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

If X and Y are independent Poisson variables such that P(X = 1)=P( X =2) and P(Y=2)= P(Y= 3).

Find the variance of X − 2Y .

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

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

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

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

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

$${e^{-\lambda_2}\lambda_2^2 \over 2!}={e^{-\lambda_2}\lambda_2^3 \over 3!}, \lambda_2>0$$ $${\lambda_2^2 \over 2}={\lambda_2^3 \over6}$$ $$\lambda_2=3.$$

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

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

Hence

$$Var(X-2Y)=Var(X)+2^2Var(Y)=$$ $$=2+4(3)=14$$

$Var(X-2Y)=14$