Convergence in probability of a sequence of Poisson random variables

Question

Can anyone prove the following problem?

Let $\lambda_{n} = \dfrac{1}{n}$ for $n = 1,2, \ldots . $ Also, let $ X_{n} \sim \text{Poisson}(\lambda_{n}) $. Prove that if $ Y_{n} = n X_{n} $, then

$$ Y_{n} \xrightarrow{\text{P}} 0.$$

Regards,

Answer 1

$P(Y_n > \epsilon) = P(n X_n > \epsilon) = P(X_n > \epsilon /n )$

Then for any $\epsilon$ and $n$ large enough this the latter is equal to $P(X_n > 0 = 1 - P(X_n = 0)$, while the latter is equal to $e^{-\frac{1}{n}} \rightarrow 1$

so we have $P(\mid Y_n \mid > \epsilon) \rightarrow 0$

$Y_n$ can't take negative values, so showing convergence of positive part to zero suffices.

Answer 2

The characteristic function of $Y_n$ is $\exp\frac{\exp(int)-1}{n}$. As $n\to\infty$, this $\to\exp0=1$ for all $t\in\Bbb R$, since $|\exp(int)-1|\le2$. Thus $Y_n$ distributionally converges to the constant random variable equal to $0$.