Variance of least square estimator I have two random variables X and Y with $X\sim Exp(a)$ and $Y \sim Exp(\frac a2)$. I have a least square estimator $a=\frac {2x +y}{2.5}$. I want to calculate the variance of the estimator and to do this I'm trying to find the joint distribution (to calculate $E[XY]$). Is there a way of knowing if they are independent? It confuses me that they have different means.I tried to use the moment generating function on E[XY] but I'm not really sure how to go about it.
 A: You only know if $X$ and $Y$ are independent from the question or the experimental setup. Do they come from the same observation or different ones? 
In general, if the variances exist, $Var(cX)=c^2Var(X)$ for a constant $c$, and if $X$ and $Y$ are independent then $Var(X+Y)=Var(X)+Var(Y)$
So what are $\qquad Var(X)$? $\qquad Var(Y)$? $\qquad Var(2X)$? $\qquad Var(2X+Y)$? $\qquad$ and $  Var\left(\dfrac{2X+Y}{2.5}\right)$?
A: I don't see how to work the problem unless $X$ and $Y$ are
independent.
Let $\hat \alpha_1 = (2X + Y)/2.5$ as stated. 
Then $E(\hat \alpha_1) = \alpha,$ so that $\hat \alpha_1$ is
an unbiased estimator of $\theta.$
Now let $\hat \alpha_2 = (X + 2Y)/2$ be another estimator of $\alpha.$
It also has $E(\hat \alpha_2) = \alpha,$ so it is also
unbiased. 
Moreover, $V(\hat \alpha_2) = \alpha^2/2.$
When you find $V(\hat \alpha_1)$ following the suggestions
in the Answer by @Henry, you will see that $V(\hat \alpha_1) > V(\hat \alpha_2).$
That is, $\hat \alpha_2$ is just as 'accurate' in
estimating $\alpha$ as is $\hat \alpha_1$ (both are aimed
at the right 'target'). However,
$\hat \alpha_2$ is more 'precise' (less variable) than $\hat \alpha_1,$
so $\hat \alpha_2$ is a 'better' estimator than $\hat \alpha_1.$
Notes: The two purposes of this comment are (i) to give you a reality
check on your value of $V(\hat \alpha_1)$ when you find it
and (ii) to give you some idea why it is important to find
variances of estimators. 
Finally, an additional clue in case
you need it.  According to your notation, if $X \sim Exp(\alpha),$
where $E(X) = \alpha$, then $V(X) = \alpha^2.$
