# Which assumption is correct about a Poisson Distribution?

Let's assume that the results of sampling over 3 months are Poisson distributed with mean $\mu$. Now we want to know the probability of an event over 1 year but we cannot resample over one year. Which is the correct way to proceed:

1. Assume that a sampling over a year is Poisson distributed and that the mean for 1 year is $4\mu$.
2. Calculate the probability for that event over the sampled 3 month period and then raise the probability to the fourth power?
• Welcome to CV! I think you should explain clearly what you mean "the probability of an event over 1 year" and "the probability for that event over the sampled 3 month period". Do you mean the probability that the event happens once? That the event happens at least once, i.e. once or more? Or perhaps the probability that the event does not happen? Could you explain why in the second option, you thought you should raise the probability to the fourth power - what rule were you attempting to use? Sep 30, 2015 at 21:19
• (I also don't understand what you mean about "sampling over 3 months". Why do we need to take a sample, if we already know $\mu$? Or are you saying that we are going to estimate $\mu$ based on our three month sample?) Oct 1, 2015 at 1:02
• @SIlverfish I think this is a pure probability self study with no actual estimation, in spite of the potential implications of "sampling". Wording aside I think the Q is really just about relationships between probabilities when observing the process for 3 months vs 12 months. Oct 1, 2015 at 1:37

Sums of independent Poisson random variables are also Poisson random variables, with rate equal to the sum of the rates of the components. So, for jointly independent random variables $$X_1,...,X_n$$ having distributions $$X_i \sim \text{Pois}(\lambda_i)$$ we have:

$$\sum_{i=1}^n X_i \sim \text{Pois} \Big( \sum_{i=1}^n \lambda_i \Big).$$

This property applies to your question, so long as the counts over the individual three-month periods are considered to be jointly independent. So long as you are willing to make this assumption, it would follow logically that the count over a full year would be a Poisson random variable with rate equal to the sum of the rates over the component periods.

There are a few minor issues you should also consider here:

• It is not ideal to frame this analysis by saying that you will assume that the count over a year is a Poisson random variable. It is much better to frame your assumptions in more fundamental operational terms. In this case you would say that you will assume independence of the counts in the three-month periods, and the Poisson distribution of the annual count follows as a logical consequence of this assumption.

• Be careful when conducting analysis over periods quantified by months. Months have different numbers of days in them, and consecutive three-month periods also have different numbers of days. When dealing with these periods it is generally preferable to have a rate that is proportional to elapsed time, so that the rate parameter for different three-month periods will be slightly different. Treating each of these periods equally will be an approximation to this, but it implicitly assumes that events occur at a slightly faster rate during periods composed of less days. That is an undesirable property of the analysis.

I think your first assumption is correct.

Lex $X_1$ be the number of cases happen in the first 1-3 month.

$X_2$be the number of cases happen in the first 4-6 month.

$X_3$be the number of cases happen in the first 7-8 month.

$X_4$be the number of cases happen in the first 9-12 month.

And assume $X_1,X_2,X_3, X_4$ are i.i.d with $Poisson(\mu)$

Then $Y=X_1+X_2+X_3+X_4$

Then $Y$ has a distribution of $Poisson(4\mu)$

I guess you are trying to do a Epidemiology incidence rate study. The incidence rate will be measured by $\frac{Number\ of\ cases}{Person-time}$, usually, person-time is person-year

Suppose at the beginning you have 100 people without disease and you follow up them for 3 month and 10 get the disease. i.e your $\hat{\mu}=10$

The incidence rate will be $\frac{10 person}{100person*3month}=\frac{1}{30}month^{-1} \tag{1}$

From above we can infer the incidence rate is also equal to $\frac{10person*4}{100person*3month*4}=\frac{40person}{100person-year}=\frac{4}{10}year^{-1}\tag{2}$

We know (1) and (2) are equal and for (2) you can treat it as a Poisson distribution with $4*\hat{\mu}$ parameter i.e 40 cases per 100 person during 1 year period.

I think this book explain the incidence rate very well.