CRLB for $\theta$ using random vectors $(x,y)$ Let $(x_1,y_1)(x_2,y_2)... (x_n,y_n)$ be independent pairs such that $y_i=\theta x_i+\epsilon_i$ where $x_i$ and $\epsilon_i$ are iid Normal (0,1) for $i=1,2..n$
It was previously computed that
$f(x,y)=\frac{1}{2\pi}e^{\frac{-1}{2}x^2-\frac12(y-\theta x)^2 }$
And that the MLE for $\theta$ is $\theta^*=\frac{\sum_{i=1}^nx_iy_i}{\sum_{i=1}^nx_i^2}$ which is unbiased for $\theta$
I need to know if this estimator achieves the CRLB.
Here is what I have so far:
For any estimator W(M) of $\theta$ using sample $M_i's$ of$M$ which are iid $i=1..n$, the CRLB is given by
$\frac{[\frac{d}{d\theta}E(W(M))]^2}{nE_\theta[(\frac{d}{d\theta}lnf(M|\theta)]^2}$
So the numerator is 1 since $E(W(M))=\theta$.
For the denominator, I worked it out as $n(\theta^2(2-\theta^2))$.
Is this correct? It seems iffy coz it can have negative values.
Another problem is in computing the variance of the estimator.
$Var(\theta^*)=Var(\frac{\sum_{i=1}^nx_iy_i}{\sum_{i=1}^nx_i^2}$). How do I go about this?
 A: The expression of your MLE isn't quite correct. The log-likelihood takes the form, 
$$ \log(L(\textbf{x},\textbf{y};\theta) = constant - \frac{1}{2}\sum\limits_{i=1}^n(y_i - \theta x_i)^2 $$
Taking the derivative and setting equal to zero gives, 
$$ \frac{d}{d\theta}\log(L(\textbf{x},\textbf{y};\theta) = \sum\limits_{i=1}^n(y_i - \theta x_i)x_i = 0 \Longrightarrow \hat{\theta} = \frac{\sum\limits_{i=1}^nx_iy_i}{\sum\limits_{i=1}^nx_i^2} $$
You can verify this gives a maximum using the second derivative. 
A: Ok. Here's my solution in determining if the estimator reaches the CRLB.
$Var(\theta^*)=Var(\frac{\sum_{i=1}^nx_iy_i}{\sum_{i=1}^nx_i^2})=Var(\frac{\sum_{i=1}^nx_i(\theta x_i+\epsilon_i)}{\sum_{i=1}^nx_i^2})$
$=Var(\frac{\sum_{i=1}^n(\theta x_i^2+x_i\epsilon_i)}{\sum_{i=1}^nx_i^2})$
$=Var(\frac{\theta\sum_{i=1}^n x_i^2}{\sum_{i=1}^nx_i^2}+\frac{\sum_{i=1}^nx_i\epsilon_i}{\sum_{i=1}^nx_i^2})$
$=Var(\theta+\frac{\sum_{i=1}^nx_i\epsilon_i}{\sum_{i=1}^nx_i^2})=Var(\frac{\sum_{i=1}^nx_i\epsilon_i}{\sum_{i=1}^nx_i^2})$
At this point, I can argue that since $x_i's$ and $\epsilon_i's$ are iid $\sim N(0,1)$ then $Var(\theta^*)$ is independent of $\theta$ which means that it cannot attain the CRLB which is dependent on $\theta$. That is, there will always be a theta that can make the CRLB smaller than $Var(\theta^*)$.
