Answer to Question #110228 in Statistics and Probability for Diksha

Question

If X and Y are independent Poisson variates such that
P(X =1) = P(X = 2) and P(Y = 2) = P(Y = 3) .
Find the variance of X− 2Y.

Answer 1

$$P(X=x)={e^{-\lambda_1}\lambda_1^x \over x!}$$

Given that $P(X=1)=P(X=2).$ Then

$${e^{-\lambda_1}\lambda_1^1 \over 1!}={e^{-\lambda_1}\lambda_1^2 \over 2!}, \lambda_1>0$$

$$\lambda_1=2$$

$$P(Y=y)={e^{-\lambda_2}\lambda_2^y \over y!}$$

Given that $P(Y=2)=P(Y=3).$ Then

$${e^{-\lambda_2}\lambda_2^2 \over 2!}={e^{-\lambda_2}\lambda_2^3 \over 3!}, \lambda_2>0$$

$$\lambda_2=3$$

For a Poisson random variable $Z,\ Var(Z)=\lambda.$ Then

$$Var(X)=\lambda_1=2, Var(Y)=\lambda_2=3$$

Hence

$$Var(X-2Y)=Var(X)+2^2Var(Y)=$$

$$=2+4(3)=14$$

Therefore

$$Var(X-2Y)=14$$