Poisson Distribution

Topics: Probability

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(definition)

Let $X$ be a discrete random variable that handles the number of events that occur a given time or space interval, knowing that these events occur with a known constant mean rate and independently of the time since the last event.

For such an $X$, we write:

$$X\sim \mathrm{Poi}(\lambda )$$

…where $\lambda$ is the parameter that represents the number of times that an event is expected to occur in a given interval ($\lambda >0$). We use this interval and its $\lambda$ as our main references when talking about a Poisson distribution.

Density Function §

(theorem)

With an $X$ such that $X\sim \mathrm{Poi}(\lambda )$, its density function is:

$$f_X(x)=\{\begin{array}{ll}\frac{{\lambda }^x{e}^{-\lambda }}{x!}& \text{if }x=0,1,\dots ,\infty \\ 0& \text{otherwise}\end{array}$$

Proof that this is a density function

Remember that $f_X(x)$ is a density function if:

$$\sum _{x\in R_X}{f}_X(x)=1$$

The density function given here is, effectively, a density function, since:

$$\begin{array}{rl}\sum _{x=0}^{\infty }\frac{{\lambda }^x{e}^{-\lambda }}{x!}& =e^{-\lambda }\sum _{x=0}^{\infty }\frac{{\lambda }^x}{x!}\\ & =e^{-\lambda }{e}^{\lambda }\\ & =1\end{array}$$

Expected Value §

(theorem)

Such an $X$ has a very simple expected value:

$$\mathbb{E}[X]=\lambda$$

Variance §

(theorem)

Such an $X$ also has a very simple variance:

$$\mathrm{Var}(X)=\lambda$$

Notice it’s the same as its expected value.

Moment-Generating Function §

(theorem)

As for the moment-generating function of such an $X$, it is:

$$M_X(t)=e^{\lambda (e^t-1)}$$

Example §

Let $X$ be the number of messages that are rejected by a web server in a second. We know that 5 messages are expected to be rejected every second (i.e. we know that $X\sim \mathrm{Poi}(5)$).

We’ll calculate:

1. The probability that exactly 3 messages are rejected in a second
2. The probability that at most 2 messages are rejected in a second.

Exactly 3 messages §

Since $X$ is discrete, we can simply evaluate the density function at $x=3$:

$$\mathbb{P}(X=3)=f_X(3)=\frac{5^3{e}^{-5}}{3!}\approx 0.1403$$

At most 2 messages §

From the idea of a distribution function, we have that:

$$\begin{array}{rl}\mathbb{P}(X\le 2)& =\mathbb{P}(X=0)+\mathbb{P}(X=1)+\mathbb{P}(X=2)\\ & =\frac{5^0{e}^{-5}}{0!}+\frac{5^1{e}^{-5}}{1!}+\frac{5^2{e}^{-5}}{2!}\\ & \approx 0.12465\end{array}$$