TAOCP 3.4.1 Exercise 26

Let $N_1$ and $N_2$ be independent Poisson random variables with means $\mu_1$ and $\mu_2$, respectively, where $\mu_1 > \mu_2 \ge 0$.

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Section 3.4.1: Numerical Distributions

Exercise 26. [M18] Let $N_1$ and $N_2$ be independent Poisson deviates with means $\mu_1$ and $\mu_2$, where $\mu_1 > \mu_2 \ge 0$. Prove or disprove: (a) $N_1 + N_2$ has the Poisson distribution with mean $\mu_1 + \mu_2$. (b) $N_1 - N_2$ has the Poisson distribution with mean $\mu_1 - \mu_2$.

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Solution

Let $N_1$ and $N_2$ be independent Poisson random variables with means $\mu_1$ and $\mu_2$, respectively, where $\mu_1 > \mu_2 \ge 0$. Recall that the probability mass function of a Poisson deviate with mean $\mu$ is

$$ \Pr{N = n} = e^{-\mu} \frac{\mu^n}{n!}, \qquad n = 0,1,2,\dots $$

(a) Sum of independent Poisson deviates

Let $N = N_1 + N_2$. For any nonnegative integer $n$,

$$ \Pr{N = n} = \sum_{k=0}^{n} \Pr{N_1 = k} \Pr{N_2 = n-k}. $$

Using independence and the Poisson pmfs,

$$ \Pr{N = n} = \sum_{k=0}^{n} e^{-\mu_1} \frac{\mu_1^k}{k!} \cdot e^{-\mu_2} \frac{\mu_2^{n-k}}{(n-k)!} = e^{-(\mu_1 + \mu_2)} \sum_{k=0}^{n} \frac{\mu_1^k \mu_2^{n-k}}{k!(n-k)!}. $$

Applying the binomial theorem,

$$ \sum_{k=0}^{n} \frac{\mu_1^k \mu_2^{n-k}}{k!(n-k)!} = \frac{1}{n!} \sum_{k=0}^{n} \binom{n}{k} \mu_1^k \mu_2^{n-k} = \frac{(\mu_1 + \mu_2)^n}{n!}. $$

Hence,

$$ \Pr{N = n} = e^{-(\mu_1 + \mu_2)} \frac{(\mu_1 + \mu_2)^n}{n!}. $$

This is exactly the Poisson pmf with mean $\mu_1 + \mu_2$. Therefore,

$$ \boxed{N_1 + N_2 \sim \text{Poisson}(\mu_1 + \mu_2)}. $$

(b) Difference of independent Poisson deviates

Let $M = N_1 - N_2$. The possible values of $M$ are all integers $\dots,-2,-1,0,1,2,\dots$. Indeed, if $\mu_2 > 0$, then

$$ \Pr(M = -1) = \Pr(N_1 = 0, N_2 = 1) + \Pr(N_1 = 1, N_2 = 2) + \cdots > 0. $$

A Poisson random variable is supported on ${0,1,2,\dots}$. Since $M$ can be negative whenever $\mu_2 > 0$, it cannot be Poisson in this case.

The probability mass function of $M$ is

$$ \Pr{M = m} = \sum_{k = \max(0,-m)}^{\infty} \Pr{N_1 = m+k} \Pr{N_2 = k} = e^{-(\mu_1 + \mu_2)} \sum_{k = \max(0,-m)}^{\infty} \frac{\mu_1^{m+k} \mu_2^k}{(m+k)! , k!}, $$

which does not reduce to a single-term Poisson pmf $e^{-\lambda} \lambda^m / m!$ for any $\lambda$. Therefore, $M$ is not Poisson for $\mu_2 > 0$.

If $\mu_2 = 0$, then $N_2 \equiv 0$ almost surely, so

$$ M = N_1 - N_2 = N_1 \sim \text{Poisson}(\mu_1 - \mu_2) = \text{Poisson}(\mu_1). $$

Thus the statement in part (b) is false in general, but true in the degenerate case $\mu_2 = 0$.

$$ \boxed{\text{For }\mu_2 > 0,; N_1 - N_2 \text{ is not Poisson; if } \mu_2 = 0, \text{ it is Poisson.}} $$

∎

This corrected solution explicitly addresses the edge case $\mu_2 = 0$, while maintaining the rigorous argument that $N_1 - N_2$ cannot be Poisson when $\mu_2 > 0$.