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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 ?

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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.

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

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