# Correct Poisson Distribution Formulation

I have a Poisson formulation with the following parameters:

• I see 90 people in a city an hour
• 1.2% of people in the city are Indian

How would I calculate:

• The maximum amount of time before I would see any Indian person with the highest probability
• The amount of Indian people I would see in an hour with the highest probability
• This seems like a typical textbook problem, and as such should have the self-study tag. It'd also help if you could tell us what you've tried so far and where you get stuck. (Right now the question has been merely copy-pasted.) See stats.stackexchange.com/tags/self-study/info – Patrick Coulombe Sep 8 '14 at 2:08
• Um, no it is not a homework problem. I'm actually in a city with these statistics and trying to figure out how often I will see an Indian person. I have no idea what to try because I haven't done a Poisson distribution since I was in undergrad and had a bad teacher. It's ironic that I formulated it so clearly that you think I'm doing a homework problem. I finished my PhD in Computer Science 2 years ago. – Chris Redford Sep 8 '14 at 2:11
• The first of those questions doesn't make sense as phrased (since, for example probability is related to interval length; you can get better probability by waiting longer). Can you clarify what you mean? The second makes sense. – Glen_b -Reinstate Monica Sep 8 '14 at 3:01
• Okay. I've tried to make it clearer. I.e. what is the time with the highest probability before I would see an Indian person, given that I see 90 people an hour. E.g. clearly it would be less probable for me to go an entire week before I saw one than it would be for me to go 5 minutes before I saw one. So, phrased that way, the higher amount of time is actually less probable. – Chris Redford Sep 8 '14 at 3:33

## 1 Answer

Let $X$ denote the the time till the first occurrence of the event of meeting an Indian person. As I understand, you are looking for the value of $t$ that would maximize the probability $P(X=t)$. Since time is a continuous variable, we can only look at $P(t < X <t+dt)$ - the probability of being close to $t$ (because $P(X=t)=0$ for continuous variables), or equivalently the mode of the random variable $X$.

You are assuming a Poisson process with rate $\lambda=1/(90*0.012)$, so the distribution of $X \sim Exp(\lambda)$. The density function is $f(t)=\lambda \exp(-\lambda t)$, which is monotone decreasing in $t$, so the mode is $0$.

So the hour with the highest probability of seeing the first Indian person is the first hour (and the instant with the highest probability is when you start the observations). However the expected number of Indian people you see in an hour does not depend on the time of observation - this is a basic assumption of the Poisson process. It will always be $\lambda$ events in one unit of time.

• Thanks. Can you give me some examples with actual values plugged in? – Chris Redford Sep 10 '14 at 22:10