How does this sequence of distributions converge?

Question

Say I have a sequence of positive parameters $\rho_1, \rho_2, \dots,$ whereby $\rho_1 > \rho_2 > \dots$ and $\lim_{n \to \infty} \rho_n = 0.$

Now for each $n \geq 1$ let $M_n \sim \hbox{Poi}(\rho_n)$ be a sequence of rvs characterised by the sequence $\{\rho_n\}_{n=1}^{\infty}$.

Clearly $\lim_{n \to \infty} \mathbb{P}(M_n = 0) = \lim_{n \to \infty} e^{-\rho_n} = 1$ since $\rho_n \to 0$. Does this imply that $M_n$ converges in distribution to the degenerate distributrion at $0$? I.e. that $M_n \overset{D}{\to} \delta_0$?

Is it true that $M_n \overset{a.s.}{\to} \delta_0$?

Any help would be much appreciated! Thanks :)

Answer 1

As Shalop mentioned, we can show that $M_n\to 0$ in probability. Indeed, let $\varepsilon\in (0,1)$. Then $$\mathbb P\left(\left\lvert M_n\right\rvert\gt \varepsilon \right)=1-\mathbb P\left(M_n =0\right)=1-e^{-\rho_n}$$ which goes to zero as $n$ goes to infinity.

In the case where $\left(M_n\right)_{n\geqslant 1}$ is independent, the use of the Borel-Cantelli lemmas shows that the sequence $\left(M_n\right)_{n\geqslant 1}$ converges to $0$ almost surely if and only if the series $\sum_{n=1}^{+\infty}\rho_n$ converges.