(a) (i) How large a sample must be taken in order that the probability will be at least 0.90 that the sample mean will be within 0.4-neighbourhood of the population mean, provided the population standard deviation is 2?

(ii) Examine whether the weak

Question

(a) (i) How large a sample must be taken in order that the probability will be at least 0.90 that the sample mean will be within 0.4-neighbourhood of the population mean, provided the population standard deviation is 2?

(ii) Examine whether the weak law of large numbers holds for the sequence {Xk} of independent random variables defined as follows:

P(Xk = -1 - 1/k) = 1/2 {1 - (1 - 1/k²)^(1/2)},

P(Xk = 1 + 1/k) = 1/2 {1 + (1 - 1/k²)^(1/2)}. (8+7=15 marks)

(b) Theoretical probabilities in the four cells of a multinomial distribution are (2+θ)/4, (1-θ)/4, (1-θ)/4 and θ/4, whereas the observed frequencies are 108, 27, 30 and 8 respectively, then estimate θ by maximum likelihood method. Also, obtain the standard error of the estimate. (20 marks)

(c) If X is a random variable with characteristic function

φ(t) = {1 - |t|, |t| ≤ 1
0, otherwise,

then obtain the corresponding probability density function. (15 marks)

UPSC Answer Check · Accepted Answer

(a) Sample size by Chebyshev's inequality and the weak law for {Xₖ}

(i) X̄ has mean μ and variance σ²/n = 4/n. By Chebyshev's inequality,
P(|X̄ − μ| < 0.4) ≥ 1 − Var(X̄)/(0.4)² = 1 − (4/n)/0.16 = 1 − 25/n.
Requiring 1 − 25/n ≥ 0.90 gives n ≥ 250. A sample of size 250 is enough for any population with σ = 2. (If X̄ may be taken as normal, n ≥ (1.645 × 2/0.4)² = 67.6, i.e. 68.)

(ii) Put aₖ = √(1 − 1/k²). Then Xₖ = −(1 + 1/k) with probability (1 − aₖ)/2 and Xₖ = 1 + 1/k with probability (1 + aₖ)/2; these add to 1.
E(Xₖ) = (1 + 1/k)aₖ → 1.
E(Xₖ²) = (1 + 1/k)², so Var(Xₖ) = (1 + 1/k)²(1 − aₖ²) = (1 + 1/k)²/k² = (k + 1)²/k⁴ ≤ 4/k².
The Xₖ are independent, so for Sₙ = X₁ + … + Xₙ,
Var(Sₙ) = Σ (k + 1)²/k⁴ ≤ Σ 4/k² ≤ 4π²/6 ≈ 6.58.
For any ε > 0, Chebyshev's inequality gives
P(|Sₙ − E(Sₙ)|/n ≥ ε) ≤ Var(Sₙ)/(n²ε²) ≤ 6.58/(n²ε²) → 0.
So (Sₙ − E Sₙ)/n → 0 in probability: the weak law of large numbers holds for {Xₖ}. Since E(Xₖ) → 1, also Sₙ/n → 1 in probability.

(b) Maximum likelihood estimate of θ and its standard error

n = 108 + 27 + 30 + 8 = 173. Up to a constant, the log-likelihood is
ℓ(θ) = 108 ln(2 + θ) + 57 ln(1 − θ) + 8 ln θ, 0 < θ < 1
(the second and third cells together have frequency 57).
Score equation: dℓ/dθ = 108/(2 + θ) − 57/(1 − θ) + 8/θ = 0.
Multiplying by θ(1 − θ)(2 + θ) gives 173θ² + 14θ − 16 = 0, so
θ̂ = [−14 + √(14² + 4 × 173 × 16)]/(2 × 173) = (−14 + √11268)/346 = (3√313 − 7)/173 ≈ 0.2663.
The other root, −0.347, lies outside (0, 1). Also d²ℓ/dθ² = −108/(2 + θ)² − 57/(1 − θ)² − 8/θ² < 0, so θ̂ gives the maximum.
For the multinomial, the Fisher information is
I(θ) = n Σ (1/pᵢ)(dpᵢ/dθ)² = n[1/(4(2 + θ)) + 1/(2(1 − θ)) + 1/(4θ)].
At θ̂ = 0.2663: I(θ̂) = 173 × (0.1103 + 0.6815 + 0.9387) = 299.4.
SE(θ̂) = 1/√I(θ̂) = 1/√299.4 ≈ 0.0578.
Hence θ̂ ≈ 0.266 with standard error ≈ 0.058.

(c) Density from the characteristic function

φ(t) = 1 − |t| for |t| ≤ 1 is absolutely integrable, so by the inversion theorem the density exists and
f(x) = (1/2π) ∫₋₁¹ (1 − |t|) e^(−itx) dt.
φ is real and even, so the sine part vanishes:
f(x) = (1/π) ∫₀¹ (1 − t) cos(tx) dt.
For x ≠ 0, integrating by parts,
∫₀¹ (1 − t) cos(tx) dt = [(1 − t) sin(tx)/x]₀¹ + (1/x) ∫₀¹ sin(tx) dt = (1 − cos x)/x².
Hence
f(x) = (1 − cos x)/(πx²) = (1/2π)[sin(x/2)/(x/2)]², x ≠ 0,
and f(0) = (1/π) ∫₀¹ (1 − t) dt = 1/(2π), the limit as x → 0.
Since f ≥ 0 and ∫ f(x) dx = φ(0) = 1, f is a valid density.

(a) (i) How large a sample must be taken in order that the probability will be at least 0.90 that the sample mean will be within…

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(a) Sample size by Chebyshev's inequality and the weak law for {Xₖ}

(b) Maximum likelihood estimate of θ and its standard error

(c) Density from the characteristic function

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