Simple question regarding the Poisson distribution

Question

I'm just trying to wrap my head around the Poisson distribution to make it more intuitive and I'm struggling with one (seemingly) paradoxical example.

Say we have a rate of 4 events in a given time period, for instance, 4 accidents in 1 week. What is the probability of exactly 6 accidents happening in 2 1-week periods, which (assuming I'm correct) is akin to an average of exactly 3 accidents per 1-week period?

Here we get:

$$P(X = 3) = \frac{\lambda^xe^{-\lambda}}{x!} = \frac{4^3e^{-4}}{3!} \approx 0.1954 $$

But say we now calculate it the following way, by estimating an average of 8 accidents per 2-week period, given our original rate of 4 accidents per 1-week period, and calculate for the original requirement (exactly 6 accidents in a 2-week period):

Here we get:

$$P(X = 6) = \frac{\lambda^xe^{-\lambda}}{x!} = \frac{8^6e^{-8}}{6!} \approx 0.1221 $$

Why are these probabilities not the same? Something tells me the original assumption of breaking down 6 accidents per 2-week period into 3 accidents per week is faulty, but I'm not really sure why. I have a few ideas but I'd prefer it if someone with expertise in this area would clarify.

Much appreciated.

Answer 1

Let's use a simpler model to illustrate the flaw in the first line of reasoning.

Suppose you have a fair coin and decide to flip it $n = 4$ times. What is the probability that you get exactly $2$ heads and $2$ tails? This is just $$\binom{4}{2} \frac{1}{2^4} = \frac{3}{8}.$$

Now what is the probability that, among the first two coin tosses, you get $1$ head and $1$ tail, and among the last two coin tosses, you again get $1$ head and $1$ tail? This is $$\left(\binom{2}{1} \frac{1}{2^2}\right)^2 = \frac{1}{4}.$$

Why are these not equal? Simply put, the second case does not count all of the outcomes in which $2$ heads and $2$ tails are possible: for instance, the outcomes $$\{HHTT, \; TTHH\}$$ are included in the former but not the latter.

By the same token, counting $3$ events in the first week and $3$ in the second week is not the same as counting $6$ events in a $2$-week period, because the former requires that $3$ events are observed in each week, whereas the latter can include such outcomes as $4$ events in the first week and $2$ in the second; or $0$ outcomes in the first week and $6$ in the second.

Therefore, if we really wanted to count the outcomes this way, we might define $X \sim \operatorname{Poisson}(\lambda = 8)$ to represent the random number of events observed in a $2$-week period, and $Y \sim \operatorname{Poisson}(\lambda = 4)$ be the random number of events in a $1$-week period. Then $$\Pr[X = 6] = \sum_{y=0}^6 \Pr[Y = y]\Pr[Y = 6-y].$$

And you can verify this numerically by calculating the sum with $$\Pr[Y = y] = e^{-\lambda} \frac{\lambda^y}{y!} = e^{-4} \frac{4^y}{y!},$$

hence $$\begin{align} \sum_{y=0}^6 \Pr[Y = y]\Pr[Y = 6-y] &= \sum_{y=0}^6 e^{-4} \frac{4^y}{y!} e^{-4} \frac{4^{6-y}}{(6-y)!} \\ &= e^{-8} \sum_{y=0}^6 \frac{4^6}{y!(6-y)!} \\ &= e^{-8} \frac{4^6}{6!} \sum_{y=0}^6 \frac{6!}{y!(6-y)!} \\ &= e^{-8} \frac{4^6}{6!} (1 + 1)^6 \\ &= e^{-8} \frac{8^6}{6!} \\ &= \Pr[X = 6]. \end{align}$$

Answer 2

If you consider a Poisson process $X(t)$ with $\lambda =4$ then $P(X(t)=k)=\frac{(\lambda t)^k e^{-\lambda t}}{k!}$. Therefore the second result is correct. $P(X(2)=6)=\frac{(4\cdot 2)^6 e^{-4\cdot 2}}{6!}$.