# Sum of Poisson process and uniform distribution

I'm studying for my final and have the following question. I'm looking for clarification of my first answer and how to approach part b) and c).

Men and women arrive at a store according to independent Poisson processes with hourly rates $$λ_M$$ = 3 and $$λ_F$$ = 4, respectively. Men shop for a time that is uniformly distributed on [0, 1], and women shop for a time that is uniformly distributed on [0, 2] (in hours).

a) What is the expected number of people who arrive at the store during the interval [0, 5]?

b) Given that 6 people arrive during the interval [2, 3], what is the probability that exactly 2 of them are men?

c) Given that Sam Smith arrived during the interval [4, 5], what is the probability that he is still in the store at time 5?

For part a I have that Y = M + F ~ $$P(7)$$ (per hour).

So $$E(5Y) = 5*E(Y) = 35$$

For part b) Binomial(6,3/7)(2) which is $${6\choose{2}}(\dfrac{3}{7})^2 (\dfrac{4}{7})^4 = 0.29$$

Seems I am answering my own question haha! If anyone could confirm this is correct that would be great, thanks.

• For b, you actually want $P(M=2|M+F=6)$,which equals $\frac{P(M=2,F=4)}{P(M+F=6)}$ Commented May 2, 2020 at 8:10

You need to remember that the sum of two independent Poisson distributed random variables also follows a Poisson distribution - you simply add their individual arrival rates to obtain the arrival rate of their sum. In other words, if $$M, F$$ denote the number of male and female customers who arrive at the store in a time interval of unit length, respectively, then the total arrivals follow

$$N = M+F = \operatorname{Poisson}(\lambda_M + \lambda_F) = \operatorname{Poisson}(7)$$

Given this, the answer to a. is clearly $$5 \cdot 7 = 35$$. For the second problem, you're being asked the conditional probability $$\mathbb P(M = 2 | N = 6)$$. You can write this as

$$\frac{\mathbb P(M = 2 \land N = 6)}{\mathbb P(N = 6)} = \frac{\mathbb P(M = 2) \cdot \mathbb P(F = 4)}{\mathbb P(N = 6)} = \frac{\lambda_M^2 \lambda_F^4}{(\lambda_M + \lambda_F)^6} \cdot \frac{6!}{2! \cdot 4!}$$

where the last step simply uses the definition of the Poisson distribution. Substituting in the values in the question gives $$\approx 0.293$$.

For c. you should remember that the arrival time of a Poisson distribution, given that an arrival has indeed happened, is uniformly distributed on the interval. In other words, an arrival process following the Poisson distribution is "translation invariant". This tells you the arrival time of Sam Smith, conditional on the fact that he arrived, is uniformly distributed on the interval $$[4, 5]$$.

Since his shopping duration is uniformly distributed on $$[0, 1]$$, the time he leaves the store will be distributed as the sum of two independent uniform distributions. The joint probability distribution function is just a uniform distribution on a square $$[4, 5] \times [0, 1]$$, and the arrival time distribution comes from integrating along the diagonals of the square. That picture clearly tells you the probability Sam Smith is still in the store by time $$5$$ is exactly $$1/2$$, since a diagonal cuts a square into two pieces of equal area.

• Oh nice! I found b a different way. Why do you raise it to the power? Commented May 2, 2020 at 8:17
• @walle_whale The definition of a Poisson distribution with arrival rate $\lambda$ is $\mathbb P(\operatorname{Poisson}(\lambda) = k) = \exp(-\lambda) \lambda^k / k!$. I'm simply substituting in the definition. Commented May 2, 2020 at 8:21
• Ok that makes sense, thanks @starfall! Commented May 2, 2020 at 8:25