# Convergence of Poisson Random Variable

For $$n \in N$$, if $$X_n \sim Poisson(\frac{1}{n})$$ then

PT: 1. $$X_n \xrightarrow[n\rightarrow \infty]{P} 0$$

1. $$nX_n \xrightarrow[n\rightarrow \infty]{P} 0$$

It says $$X_n$$ converges to 0 in probability.

Attempt:

$$\forall \epsilon >0,$$

$$Pr(|X_n - 0| > \epsilon ) = 1 - e^{-\lambda} \Sigma_{i=0}^{\lfloor{\epsilon}\rfloor} \frac{\lambda^i} {i!}$$

Putting $$\lambda = \frac{1}{n}$$ in the above equation, 1 is proved by saying as limit of n approaches $$\infty$$, the summation is a finite sum of quantities approaching 0 therefore it is 0.

Q1. Is there a better way to prove ?

Q2. How can we prove 2 ?

Since $$X$$ is discrete, you can simplify a little:

$$\lim_{n\to\infty}p(X_n=0) = \lim_{n\to\infty}\text{e}^{-{1 \over n}} = \text{e}^{\lim_{n\to\infty}{-{1\over n}}} = \text{e}^0=1$$

where we can go from the second to the third term by the continuity of the exponentiation function.

The second statement follows from the first, as $$n\cdot0 = 0$$ and $$n\cdot X \neq 0$$ if $$X \neq 0$$, so $$p(nX_n=0) = p(X_n=0)$$, and since they are equal $$\forall n$$, their limits are equal too.

• I was focused on $\epsilon >0$ and totally lost the point of the question. Feb 23, 2020 at 21:51
• Well $\epsilon > 0$ is the way to go with continuous distributions, so understandable! Feb 23, 2020 at 22:58

Let $$X = \lim_{n\rightarrow\infty}X_n$$, then we have
\begin{align*} P(X=k) & = \lim_{n\rightarrow\infty}\,P(X_n=k) \\ & = \lim_{n\rightarrow\infty}\frac{1}{e^{\frac{1}{n}}n^{k}k!} & = \begin{cases} & 1;\qquad X_n = 0 \\ & 0;\qquad \text{otherwise} \end{cases} \end{align*}
Now, using this definition of $$X$$, we get
\begin{align*} \lim_{n\rightarrow \infty} P(|nX_n - X| > \epsilon) & = \lim_{n\rightarrow \infty} P(|nX_n|>\epsilon) \\ & = \lim_{n\rightarrow\infty} P(X_n > \frac{\epsilon}{n}) \\ & = 1 - \lim_{n\rightarrow\infty} P(X_n \leq \frac{\epsilon}{n}) \\ & = 1 - 1 = 0\qquad\text{since, }P(X\leq 0) = 1 \end{align*}