Adjusting poisson distribution lambda

If I have a poisson distribution for some variable x such that $\lambda = 100$, is it possible to linearly adjust that variable to generate probabilities for smaller increments?

For example, if I have on average 100 customers arriving an hour ($\lambda = 100$) can I generate probabilities for a half-hour period by simply halving lambda ($\lambda = 50$)?

If not, how can lambda be adjusted to allow for calculations in a different time frame?

Solution 1:

Let us look at your example. You have a random variable $X$, which has Poisson distribution, with parameter $\lambda=100$. The probability that there will be $k$ "events" in time $t$, where $t$ is measured in hours, is
$$e^{-100t}\frac{(100t)^k}{k!}.$$

Let $Y$ be a new random variable, the number of events in time $s$, where $s$ is time measured in units of $1/4$ hour. I changed the letter used for time, so as to avoid confusion with time as measured in hours. And I am using a new letter for the random variable, to prevent confusion with $X$.

The random variable $Y$ has Poisson distribution, with parameter $\theta=100/4=25$. In other words, the parameter scales as one would expect. After all, for the Poisson, the parameter is the mean.

Then for example the probability that there will be $k$ events in time interval of length $s=3$ (quarter-hours) is given by

$$e^{-75}\frac{(75)^k}{k!}.$$

(We did not need to invent the new $Y$ to find this probability. In the original $X$, we could simply let $t=0.75$.)

The probability of $k$ events in time interval $s=4$ (aka $1$ hour) is also easily computed. Here $\theta s=100$, and you get probability $$e^{-100}\frac{(100)^k}{k!}.$$ Not exactly a big surprise! But it is a partial check on the correctness of the assertion.

So yes, you can "linearly" adjust, as long as you know what you are doing. That fact is built into the fact that everywhere that $t$ occurs in the formula, it comes as the part of the "package" $\lambda t$.

Solution 2:

(Original answer deleted.)

EDIT: If $X = \{X_t: t \geq 0\}$ is a Poisson process with rate $\lambda$, then $$ {\rm P}(X_t=k)=\frac{{e^{ - \lambda t} (\lambda t) ^k }}{{k!}}. $$

Suppose that $\lambda=100$. If $t=1$, then $$ {\rm P}(X_t=k)=\frac{{e^{ - 100} (100) ^k }}{{k!}}, $$ while if $t=1/2$, then $$ {\rm P}(X_t=k)=\frac{{e^{ - 100/2} (100/2) ^k }}{{k!}}. $$