Why does $\frac{Y_1+...+Y_n-n}{\sqrt{n}}\Rightarrow Z$ implies $\Bbb{P}\left(\frac{Y_1+...+Y_n-n}{\sqrt{n}}\leq 0\right)\rightarrow \Bbb{P}(Z\leq 0)$?

Question

I have a general question which appears in the following example I will describe. Let $Y_1,...$ be a sequence of independent poisson random variable with parameter $1$, i.e. $Y_k\sim Poi(1)$ for all $k$. Then we know that $$\frac{Y_1+...+Y_n-n}{\sqrt{n}}\stackrel{(d)}{\rightarrow} Z\sim N(0,1)~~~~~~~~~~~~~~~~~~~~~~(1)$$where $\stackrel{(d)}{\rightarrow}$ denotes convergence in distribution and $N(0,1)$ is a normal random variable with expectation value $0$ and variance $1$.

In the solutions they deduced from $(1)$ that $$\Bbb{P}\left(\frac{Y_1+...+Y_n-n}{\sqrt{n}}\leq 0\right)\rightarrow \Bbb{P}(Z\leq 0)~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)$$

I don't see why the convergence in distribution in $(1)$ implies $(2)$.

I saw this step many times but this is the last example I can remember.

Thanks for your help.

Answer 1

Convergence in distribution of $X_{n}\to X$ in particular implies that $$F_{n}(x)\to F(x)$$ pointwise(where $F$ is continuous which is everywhere in this case) where $F$ is the cdf of $X$ and $F_{n}$ is the cdf of $X_{n}$

Denote the random variable $X_{n}=\frac{Y_1+...+Y_n-n}{\sqrt{n}}$ and let it have cdf $F_{n}$.

So $$\Bbb{P}\left(\frac{Y_1+...+Y_n-n}{\sqrt{n}}\leq 0\right)=F_{n}(0)\to F(0)=\Bbb{P}(Z\leq 0)$$