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I am trying to find the UMVUE for the parameter $p$ for an n i.i.d geometric distribution:

$(1-p)^{x-1}p$ for $x=1,2,…$ and $0<p<1$

and found that:

$P(X_1=1)$ is an unbiased estimator , so let $w=I[X_1=1]$ be my unbiased estimator and since $sum_i X_i=t$ is complete and sufficient statistic for geometric distribution, I can improve my unbiased estimator as follows:

$E[wmidsum_i X_i=t] = P(X_1=1midsum_i X_i=t) = frac{P(X_1=1,sum_i X_i=t-1)}{P(sum_i X_i=t)}$

So I have two questions now:
what is the pdf for $sum_i X_i=t-1$ ? .. I know it is negative binomial but can't write it correctly
and my second question is what is the variance of this modified unbiased estimator and does it achieve the Cramer-Rao lower bound ?

For the first question only:

$P(X_1=1)=p$

$P(sum_{i=2}^{n}X_i=t-1)={t-2choose n-2}p^{n-1}(1-p)^{t-n}$, $t=n,n+1...$

$P(sum_{i=1}^{n}X_i=t)={t-1choose n-1}p^{n}(1-p)^{t-n}$, $t=n,n+1...$

So the UMVUE is $hat p=frac{n-1}{sum_{i=1}^{n} X_i-1}$

For CRLB you may look here.

But for the variance of the UMVUE:

$Var(hat p)=sum_{t=n}^infty left(frac{n-1}{t-1}-pright)^2 {t-1choose n-1}p^n(1-p)^{t-n}$

I'm afraid I was not able to get a closed form. Neither it worked for $E(hat p^2)$

Maybe somebody else can step in.