1
$\begingroup$

Consider the random variable $N_n$ drawn from a Poisson distribution with intensity parameter $n$ so that $E(N_n)=n$.

Could you help me to show that $\frac{N_n}{n}\rightarrow_p 1$ as $n\rightarrow \infty$?

My doubts: I do not understand what happens to $N_n$ as $n\rightarrow \infty$. If I consider the probability distribution of the Poisson, when $n\rightarrow \infty$ the probability of observing $X_n=x$ goes to zero. Is this somehow related to the statement above?

$\endgroup$
3
$\begingroup$

It's true that the probability of $N_n$ taking any fixed value goes to zero as $n\to\infty$ but that has little to do with what it's value will tend to be. It is simply a reflection of the fact that there are a wider range of likely values, so the probability of any one of them must be very small.

The slickest way to prove your statement is probably to show that a Poission with mean $n$ is equal in distribution to the sum of $n$ independent Poissons with mean $1$ and then use the law of large numbers.

$\endgroup$
  • $\begingroup$ Thanks. So $N_n\sim\sum_{i=1}^n Y_i$, where $Y_i$ is distributed as a Poisson with mean $1$ . Hence $plim\Big(\frac{N_n}{n}\Big)=plim\Big( \frac{n\frac{1}{n}\sum_{i=1}^n Y_i}{n}\Big)=plim\Big( \frac{1}{n}\sum_{i=1}^n Y_i\Big)\overbrace{=}^{\text{LLN}} E(Y_i)=1$. $\endgroup$ – STF Nov 14 '17 at 10:20
  • $\begingroup$ Yes that is right $\endgroup$ – spaceisdarkgreen Nov 14 '17 at 13:25

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.