2
$\begingroup$

In Poisson point processes, we first let $N$ be a random Poisson random variable, with parameter $\lambda$: $$ P(N=n)=\frac{e^{-\lambda}\cdot\lambda^n}{n!},\qquad n=0,1,\ldots $$ Without loss of generality, we let $Y = (Y_1, Y_2, \dots, Y_n,\dots)$ be an i.i.d. sequence (infinite) of standard normal random variables. In addition, assume $N$ is independent of $Y$. $$ f_{Y_i}(y_i)=\frac{1}{\sqrt{2\pi}}e^{-\frac{y_i^2}{2}},\qquad -\infty<y_i<\infty $$ Now we take an arbitrary interval $[a, b]$ and let $X$ be the number of $Y_i$s landing on the interval $[a,b]$, I want to know what is the distribution of $X$ here, I know it is a little bit complicated...

I searched Wikipedia and read the section on the uniformity of the points, but I still not getting it.

$\endgroup$
4
$\begingroup$

I think you're confusing things here because this has nothing to do with the distribution of points of the Poisson Process. Fix $N$ for a moment. Then the number of $Y_i$'s that fall within $[a,b]$ is binomially distributed: define $p:=p({[a,b]})=P(Y_1\in [a,b])=F(b)-F(a)$, where $F$ is the CDF of the standard normal. Then $X$ is binomially distributed (conditional on $N$ fixed): $P(X|N)=\binom{N}{X}p^X(1-p)^{N-X}$.

You're interested in $N$ being Poisson distributed. Since you're allowed $N=0$ with positive probability, lets define $P(X=0|N=0)=1$. Above we calculated the conditional distribution of $X$ so:

$$P(X)=\sum_{N=0}^\infty P(X|N)P(N),$$

which looks like it can be expressed in terms of generalized gamma functions here.

$\endgroup$
4
  • $\begingroup$ Thanks so much for your help. It is really a nice point that we define p as the probability that $Y_i$ is in [a,b] so $p = F(b)-F(a)$ But I still wonder whether the binomial result is the probability mass function, or the conditional distribution..... Again thanks so much! $\endgroup$
    – son520804
    Dec 1 '17 at 18:59
  • 1
    $\begingroup$ I'm not sure I understand your question. $X$ is binomially distributed, conditional on the value of $N$. $\endgroup$
    – Alex R.
    Dec 1 '17 at 21:04
  • $\begingroup$ Thanks, I was asking the wrong question...... X|N should be binomially distributed but ..... X itself should incorporate the poisson process, to get the poisson distribution.~~ You help~ $\endgroup$
    – son520804
    Dec 1 '17 at 21:43
  • 1
    $\begingroup$ If you do the algebra correctly, you will see that the unconditional distribution of X is also Poisson, with a parameter λp = λ[F(b)-F(a)] $\endgroup$
    – Zahava Kor
    Dec 1 '17 at 21:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.