$X=10(Y-1) \forall y \ge 2$. Otherwise $X=0$. $Y$ follows a Poisson distribution with mean $0.3$. What is $Var(X)$?

Question

$X=\begin{cases} 0: &y=0,1 \\10(Y-1): &y=2,3,4,... \end{cases}$

$E[X] = E[10(Y-1)] -(10(0)-1)P(Y=0)-(10(1)-1)P(Y=1)$

$E[X]=10E[Y]-10+P(Y=0)-9P(Y=1) = 10(0.3)-10+e^{-0.3}-9e^{-0.3}(0.3)$

$E[X] = -12.9265457655$ (Uh-oh)

$E[X^2] = E[(10(Y-1))^2] -(10(0)-1)^2P(Y=0)-(10(1)-1)^2P(Y=1)$

$E[X^2] = E[100(Y^2-2Y+1)]-P(Y=0)-81P(Y=1)$

$E[X^2]= 100E[Y^2]-200E[Y]+100-e^{-0.3}-81e^{-0.3}(0.3)$

$E[Y^2] = Var(Y)+E[Y]^2 = 0.3 + (0.3)^2 = 0.39$

$\therefore E[X^2] = 100(0.39)-200(0.3)+100-e^{-0.3}-81e^{-0.3}(0.3) = 60.2572990168$

$Var(X) = 60.257 - (-12.927)^2 = -106.85$

Clearly this is wrong.

Answer 1

Let $Z=10X$

What is the difference with $Y$?

For $x\geq 2$, $(Z=10x)$ if and only if $(Y=10x)$

$(Z=0)$ or $(Z=10)$ if and only if $(Y=0)$

So $E(Y)=E(Z)-0P(Z=0)-10.P(Z=10)$

then $E(Y)=10E(X)-10.P(X=1)$

And you know $E(X)$ ...