Finding $E(X(X-1))$ where $X$ is a Poisson random variable such that $P(X=2) = \frac{2}{3} P(X=1)$

Question

Let $X$ be a Poisson random variable such that $P(X = 2) = \frac{2}{3} P (X = 1)$. Find $E(X(X − 1))$.

I know $E(X(X-1)) = E(X^2) - E(X)$, but I don't know where to go from here.

Answer 1

From the identity \begin{equation} \text{Var}(X) = E(X^2) - (E(X))^2 \end{equation} you can get \begin{equation} E(X^2) = \text{Var}(X) + (E(X))^2 \end{equation} remember that $E(X)$ is just the mean, and the varriance of a poisson distribution is the same as the mean($E(X) = \text{Var(X)})$. This yields the final equation \begin{equation} E(X(X - 1)) = E(X^2) - E(X) = E(X) + (E(X))^2 - E(X) = (E(X))^2 \end{equation} So from your first equation solve for the mean. Your answer will be the square of your mean.

Answer 2

First we find the parameter $\lambda$ of the Poisson. We have $\Pr(X=2)=e^{-\lambda}\frac{\lambda^2}{2!}$ and $\Pr(X=1)=e^{-\lambda}\frac{\lambda}{1!}$. From the given equation we conclude that $\frac{\lambda^2}{2}=\frac{2}{3}\lambda$, so now we know $\lambda$, since it cannot be $0$.

The expectation of $X(X-1)$ is equal to $$\sum_{k=0}^\infty k(k-1)e^{-\lambda}\frac{\lambda^k}{k!}.$$ The first two terms are $0$, and for $k\ge 2$ we have $\frac{k(k-1)}{k!}=\frac{1}{(k-2)!}$, so our sum is $$\sum_{k=2}^\infty e^{-\lambda}\frac{\lambda^k}{(k-2)!}.$$ Replace $k-2$ by $n$. Our sum is $$\sum_{n=0}^\infty e^{-\lambda}\frac{\lambda^{n+2}}{n!}.$$ Bring out a $\lambda^2$. We end up with $$\lambda^2\sum_{n=0}^\infty e^{-\lambda}\frac{\lambda^{n}}{n!}.$$ But the sum after the $\lambda^2$ is $1$, for it is the sum of all the Poisson probabilities. We conclude that $E(X(X-1))=\lambda^2$.

Remark: In a similar way, we could compute $E(X(X-1)(X-2))$, and other similar expressions. This sort of expectation is easier to get at than things like $E(X^2)$ and $E(X^3)$. In fact, finding the expectation of $X(X-1)$ is one of the standard paths for finding the variance of $X$.

Answer 3

$$ \frac{\lambda^2 e^{-\lambda}} 2 = \Pr(X = 2) = \frac 2 3 \Pr(X = 1) = \frac 2 3 \cdot \frac{\lambda e^{-\lambda}}{1}. $$ Multiplying both sides by $6e^\lambda/\lambda$, we get $$ 3\lambda = 4 $$ so $\lambda = 4/3$.

So $$ \operatorname{E}(X(X-1)) = \operatorname{E}(X^2) - \operatorname{E}(X) = \Big( (\operatorname{E} X)^2 + \operatorname{var}(X) \Big) - \lambda = (\lambda^2 + \lambda) - \lambda = \lambda^2 = \frac{16} 9. $$