Have I applied the delta method correctly? I have UMVUE 
$$\tilde\theta = \frac{(n-1)(U-n)}{(U-1)(U-2)}$$ for $P(Y=2)=\theta(1-\theta)$
where $U=\sum_{i=1}^n Y_i$
$Y_i \sim \text{geometric} (\theta)$
I am using delta method to find the variance of $\tilde\theta$
So far, I have defined:
$$g(y)=\frac{(n-1)(y-n)}{(y-1)(y-2)}$$
$$g'(y) = \frac{(n-1)[n(2y-3)-y^2+2]}{(y-2)^2(y-1)}$$
Replacing $y$ with $\theta$ for $g'(\theta)$
Since $\tilde \theta$ is unbiased estimator, $E[\tilde \theta] = \theta$
$Var(\sum Y_i)=n^2(1-\theta)/\theta^2 $ [ This is the step where I am confused, do I have to take the variance of one of the $Y_i$'s or sum of $Y_i'$s
So, $Var(\tilde \theta) = [g'(\theta)]^2Var (\sum Y_i)$
Is this approach is correct? Any pointers would be helpful.
 A: How did you calculate the UMVUE? Looks pretty intense...
A few tips:
Here is how the delta method goes:
If you have some random variable $Y$ such that it is asymptotically normal e.g. 
$$\sqrt{n}(Y-\theta) \overset{D}{\longrightarrow} Z, \quad Z\sim N(0, \sigma^2)$$
then $$\sqrt{n}(g(Y)-g(\theta)) \overset{D}{\longrightarrow} g'(\theta)Z, \quad Z\sim N(0,\sigma^2)$$
Things to note: You can typically find a $Y$ by using the central limit theorem which often gives an asymptotically normal result about an average (e.g. $\bar{Y}$). 
Note that it is $g'(\theta)$ where $\theta$ is what you have to subtract from $Y$ so that it is asymptotically $N(0,\sigma^2)$. So for the geometric you might want to start off by noting that by the CLT
$$ \sqrt{n}(\bar{Y} - 1/\theta) \overset{D}{\longrightarrow} Z, \quad Z\sim N\left(0,\dfrac{1-\theta}{\theta^2}\right)$$
Now note that if you wanted to get this to be a statement about your estimator $\tilde{\theta}$ your $g$ function will have to change $\bar{Y}$ to $\sum_{i=1}^n Y_i$. Also note that you will plug $1/\theta$ into $g$. 
I hope this helps. 
