Need explanation of this result: $\frac{n}{\sum_n x_i Y_i} \sim N(\lambda,\frac{\lambda^2}{n})$ I have a distribution of an iid sample given by $$Y_i \sim Exp(\lambda x_i), \quad i=1\dots n$$
I have an estimator $$\hat\lambda = \frac{n}{\sum_n x_i Y_i}$$ and I am trying to figure out how the book arrived to $\hat\lambda \sim N(\lambda,\frac{\lambda^2}{n})$
The only explanation is something scribbled on the side on the side $$\frac{d^2<something>}{d\lambda^2} = \frac{-n}{\lambda}$$
but I cannot understand it's significance.
 A: I believe the answer is, that the author used Cramer Rao to calculate a lower bound for the variance and then, using asymptotic normality, the variance should achieve that lower bound.
A: I did not realize this question was both asked and answered by berrygreen!
You are right, the followings are the detailed answer
This formula:
$$\frac{d^2<something>}{d\lambda^2} = \frac{-n}{\lambda}$$  I believe it was  used to calculate the Fisher information.
You pdf of $Y_i$ is
$$f(y_i;\lambda x_i)=(\lambda x_i)e^{-(\lambda x_i)y_i} \tag{1}$$ 
Your likelihood function is :
$$L(\lambda)=\lambda^n\prod_{i=1}^nx_ie^{-\lambda\sum_{i=1}^nx_iy_i}$$
The log likelihood function is:
$$l(\lambda)=nlog\lambda+\sum_{i=1}^nx_i-\lambda\sum_{i=1}^n x_iy_i$$
Take derivative of $\lambda$ and set to $0$
$$\frac{n}{\lambda}-\sum_{i=1}^n x_iy_i=0$$
$$\hat{\lambda}=\frac{n}{\sum_{i=1}^n x_iy_i}$$
Which is correct.
Next you need calculate the Fisher information.
$$I(\lambda)=-\int_{-\infty}^{\infty}\frac{\partial log^2 f(y_i;\lambda x_i)}{\partial^2\lambda}f(y_i;\lambda x_i)dy_i \tag{2}$$
Now from $(1)$
$$\frac{\partial log^2 f(y_i;\lambda x_i)}{\partial^2 \lambda}=-\frac{1}{\lambda^2}$$
From $(2)$
$$I(\lambda)=\frac{1}{\lambda^2} \tag{3}$$
Next you need use the theorem
$$\sqrt{n}(\hat{\lambda}-\lambda)\overset{D}{\rightarrow}N(0,\frac{1}{I(\lambda)})=N(0,\lambda ^2)  \tag{4}$$
Rearrange $(4)$  you will get
$$\hat{\lambda}\sim N(\lambda,\frac{\lambda^2}{n})$$
This arrives your book's conclusion.
