Poisson distribution and the choice of unit

Question

Suppose that the average number of cars passing by a building, $X$, is 3 cars per minute. Assuming that $X$ follows the Poisson distribution with mean $\lambda=3$, the probability that 1 car passes per minute is $$P(X=1)=e^{-3}\frac{3^1}{1!}\approx0.15.$$ On the other hand, let $Y$ be the number of cars passing by the building in an hour. Then $Y$ follows the Poisson distribution with mean $60\lambda=180$, so the probability that 60 cars pass by the building in an hour is $$P(Y=60)=e^{-180}\frac{180^{60}}{60!}\approx1.67\times10^{-25},$$ which is much smaller than $P(X=1)$. But all I did was to change the unit from minute to hour, and intuitively, the probability of 1 car per minute should be the same as the probability of 60 cars per hour. Somehow the choice of unit seems to affect the Poisson process significantly, which is odd. I suspect that I'm not fully understanding the Poisson process. Can anyone explain in layman's term?

Answer 1

intuitively, the probability of 1 car per minute should be the same as the probability of 60 cars per hour.

This intuition is incorrect. When $\lambda = 3$ this means that the mean number of cars per minute is $3$, and so having only $1$ car pass per minute is somewhat unusual; it means fewer cars than usual passed in a minute. When you go up to an hour, the mean number of cars is $180$, and having only $60$ cars pass during that time is much more unusual; it means that there had to be unusually few cars for sixty minutes in a row, which is much less likely than just having unusually few cars for a single minute. In fact

$$ \left( 3e^{-3} \right)^{60} \approx 2.8 \times 10^{-50}$$

so the second probability is a bit higher than the first one to the power of sixty (since there are more ways to get $60$ cars in an hour than just to have one car per minute; some of those minutes there could be fewer cars and other times there could be more). But you can see it's roughly-ish the same ballpark.

The question is a bit like asking what the difference is between flipping heads once and flipping heads $60$ times in a row. The answer is to flip heads $60$ times in a row you have to get very very lucky for a long time! These just aren't the same distribution and the Poisson distribution for larger values of $\lambda$ isn't a scaled version of the Poisson distribution for smaller values of $\lambda$ either.

Edit: Some comment about "units" is also in order. I think your intuition is that the parameter $\lambda$ is like a velocity, so it should be expressible in different units (cars per minute, cars per hour) without changing what's going on. This is not correct! Actually the parameter $\lambda$ is and must be dimensionless or else the $e^{-\lambda}$ in the definition of the Poisson distribution doesn't make sense. $\lambda$ is just a number, and the Poisson distribution for small values of $\lambda$ behaves quite differently than for large values of $\lambda$. You can see this concretely in the fact that the mean is $\lambda$ but the standard deviation is $\sqrt{\lambda}$ (which also would not make sense if $\lambda$ had units).

So how do we square this with the impression that "cars per minute" sounds like a quantity with units? The answer is that to get a specific Poisson distribution we have to pick a specific time window; if $3$ cars pass per minute then we can consider the number of cars that pass in a minute and get $P(3)$ or the number of cars that pass in an hour and get $P(180)$, which means that the rate of $3$ cars passing per minute is being multiplied by a number $t$ of minutes, so the time units cancel. The quantity $3t$ is then dimensionless and this is the actual parameter of the Poisson distribution.

Answer 2

A simple analogy: Roll a single fair (cubic) die. There are 6 possible outcomes (1, 2, 3, 4, 5, 6), each with probability $\frac{1}{6} = 0.1666666...$.

Now roll 60 dice. Their total can be anything between 60 and 360. That's 301 distinct values. Furthermore, their distribution is no longer uniform.

• The lowest total value (60) can be achieved only by having 1's on all 60 dice, which has a probability of $(\frac{1}{6})^{60} \approx 2.046 \times 10^{-47}$. It's the same for the highest total value (360), which can be achieved only by rolling 6's on all 60 dice.
• The most likely total value (210) has a probability of approximately $0.030077$. But this is only about 18% of the corresponding maximum probability when rolling 1 die. The more dice you have, the probability of achieving any exact sum will get lower.
• However, the probability of getting a sum near 210 — between 180 and 240, inclusive — rises to $0.50462$. Contrast with the $\frac{1}{3}$ probability of rolling either 3 or 4 on a single die.

The more dice you roll, the less likely any one exact total gets, and the more the distribution clusters towards the center and away from the extremes. So, you can't just assume that conducting a probability trial 60 times will have the same probabilities as conducting it once.

Answer 3

If the expected rate of passage is $\lambda$ cars per unit of time, then the probability of $n$ cars passing in $t$ units of time is $$\mathbb P(N=n)=e^{-\lambda t}\frac{(\lambda t)^n}{n!}$$

so it does not matter whether

• you say $\lambda=3$ cars per minute and $t=60$ minutes,
• or you say $\lambda=180$ cars per hour and $t=1$ hour.

In both cases you have $\lambda t=180$ cars expected to pass in the time period and so $$\mathbb P(N=n)=e^{-180}\frac{180^n}{n!}.$$