# 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:
$$\lim_{n\to\infty}p(X_n=0) = \lim_{n\to\infty}\text{e}^{-{1 \over n}} = \text{e}^{\lim_{n\to\infty}{-{1\over n}}} = \text{e}^0=1$$
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. – whisperer Feb 23 at 21:51
• Well $\epsilon > 0$ is the way to go with continuous distributions, so understandable! – jbowman Feb 23 at 22:58