Distribution Poisson and $\lim_{n \rightarrow +\infty} \mathbb{P}(Y_n \geq \lambda n + x \sqrt{n})$ If $Y_n$ follows a distribution Poisson $(\lambda n)$ with $n \in \mathbb{N}$ and $x \in \mathbb{R}$.
We have to determine $\lim_{n \rightarrow +\infty} \mathbb{P}(Y_n \geq \lambda n + x \sqrt{n})$.
Is there a method to find this ? Could somoeone help me (I work as an autodidact and haven't classes). Thank you in advance.
 A: Let $X_1, X_2, X_3, \dots$ are i.i.d. Poisson$(\lambda)$. Let $Y_n=X_1+X_2\dots +X_n$.
Using CLT $$\frac{Y_n-n\lambda}{\sqrt{n\lambda}}\Rightarrow Z\equiv N(0,1),\text{as}\space n\to\infty$$
Thus $\lim_{n \rightarrow +\infty} \mathbb{P}(Y_n \geq \lambda n + x \sqrt{n})=\lim_{n \rightarrow +\infty} \mathbb{P}(\frac{Y_n-n\lambda}{\sqrt{n\lambda}}\geq \frac{x}{\sqrt{\lambda}})=\mathbb{P}(Z\geq \frac{x}{\sqrt{\lambda}})=\Phi(-\frac{x}{\sqrt{\lambda}})$.
A: The fact that it's about a limit as $n\to\infty$ is one indication that the central limit theorem may be involved.
The expected value is $n\lambda$ and the standard deviation is $\sqrt{n\lambda}.$
You have $\dfrac{x\sqrt n}{\sqrt{n\lambda}} = \dfrac x {\sqrt\lambda}.$
\begin{align}
& \Pr(Y_n \ge \lambda n + x\sqrt n) = \Pr\left( \frac{Y_n-n\lambda}{\sqrt{n\lambda}} \ge \left( \frac x {\sqrt\lambda} \cdot 1\text{ standard deviation} \right) \right) \\
\approx {} & \Pr\left( Z \ge \frac x {\sqrt\lambda} \right) \text{ where } Z \text{ has a standard normal distribution.}
\end{align}
