Question

Let \(X\) and \(Y\) be independently distributed Poisson random variables such that \(P[X=1]=P[X=2]\), \(P[Y=3]=P[Y=4]\). Then the variance of \((2X-Y)\) is:

• A. 6
• B. 12
• C. 8
• D. 16

Hint:

Equal consecutive Poisson probabilities pin down the parameter, since P(X=k)=P(X=k+1) gives lambda=k+1. Then use Var(aX+bY)=a^2Var(X)+b^2Var(Y) for independent X, Y.

Answer 1

The Correct Option is B

Step 1: For \(X \sim Poisson(\lambda)\), \(P(X=1)=\lambda e^{-\lambda}\) and \(P(X=2)=\frac{\lambda^2}{2} e^{-\lambda}\). Setting them equal: \[\lambda = \frac{\lambda^2}{2} \Rightarrow \lambda = 2\] (since \(\lambda \ne 0\)). So \(Var(X) = \lambda = 2\).
Step 2: For \(Y \sim Poisson(\mu)\), \(P(Y=3)=\frac{\mu^3}{6} e^{-\mu}\) and \(P(Y=4)=\frac{\mu^4}{24} e^{-\mu}\). Setting them equal: \[\frac{\mu^3}{6} = \frac{\mu^4}{24} \Rightarrow \mu = 4\] So \(Var(Y) = \mu = 4\).
Step 3: Since \(X, Y\) are independent: \[Var(2X-Y) = 4\,Var(X) + Var(Y) = 4(2) + 4 = 12\]
The variance of \((2X-Y)\) is 12.