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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?

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  • $\begingroup$ That's the answer. (You might also mention that $x$ is any nonnegative integer.) $\endgroup$ – littleO Nov 1 '18 at 9:23
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    $\begingroup$ The pmf is more correctly written as $$P(X=x)=\begin{cases}\frac{e^{-\lambda}\lambda^x}{x!}&,\text{ if }x=0,1,2,\ldots\\\quad0&,\text{ otherwise }\end{cases}\quad,\,\lambda>0$$ $\endgroup$ – StubbornAtom Nov 1 '18 at 9:47
  • $\begingroup$ 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. $\endgroup$ – Aditya Dua Nov 1 '18 at 22:04
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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 $\lambda > 0$.

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