(a) Let E, F and G be three pairwise independent events such that P(E∩F) = 0·1 and P(F∩G) = 0·3. Prove that P(Eᶜ∪G) ≥ 11/12. 10 marks

(b) If X and Y are non-negative independent random variables and their joint moment generating function is given by

Question

(a) Let E, F and G be three pairwise independent events such that P(E∩F) = 0·1 and P(F∩G) = 0·3. Prove that P(Eᶜ∪G) ≥ 11/12. 10 marks

(b) If X and Y are non-negative independent random variables and their joint moment generating function is given by M_X,Y(t₁,t₂) = e^t₁+2e^t₂-3; t₁ > 0, t₂ > 0, then show that 2P(X+Y=2) = 9P(X+Y=0). 10 marks

(c) If X₁, X₂, ... be a sequence of i.i.d. U(0, 1) random variables, then find the value of Limₙ→∞ P(∑ᵢ₌₁ⁿ Xᵢ ≤ n/2 + √(n/144)) [use √3 = 1·74, Φ(0·29) = 0·6141, Φ(1) = 0·8413]. 10 marks

(d) Let X₁, X₂, ..., Xₙ be a random sample from normal N(μ, σ²) distribution. Obtain sufficient statistic for parameters (μ, σ²) when both the parameters are unknown. If σ² is known, what will be sufficient statistic for parameter μ ? 10 marks

(e) A random variable X has the following distribution under H₀ and H₁ :
x : 1 2 3 4 5 6
f₀(x) : 0·01 0·01 0·01 0·01 0·01 0·95
f₁(x) : 0·05 0·04 0·03 0·02 0·01 0·85
Find the best test of size 0·03 and its probability of type-II error for testing H₀ : f = f₀ versus H₁ : f = f₁. Is it unbiased test ? Why ? 10 marks

UPSC Answer Check · Accepted Answer

(a) Let p = P(E), q = P(F), r = P(G). Pairwise independence gives P(E∩F) = pq = 1/10 and P(F∩G) = qr = 3/10. Hence p = 1/(10q) and r = 3/(10q).

Now Eᶜ ∪ G is the complement of E ∩ Gᶜ. Therefore
P(Eᶜ ∪ G) = 1 − P(E ∩ Gᶜ).

Because E and G are independent, P(E∩G) = pr. So
P(E ∩ Gᶜ) = P(E) − P(E∩G) = p − pr = p(1−r).
Thus P(Eᶜ ∪ G) = 1 − p(1−r).

We need p(1−r) ≤ 1/12. Substitute p and r:
p(1−r) = [1/(10q)] [1 − 3/(10q)] = 1/(10q) − 3/(100q²).
Then
p(1−r) ≤ 1/12
⇔ 1/(10q) − 3/(100q²) ≤ 1/12
⇔ multiply by 300q²:
30q − 9 ≤ 25q²
⇔ 25q² − 30q + 9 ≥ 0
⇔ (5q − 3)² ≥ 0,
which is always true. Hence p(1−r) ≤ 1/12. Therefore
P(Eᶜ ∪ G) = 1 − p(1−r) ≥ 1 − 1/12 = 11/12.

Therefore P(Eᶜ ∪ G) ≥ 11/12.

(b) As printed, M(t₁,t₂) = e^(t₁+2e^(t₂)−3) gives M(0,0) = e^(−1) ≠ 1, so it cannot be a joint moment generating function. The intended valid form that matches the conclusion is
M(t₁,t₂) = exp(e^(t₁) + 2e^(t₂) − 3).
Assume this standard correction.

Since X and Y are independent,
M(t₁,t₂) = M_X(t₁) M_Y(t₂).
Now
exp(e^(t₁) + 2e^(t₂) − 3)
= exp(e^(t₁) − 1) · exp(2e^(t₂) − 2).

The mgf of Poisson(λ) is exp(λ(e^t − 1)). Hence
M_X(t₁) = exp(1·(e^(t₁) − 1)) ⇒ X ∼ Poisson(1),
M_Y(t₂) = exp(2·(e^(t₂) − 1)) ⇒ Y ∼ Poisson(2).

Therefore X + Y ∼ Poisson(1+2) = Poisson(3). So for S = X + Y,
P(S = k) = e^(−3) 3^k / k! , k = 0,1,2,...

