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?
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?
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 $Z \sim {\rm N}(0,1)$.