# The Poisson pmf

I have this the problem below, where I have to find the pmf.

Let $$X \sim \mathcal{Poiss}(\lambda)$$. Write down the pmf $$p\text{x}$$.

I know that a Poisson random variable has a PMF given by: $$P(X = x) = \frac{\lambda^xe^{-\lambda}}{x!}$$

So is this the answer to the problem, or am I missing something?

• That's the answer. (You might also mention that $x$ is any nonnegative integer.) – littleO Nov 1 '18 at 9:23
• The pmf is more correctly written as $$P(X=x)=\begin{cases}\frac{e^{-\lambda}\lambda^x}{x!}&,\text{ if }x=0,1,2,\ldots\\\quad0&,\text{ otherwise }\end{cases}\quad,\,\lambda>0$$ – StubbornAtom Nov 1 '18 at 9:47
• As a nitpick, avoid using $x$ for a discrete valued random variable. $P(X=k)$ or $P(X=n)$ would be a more typically used notation. As mentioned, it's a nitpick. – Aditya Dua Nov 1 '18 at 22:04

Yes, you have the correct equation for all nonnegative integer inputs of the PMF. However, you can improve your answer. A Poisson PMF with parameter $$\lambda$$ is more correctly written as: $$P(X = k) = \begin{cases} \frac{e^{-\lambda}\lambda^{k}}{k!} & k = 0,1,2,...\\ 0 & \text{otherwise} \end{cases}$$ It is also important to note that $$\lambda > 0$$.