Asymptotic standard normal distribution I need to solve the following exercise.


  Assume that $X_\lambda$ is Poisson distributed with mean $\lambda$ . Show that $Y(\lambda) = \frac{X_\lambda - \lambda}{\sqrt{\lambda}}$ is asymptotic standard normal distributed for $\lambda \to \infty$.


Any help would be really appreciated.
Not quite sure if the one on Wiki is right or if I understand it, but anyway I'm basing my answer on CLT in Larsen and Marx (an elementary probability book):





Anyway, this answer might be weird or very wrong or stupid but...

Consider iid $X_i$ ~ $Poi(\lambda/n)$.

Then $X := \sum_{i=1}^{n} X_i$ ~ $Poi(\lambda)$

Then

$$(\frac{1}{n} \sum_{i=1}^{n} X_i - \mu) / (\sigma/\sqrt{n})$$

$$= (\frac{1}{n} \sum_{i=1}^{n} X_i - \lambda/n) / (\sqrt{\lambda}/n)$$

$$= (\sum_{i=1}^{n} X_i - \lambda) / (\sqrt{\lambda})$$

$$= Y(\lambda)$$



Now I'm thinking we have to come up w/ some kind of relationship between n and $\lambda$ s.t. as $n \to \infty$, $\lambda \to \infty$



Guess 1:

$$= (\sum_{i=1}^{(\frac{\lambda}{E[X_i]})} X_i - \lambda) / (\sqrt{\lambda})$$

I was thinking as $\lambda \to \infty \iff n \to \infty$ as $\lambda = n E[X_i]$ but then again $E[X_i]$ is a function of both n and $\lambda$



Guess 2:

$P(X_i = k) \ge 0 \to (\frac{\lambda}{n})^k \ge 0$

Not sure how to proceed. I guess $\exists a &gt; 0$ s.t. $\lambda^k = a n^k$ ?
Not quite sure if the one on Wiki is right or if I understand it, but anyway I'm basing my answer on CLT in Larsen and Marx (an elementary probability book):





Anyway, this answer might be weird or very wrong or stupid but...

Consider iid $X_i$ ~ $Poi(\lambda/n)$.

Then $X := \sum_{i=1}^{n} X_i$ ~ $Poi(\lambda)$

Then

$$(\frac{1}{n} \sum_{i=1}^{n} X_i - \mu) / (\sigma/\sqrt{n})$$

$$= (\frac{1}{n} \sum_{i=1}^{n} X_i - \lambda/n) / (\sqrt{\lambda}/n)$$

$$= (\sum_{i=1}^{n} X_i - \lambda) / (\sqrt{\lambda})$$

$$= Y(\lambda)$$



Now I'm thinking we have to come up w/ some kind of relationship between n and $\lambda$ s.t. as $n \to \infty$, $\lambda \to \infty$



Guess 1:

$$= (\sum_{i=1}^{(\frac{\lambda}{E[X_i]})} X_i - \lambda) / (\sqrt{\lambda})$$

I was thinking as $\lambda \to \infty \iff n \to \infty$ as $\lambda = n E[X_i]$ but then again $E[X_i]$ is a function of both n and $\lambda$



Guess 2:

$P(X_i = k) \ge 0 \to (\frac{\lambda}{n})^k \ge 0$

Not sure how to proceed. I guess $\exists a &gt; 0$ s.t. $\lambda^k = a n^k$ ?