0
$\begingroup$

Let $N$ be the pmf of a Poisson distribution, $Z$ be a nonnegative random variable, I read in the answer to another question that \begin{align} \mathbb{P}(N(Z)=0) &= \mathbb{E}\mathbb{P}(N(z)=0|Z=z)\\ &= \mathbb{E}e^{-Z}. \end{align}

I get the first equal sign is by Law of total expectation, but I don't understand the second equal sign. Is $\mathbb{P}(N(Z)=0|Z = z) = e^{-Z}$? Intuitively, I can see this equation is correct, but looking at the conditional part, I want to substitute the random variable $Z$ with the given constant $z$ and get $e^{-z}$. Also I tried to use the Bayes theorem, apparently it's wrong but I don't know where I got it wrong, \begin{align} \mathbb{P}(N(Z)=0 | Z=z) &= \frac{\mathbb{P}(N(Z)=0, Z=z)}{\mathbb{P}(Z=z)}\\ &=\frac{\mathbb{P}(N(z)=0)}{\mathbb{P}(Z=z)}\\ &=\frac{e^{-z}}{f(z)} \end{align} Can someone help me understand this concretely? That would be much appreciated.

$\endgroup$

1 Answer 1

Reset to default
0
$\begingroup$

That other post defines $N(\,.)$ as a Poisson process and not as the PMF of a Poisson distribution. We know that $$ \mathbb P(N(t)=n)=\frac{\lambda^nt^n}{n!}e^{-\lambda t}\,. $$ In particular, when the rate $\lambda$ is one, $$ \mathbb P(N(t)=0)=e^{-t}\,. $$ When $Z$ is a nonnegative random variable that is independent of $N$ then $$ \mathbb P(\,N(Z)=0\,|\,Z\,)=e^{-Z}\,. $$ From this $$ \mathbb P(N(Z)=0)=\mathbb E[e^{-Z}] $$ follows trivially by the tower property of conditional expectations.

The other post seems to neglect that $Z$ needs to be independent of $N$ and uses a confusing notation: $\mathbb P(\,N(Z)=0\,|\,Z\color{red}{=z}\,)$ is the deterministic function $e^{-\color{red}{z}}$ of which we do not have to take an expectation.

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.