Computing $\ E[Y^2] $ when $Y$ is a piecewise function of $X \sim Pois(2) $

Question

$\ X \sim Pois(2) $ and let$\ Y $ be a random variable : $\ Y = \begin{cases} 2X , X \le 3 \\ X , X \ge 4 \end{cases} $

compute $\ E[Y^2] $ ?

I have calculated that $\ E[Y] = E[X] + 10e^{-2} $ and I have the identity $\ Var(X) = E[X^2] - (E[X])^2 $ and because $\ X $ is poisson then $\ Var(X) = E[X] $. but I'm not sure if $\ Y $ is also a poisson variable?

Any suggestions?

Answer 1

$Y$ cannot be Poisson, because for example, $$\Pr[Y = 3] = \Pr[2X = 3] = \Pr[X = 3/2] = 0.$$

If we define $Y$ as in the body of the post, i.e. $$Y = \begin{cases} 2X, & X \le 3 \\ X, & X \ge 4, \end{cases}$$ then we note that $$\begin{align*}\operatorname{E}[Y^2] &= \sum_{x=0}^3 (2x)^2 e^{-\lambda} \frac{\lambda^x}{x!} + \sum_{x=4}^\infty x^2 e^{-\lambda} \frac{\lambda^x}{x!} \\ &= 4 \sum_{x=0}^3 x^2 e^{-\lambda} \frac{\lambda^x}{x!} + \sum_{x=4}^\infty x^2 e^{-\lambda} \frac{\lambda^x}{x!} \\ &= 3 \sum_{x=0}^3 x^2 e^{-\lambda} \frac{\lambda^x}{x!} + \left(\sum_{x=0}^3 x^2 e^{-\lambda} \frac{\lambda^x}{x!} + \sum_{x=4}^\infty x^2 e^{-\lambda} \frac{\lambda^x}{x!} \right) \\ &= 3 \sum_{x=0}^3 x^2 e^{-\lambda} \frac{\lambda^x}{x!} + \sum_{x=0}^\infty x^2 e^{-\lambda} \frac{\lambda^x}{x!} \\ &= e^{-\lambda} (3\lambda + 6 \lambda^2 + \tfrac{9}{2} \lambda^3) + \operatorname{E}[X^2] \\ &= 66e^{-2} + \operatorname{E}[X]^2 + \operatorname{Var}[X] \\ &= 66e^{-2} + \lambda^2 + \lambda \\ &= 6 + 66e^{-2}. \end{align*}$$

Answer 2

\begin{align} E[Y^2] &= \sum_{x=0}^3 (2x)^2 Pr(X=x)+ \sum_{x=4}^\infty x^2 Pr(X=x)\\ &=\sum_{x=0}^3 4x^2 Pr(X=x)+ \sum_{x=4}^\infty x^2 Pr(X=x)\\ &=\sum_{x=0}^3 3x^2 Pr(X=x)+ \sum_{x=0}^\infty x^2 Pr(X=x)\\ &=3\sum_{x=1}^3 x^2 Pr(X=x)+ E[X^2]\\ &= 3\left( Pr(X=1) + 4Pr(X=2)+9Pr(X=3) \right) + E[X^2] \end{align}