Asymptotic Normality of MLE Poisson Parameter (Lambda)

2.2k Views Asked by At

I'm need help with asymptotic normality of MLE.

Example:

$X_1,..., X_n$ iid $X_i$ ~ $Poisson(\lambda) $

Likelihood:

$L(\lambda)=\prod_{i=1}^{n} \frac{e^{\lambda}\lambda^{x_i}}{x_i!}$

Log-Likelihood:

$l(\lambda) = -n\lambda + \sum_{i=1}^{n}x_iln(\lambda)-\sum_{i=1}^{n}ln(x_i!)$

Score Function:

$U(\lambda)= -n +\frac{ \sum_{i=1}^{n}x_i}{\lambda}$

Fisher Information:

$I_{e}^{}(\lambda) = \frac{n}{\lambda}$

and

$ \frac{1}{I_{e}^{}} = I_{e}^{-1}(\lambda) = \frac{\lambda}{n}$

MLE:

$\hat{\lambda} = \sum_{i=1}^{n} \frac{x_i}{n}$

So far I understand and I know do the calculations. But for asymptotic normality I don't know if this it correct.

Asymptotic Normality:

$\sqrt{n}(\frac{\hat{\lambda}-\lambda}{Var(\hat{\lambda})})$ ~ $N(0, 1)$

$\sqrt{n}(\hat{\lambda}-\lambda)$ ~ $N(0, Var(\hat{\lambda}))$

$(\hat{\lambda}-\lambda)$ ~ $N(0, I_{e}^{-1}(\lambda))$

$(\hat{\lambda}-\lambda)$ ~ $N(0, \frac{\lambda}{n})$

$\hat{\lambda}$ ~ $N(\lambda,\frac{\lambda}{n})$

That's enough to show the asymptotic normality of $\lambda$?