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.

  • 1
    $\begingroup$ @Glen_b Corrected the question! $\endgroup$
    – user30438
    Dec 18, 2013 at 23:38

1 Answer 1


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.


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