θ − S/n = 0.
Hence
θ = [−1 ± √(1+4S/n)]/2.
Since θ > 0, the only positive solution is
θ̂ = [−1 + √(1+4S/n)]/2.
For S > 0, which occurs with probability 1, this is positive. Also l′(θ) has the sign of A − θ − θ², where A = S/n > 0. This expression decreases strictly from A > 0 at θ = 0 to −∞ as θ → ∞. Thus l′(θ) changes sign once from positive to negative, so θ̂ is the unique global maximum. Hence θ̂ is the unique MLE.

For asymptotic distribution, the per-observation log-density is
log f = −1/2 log(2πθ) − (z−θ)²/(2θ).
The score is
U(θ) = z²/(2θ²) − 1/(2θ) − 1/2.
Then
U′(θ) = −z²/θ³ + 1/(2θ²).
Fisher information is
I(θ) = −E[U′(θ)] = E[z²]/θ³ − 1/(2θ²).
Since Z ~ N(θ, θ), E[z²] = θ + θ². Hence
I(θ) = (θ+θ²)/θ³ − 1/(2θ²)
= (1+2θ)/(2θ²).
By asymptotic normality of the MLE,
√n(θ̂ − θ) → N(0, 2θ²/(1+2θ)).

(c)(i) By Bochner’s theorem, sufficient conditions for φ(t) to be a characteristic function are:
φ(0) = 1;
φ is continuous at 0;
φ is positive definite, i.e. for all n, real t₁,…,tₙ and complex a₁,…,aₙ,
ΣⱼΣₖ φ(tⱼ − tₖ)aⱼ conj(aₖ) ≥ 0.
These conditions are also necessary.

(c)(ii)
1. Consider φ(t) = e^(−t⁴). If it were a characteristic function of X, then φ′(0) = 0 and φ″(0) = 0, since e^(−t⁴) = 1 − t⁴ + ⋯. But for a characteristic function with second derivative at 0,
φ″(0) = −E[X²].
Thus E[X²] = 0, so X = 0 almost surely, giving φ(t) = 1 for all t. This contradicts e^(−t⁴) ≠ 1 for t ≠ 0. Hence e^(−t⁴) is not a characteristic function.
2. Consider φ(t) = 1/(1+|t|). Let f(t) = 1/(1+t), t ≥ 0. Then f(0) = 1, f is continuous, even, decreasing, and
f′(t) = −1/(1+t)² < 0,
f″(t) = 2/(1+t)³ > 0.
So f is convex on [0, ∞) and f(t) → 0 as t → ∞. By Polya’s criterion, an even, continuous function with these properties is a characteristic function. Therefore 1/(1+|t|) is a characteristic function.

(a) Let Y₁, Y₂, Y₃, ... be independent and identical Poisson random variables with parameter 1. Use central limit theorem to…

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