[Math] How to use the Cramer-Rao lower bound (CRLB)to show that $\bar{Y}$ is the best unbiased estimator of $\lambda$

Question

Let $Y_1,\ldots,Y_n$ be a random sample from Poisson ($\lambda$).

Derive the Cramer-Rao lower bound (CRLB) for the variance of any unbiased for estimator of $\lambda$.

MY APPROACH:

$$
L(\lambda)=\left(\prod_{i=1}^n \frac{1}{y_i!}\right)\lambda^{\sum_{i=1}^n y_i}e^{-\lambda n}
$$

$$
l(\lambda)=-\lambda n+\log(\lambda)\sum_{i=1}^n y_i-\sum_{i=1}^n \log(y_i!).$$

$$\frac{d \ln L(\lambda)}{\lambda} = \frac{1}{\lambda} \sum_{i=1}^n y_i – n$$

$$ \frac{d^2 \ln L(\lambda)}{\lambda} = \frac{-1}{\lambda^2} \sum_{i=1}^n y_i$$

$$E\left[\frac{-1}{\lambda^2} \sum_{i=1}^n y_i\right] = \frac{-1}{\lambda^2} n \bar{y}$$.

Can anyone please confirm what I am doing is correct? If so , how do I use this result to show that $\bar{Y}$ is the best unbiased estimator of $\lambda$?

Answer 1

The last equation confuses $\bar{y}$ with $\lambda$. Actually, $E(\sum y_i) = \sum E(y_i) = n \lambda$

With this correction, you get that the CRB (for any unbiased estimator of $\lambda$) is $\lambda/n$

Now, verify that the estimator (of $\lambda$) $\bar{y} = \sum y_i/n$ is unbiased, compute its variance, and compare it with the CRB.