ISS PREVIOUS YEAR PAPER-1 2016 SOLUTION Q.NO. 4

ISS PREVIOUS YEAR PAPER-1 2016 SOLUTION Q.NO. 4

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Question

X₁ and X₂ are independent Poisson random variables such that:

P(X₁ = 2) = P(X₁ = 1)P(X₂ = 2) = P(X₂ = 3)

Find the variance of (X₁ − 2X₂).

Options:(a) 14(b) 4(c) 3(d) 2

Solution

Step 1: Use Poisson Distribution Formula

For a Poisson random variable with parameter λ:

P(X = k) = (e^(−λ) × λ^k) / k!

Step 2: Find λ₁ for X₁

Given: P(X₁ = 2) = P(X₁ = 1)

Using formula:

(e^(−λ₁) × λ₁²) / 2! = (e^(−λ₁) × λ₁) / 1!

Cancel e^(−λ₁):

λ₁² / 2 = λ₁

Multiply both sides by 2:

λ₁² = 2λ₁

⇒ λ₁ = 2

Step 3: Find λ₂ for X₂

Given: P(X₂ = 2) = P(X₂ = 3)

(e^(−λ₂) × λ₂²) / 2! = (e^(−λ₂) × λ₂³) / 3!

Cancel e^(−λ₂):

λ₂² / 2 = λ₂³ / 6

Multiply both sides by 6:

3λ₂² = λ₂³

⇒ λ₂ = 3

Step 4: Use Variance Property

For independent variables:

Var(aX + bY) = a²Var(X) + b²Var(Y)

For Poisson distribution:

Var(X₁) = λ₁ = 2Var(X₂) = λ₂ = 3

Step 5: Calculate Required Variance

Var(X₁ − 2X₂) = Var(X₁) + (−2)²Var(X₂)

= 2 + 4 × 3

= 2 + 12

= 14

Final Answer:- 14

ISS PREVIOUS YEAR PAPER-1 2016 SOLUTION Q.NO. 4 :- CLICK HERE