Thus
P(X+Y = 0) = P(S=0) = e^(−3),
P(X+Y = 2) = P(S=2) = e^(−3) 3² / 2! = 9e^(−3)/2.

Hence
2P(X+Y = 2) = 2 · 9e^(−3)/2 = 9e^(−3) = 9P(X+Y = 0).

Therefore 2P(X+Y = 2) = 9P(X+Y = 0).

(c) Let S_n = ∑ (i=1 to n) X_i, where X_i ∼ U(0,1). Then
E(X_i) = 1/2, Var(X_i) = 1/12.
So E(S_n) = n/2, Var(S_n) = n/12, standard deviation √(n/12).

By the Lindeberg–Lévy Central Limit Theorem,
(S_n − n/2) / √(n/12) → N(0,1) in distribution.

Now
P(S_n ≤ n/2 + √(n/144))
= P( (S_n − n/2)/√(n/12) ≤ √(n/144)/√(n/12) ).

The standardized upper limit is
√(n/144)/√(n/12) = √(12/144) = √(1/12) = 1/√12 = √3/6.

Given √3 = 1·74,
√3/6 = 1·74/6 = 0·29.

Therefore the limiting probability is
Φ(0·29) = 0·6141.

Hence Lim_n→∞ P(∑ (i=1 to n) X_i ≤ n/2 + √(n/144)) = 0·6141.

(d) 
(i) Both μ and σ² unknown.
The joint density of X₁, X₂, ..., Xₙ is
f(x₁,...,xₙ; μ,σ²) = (2πσ²)^(−n/2) exp[ −(1/(2σ²)) ∑ (i=1 to n) (x_i − μ)² ].

Expand the exponent:
∑ (x_i − μ)² = ∑ x_i² − 2μ ∑ x_i + nμ².
Thus
f = (2πσ²)^(−n/2) exp[ −∑ x_i²/(2σ²) + μ∑ x_i/σ² − nμ²/(2σ²) ].

By the Neyman–Fisher factorization theorem, the density depends on the sample only through
∑ X_i and ∑ X_i².
Therefore a sufficient statistic for (μ,σ²) when both are unknown is
T = (∑ X_i, ∑ X_i²).
Equivalently, T = (X̄, S²), where X̄ is the sample mean and S² is the sample variance.

(ii) If σ² is known.
Then the joint density becomes
f = (2πσ²)^(−n/2) exp[ −∑ (x_i − μ)²/(2σ²) ]
= (2πσ²)^(−n/2) exp[ −∑ x_i²/(2σ²) ] exp[ μ∑ x_i/σ² − nμ²/(2σ²) ].

The factor depending on both μ and the data is exp[ μ∑ x_i/σ² ].
Hence by factorization, the sufficient statistic for μ is
∑ X_i, or equivalently X̄.

Therefore, for known σ², ∑ X_i is sufficient for μ.

(e) By the Neyman–Pearson Lemma, the best test rejects H₀ for large values of the likelihood ratio
Λ(x) = f₁(x)/f₀(x).

Compute Λ(x):
x=1: 0·05/0·01 = 5
x=2: 0·04/0·01 = 4
x=3: 0·03/0·01 = 3
x=4: 0·02/0·01 = 2
x=5: 0·01/0·01 = 1
x=6: 0·85/0·95 = 17/19 < 1.

The largest likelihood ratios occur at x = 1, 2, 3, 4, 5.
We need size 0·03 under H₀. Under H₀,
P(X=1)=0·01, P(X=2)=0·01, P(X=3)=0·01.
So rejecting H₀ when X ∈ {1,2,3} gives size
α = 0·01 + 0·01 + 0·01 = 0·03.

Thus the best critical region is
C = {1,2,3}.
Reject H₀ if X = 1, 2, or 3.

Type-II error is the probability of accepting H₀ when H₁ is true:
β = P_H₁(X ∈ {4,5,6})
= 0·02 + 0·01 + 0·85
= 0·88.

The power of the test is
1 − β = 0·12.

Since power = 0·12 > size = 0·03, the test is unbiased.

Therefore the best test of size 0·03 rejects H₀ for X = 1,2,3; its type-II error is 0·88; and it is unbiased because its power exceeds its size.

(a) Let E, F and G be three pairwise independent events such that P(E∩F) = 0·1 and P(F∩G) = 0·3. Prove that P(Eᶜ∪G) ≥ 11/12. 10…

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