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AI HL · 13.2

Probability Distributions

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13.2.12

The sum of two independent Poisson variables

13.2.12

The sum of two independent Poisson variables

Target: I can state that if X ~ Po(λ) and Y ~ Po(μ) are independent, then X + Y ~ Po(λ + μ). (+3 more)

By the end of this section

Key points

Adding Poisson variables

  • If \(X \sim \text{Po}(\lambda_X)\) and \(Y \sim \text{Po}(\lambda_Y)\) are independent and \(Z = X + Y\), then \(Z \sim \text{Po}(\lambda_X + \lambda_Y)\).
  • The means add. Both variables must count events in the same period, so rescale first if they do not.
  • Define Z in words ("the total number of hits per hour"), write its distribution, then use the calculator.
  • This is for a sum only. Do not use it for a difference \(X - Y\).

Questioning the model

  • Two assumptions are often unrealistic: a constant average rate, and independence.
  • Constant rate: say when the rate changes in the context (busier at some times of day or on some days).
  • Independence: say how one count could be linked to the other (the same people, the same weather, the same cause).
  • Answer in the context. "The assumptions are unrealistic" alone earns nothing.

On the calculator

Poisson cdf for a total

  • Add the two means to get λ for Z.
  • Enter λ, lower and upper, as for any Poisson cdf.
  • For P(Z > k), find P(Z ≤ k) and take it from 1.

Screenshot and video to come

Explore: the Poisson distribution

X counts the events in a period, at a constant average rate. Set the rate and the number of periods, then choose which values of X to add up, or press a bar.

=

λ = m × t
E(X) = λ
Var(X) = λ
Standard deviation, √λ

Move a slider to change the distribution.

  1. Set m = 4, then change t from 1 to 2 to 3. What does t do to the mean? What happens to the shape?
  2. Compare the mean with the variance each time you move a slider. What do you notice? Is that true for a binomial distribution?
  3. Move the upper slider to its largest value (∞). Which probability do you take away from 1 to get the same answer?
  4. Add the second variable. Where does the gold curve for X + Y sit compared with the two means? Why must the means add?

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.
Two independent sources

A weather station records lightning strikes over a lake. Strikes on the north shore occur at a constant average rate of 5.2 per hour and strikes on the south shore at 4.7 per hour. Assume the two counts are independent Poisson variables. Find the probability that more than 12 strikes are recorded in one hour.

Working
  1. Define Z

    Let Z be the total number of strikes per hour. Z = X + Y

    X: north shore, Y: south shore.

  2. Add the means

    Z ~ Po(5.2 + 4.7) = Po(9.9)

    The sum of two independent Poisson variables is Poisson, with the means added.

  3. Complement

    P(Z > 12) = 1 − P(Z ≤ 12)

  4. Poisson cdf

    From the calculator: Poisson cdf with λ = 9.9, lower = 0, upper = 12

  5. Answer

    P(Z ≤ 12) = 0.8009, so P(Z > 12) = 0.199

    3 significant figures at the end.

Video walkthrough coming
Why the model may fail

A school reception receives calls from parents at a constant average rate of 3.5 per hour and calls from suppliers at 1.5 per hour. Assume the two counts are independent Poisson variables. Find the mean number of calls per hour in total. Then explain why the assumption that these are independent Poisson variables is unlikely to hold.

Working
  1. The total

    Z ~ Po(3.5 + 1.5) = Po(5)

    Add the means. The mean of Z is 5.

  2. Constant rate

    The rate is unlikely to be constant: parents call before and after school, suppliers call in the morning.

    Say when and why the rate changes in this context.

  3. Independence

    The counts are unlikely to be independent: on a day with a school trip or a closure, both parents and suppliers call more.

    Name a cause that links the two counts.

Video walkthrough coming
Different periods

A kiosk sells coffees at a constant average rate of 8 per hour and teas at a constant average rate of 3.5 per 30 minutes. Assume the two counts are independent Poisson variables. Find the probability that at least 18 hot drinks are sold in one hour.

Working
  1. Same period

    Teas for 1 hour: λ = 3.5 × 2 = 7

    Both means must be for the same period before they are added.

  2. Add the means

    Let Z be the total number of drinks per hour. Z ~ Po(8 + 7) = Po(15)

  3. Complement

    P(Z ≥ 18) = 1 − P(Z ≤ 17)

    At least 18 starts at 18, so the cdf stops at 17.

  4. Poisson cdf

    From the calculator: Poisson cdf with λ = 15, lower = 0, upper = 17

  5. Answer

    P(Z ≤ 17) = 0.7489, so P(Z ≥ 18) = 0.251

Video walkthrough coming