I am struggling to find a common mean of 2 Poisson distributions. For example, suppose the number of customers entering a shop follows a Poisson distribution. We are given that the mean number is 8 per hour everyday except Sundays and 3 per hour on Sundays. The staffing would be decided according to the mean number of customers per hour over 6 months. (So roughly 26 days where the mean is 3, and 154 days where the mean is 8).

Is it possible to find a common mean for the Poisson distribution for this 6 month period, and how could I go about this?



1 Answer 1


Let's look at one week, assuming 8 hours of business each day of the week. Then the number of customers on Monday will have a Poisson distribution with $\lambda_1 = 64$ customers. Similarly, Tuesday through Saturday will have $\lambda_i = 64$ per day.

The sum of two Poisson random variables is Poisson, where the rates add. So the total customers Monday through Saturday will be Poisson with rate 384, and the total customers per week will be Poisson with rate $384+24 = 408.$ That is $T \sim \mathsf{Pois}(\lambda = 408)$ and $E(T) = Var(T) = 408.$

However, the average per hour will not be Poisson because the average can take non-integer values. The average per hour will be a non-Poisson random variable $A$ with $$E(A) = E\left(\frac{1}{56}T\right) = \frac{1}{56}E(T)= \frac{408}{56} = 7.2857,$$

$$Var(A) = Var\left(\frac{1}{56}T\right) = \frac{1}{56^2}Var(T)= \frac{408}{56^2}$$

The random variable $A,$ although not Poisson, is approximately normal, so you can find approximate probabilities that $A$ lies in various intervals.

The problem for six months can be solved similarly.

Addendum: Here is a simulation for the weekly version of the problem discussed above; the normal approximation will be better for 6 months.

t = replicate(10^6, sum(rpois(6, 64)) +rpois(1,24))
mean(t);  var(t)
[1] 408.0416     # aprx E(T) = 408
[1] 408.9242     # aprx Var(T) = 408
a = t/56
mean(a);  var(a)
[1] 7.286458
[1] 0.1303967

hist(a, prob=T, br=20, col="skyblue2")
 curve(dnorm(x, mean(a), sd(a)), add=T, col="red")

enter image description here

  • 1
    $\begingroup$ Thank you; this is very helpful. I couldn't get my head around what distribution the average would follow. $\endgroup$
    – Matt
    Commented Sep 9, 2020 at 7:49
  • $\begingroup$ Just a follow-up- suppose we want to adequate staffing (say 1 staff to 1 customer) with say 95% confidence. Using the Poisson with a mean of 8, 15 staff would be required (from the Poisson table); however, if I use a normal distribution with mean 7.2857 and SD=sqrt of 0.130102, I get 8. I was wondering if one would expect that much a difference between the Poisson and the normal approximation? $\endgroup$
    – Matt
    Commented Sep 9, 2020 at 8:06
  • $\begingroup$ The $A$ above is a weekly average. If you would consider a monthly or a yearly average, $E[A]$ would be the same, but $Var[A]$ would be even further reduced. So you are basing your staffing decision on a quantity that has vanishing variance. That is not a good idea because you are not reserving any staff at all to cope with the random fluctuations in hourly customer numbers. $\endgroup$ Commented Sep 9, 2020 at 8:23
  • $\begingroup$ I don't think you're getting the same kind of information looking at the Poisson table as using the normal mean and variance. Poisson arrivals have exponential interarrival times so (without using a queueing theory approach) it will be difficult to see how to ensure that no customer would need to wait for attention. And it gets even more complicated if you have to take fluctuations in arrival rates during the day. Best strategy in practice may be to use analysis such as in this problem to guess at adequate staffing and then add or overlap staff at times that prove to be busy. $\endgroup$
    – BruceET
    Commented Sep 9, 2020 at 8:27

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