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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?

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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.

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  • $\begingroup$ 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

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