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I have this problem:

Let $X_n \sim \mathrm{Poisson}(1/n)$. Show that $X_n \to 0$ and $Y_n = n X_n \to 0$?

I already proved the first statement, but every time I try to show the second one I get a divergent result. Can somebody help me with this?

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    What sort of convergence are you interested in? The $\to$ can mean at least a few different things.2011-03-20

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It is easy to show that $Y_n \to 0$ in probability (hence also in distribution). Indeed, for any fixed $\varepsilon > 0$ and as $n \to \infty$, $$ {\rm P}(nX_n > \varepsilon ) = {\rm P}(X_n > \varepsilon /n) = 1 - {\rm P}(X_n \le \varepsilon /n) = 1 - {\rm P}(X_n = 0) = 1 - e^{-1/n} \to 0. $$

Exercise: Prove this using characteristic functions.

EDIT: Actually, for any sequence $(a_n)$ it trivially holds that $a_n X_n \to 0$ in probability, since $$ {\rm P}(a_n X_n = 0) \ge {\rm P}(X_n = 0) = e^{ - 1/n} \to 1. $$ As a further exercise, let's consider the case where $X_n \sim {\rm N}(0,1/n)$. Then, the characteristic function of $a_n X_n$ is given by $$ {\rm E}[e^{{\rm i}t(a_n X_n) } ] = {\rm E}[e^{{\rm i}(a_n t)X_n } ] = e^{ - n^{ - 1} a_n^2 t^2 /2}. $$ Thus we need $a_n^2 /n \to 0$ in order to have that $a_n X_n \to 0$ in distribution, hence also in probability (note that convergence in distribution to a constant $c$ implies convergence in probability to $c$; here $c=0$).

Remark. For the last example, rather than using characteristic functions, just notice that $a_n X_n$ is equal in distribution to $(a_n / \sqrt{n}) Z$, where $Z \sim {\rm N}(0,1)$.

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    I was interested in probability, the way I was doing it was squaring both sides inside the probability and using Markov inequality and then trying to show E(Y_n^2) \rightarrow 0, which is much more difficult than that...2011-03-20
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    I was interested in probability, the way I was doing it was P(|Yn|>epsilon)=P(|Yn|^2>epsilon^2) using Markov inequality and then trying to show E(Yn^2) -> 0, which envolve to show that E(Yn^2) +E(Yn)^2 ->0. But as always simplicity is the way!!2011-03-20
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    Indeed, expectations are not useful here: ${\rm E}(nX_n) = n(1/n)=1$, ${\rm Var}(nX_n) = n^2 (1/n) \to \infty$.2011-03-20
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    Yes I found that, thanks2011-03-20