# Why does the expected value and variance differ for $|X - Y|$?

Suppose Antarctica has $N \sim \mathrm{Poisson}(\lambda)$ penguins. Each penguin is independently cute with probability $p$ (and not cute with probability $1-p$). Let $X$ and $Y$ be the numbers of cute and not cute penguins. Let $D = X - Y$. Find $E(D)$ and $\mathrm{Variance}(D)$.

For this problem, I thought that $D$ could be modeled by $\sim \mathrm{Poisson}(\lambda p - \lambda (1 - p))$ or $\sim \mathrm{Poisson}(\lambda (2p - 1))$. That would male $E(D) = \lambda (2p - 1)$.

However, apparently, the variance of $D$ is $\lambda$, not $\lambda (2p - 1)$ as one would expect from a Poisson random variable. Why is that?

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Off topic, but I think $p=1$ :) –  Hagen von Eitzen Dec 13 '12 at 17:24
Are you asking about $|X-Y|$, or $X-Y$? –  Chris Eagle Dec 13 '12 at 17:25
Also, why do you think $D$ is Poisson? –  Chris Eagle Dec 13 '12 at 17:26
Just $X-Y$. I think $D$ is Poisson because a Poisson random variable is itself the sum of many random variables. I thus felt that the size of the subset of cute penguins would also be distributed Poisson. –  John Hoffman Dec 13 '12 at 20:42

$D=X-Y$ is not Poisson unless $p=1$. In particular $P(D \lt 0) \gt 0$ and in fact $P(D \lt -k) \gt 0$ for all $k$.
$X$ is Poisson with parameter $\lambda p$, and $Y$ is Poisson with parameter $\lambda (1-p)$ and it turns out they are independent. This leads to the expected value and variance you quote:
$$E(D) = E(X)-E(Y) = \lambda p - \lambda (1-p) = \lambda (2p-1)$$
$$Var(D) = Var(X)+Var(Y) = \lambda p + \lambda (1-p) = \lambda$$
which is another indication $D$ is not Poisson.