Use Rao-Blackwell Theorem on $N(0,\sigma^2)$ distribution Suppose that $X_1,X_2,...,X_n$ is a random sample from a normal distribution, $X_i\sim N(0,\sigma^2)$.
Find the UMVUE of  $\phi(\sigma)=2\mathbb{P}_{\sigma}(X>c)$ where $c>0$. 
My try: $Y=\mathbb{1}_{\{|X_1| >c\}}$ be unbiased estimator of $\phi(\sigma)$.
I guess $T=\sum_{i=1}^{n} {X_i^2}$ be the complete sufficient statistic for $(X_i)$. I only proved $T$ be sufficient statistic, completeness still remains. I try to find the density function of $T$. I know that fact if $X_i\sim N(0,\sigma^2)$ then $\sum_{i=1}^{n} \left(\dfrac{X_i}{\sigma}\right)^2 \sim \chi^2_{n}$ imples $T \sim \sigma^2 \chi^2_n$ , $\sigma$ is variable not a fixed number.
Lats step I calculate $$E(\mathbb{1}_{\{|X_1| >c\}}  \ | \ \sum_{i=1}^{n} {X_i^2}) =\mathbb{P} (|X_1| >c\ \ | \ \sum_{i=1}^{n} {X_i^2})\\=\mathbb{P} (X_1^2 >c^2\ \ | \ \sum_{i=1}^{n} {X_i^2})= \mathbb{P} (\frac{X_1^2}{\sum_{i=1}^{n} {X_i^2}} >\frac{c^2}{\sum_{i=1}^{n} {X_i^2}}\ \ | \ \sum_{i=1}^{n} {X_i^2})$$
We know $X_1^2 \sim \chi_1^2$. I hope $\dfrac{X_1^2}{\sum_{i=1}^{n} {X_i^2}}$ and $\sum_{i=1}^{n} {X_i^2}$ are independent (it look like in case $X_i^2 \sim \text{Exp}(n)$)
 A: That $T=\sum\limits_{i=1}^n X_i^2$ is complete follows from the fact that the $N(0,\sigma^2)$ density is a member of a (regular) one-parameter exponential family. Completeness can be shown directly also.
It is well-known that if $U$ and $V$ are independent Gamma variables with a common rate/scale parameter, then $\frac{U}{U+V}$ has a Beta distribution independent of $U+V$. In particular, this holds if $U$ and $V$ have a chi-square distribution.
So you have $$\frac{X_1^2}{\sum\limits_{i=1}^n X_i^2}=\frac{X_1^2/\sigma^2}{X_1^2/\sigma^2+\sum\limits_{i=2}^n X_i^2/\sigma^2}\sim \mathsf{Beta}\left(\frac12,\frac{n-1}{2}\right)\,,$$
which is indeed independent of $T$. 
Hence $E_{\sigma}\left[I_{|X_1|>c}\mid T=t\right]$ reduces to
\begin{align}
P_{\sigma}\left[\frac{X_1^2}{\sum_{i=1}^n X_i^2}>\frac{c^2}{t}\,\,\Big\lvert\,\, T=t\right]&=P_{\sigma}\left[\frac{X_1^2}{\sum_{i=1}^n X_i^2}>\frac{c^2}{t}\right]
\\&=\frac{\sqrt \pi \Gamma\left(\frac{n-1}{2}\right)}{\Gamma(\frac n2)}\int_{c^2/t}^1 \frac{(1-x)^{(n-3)/2}}{\sqrt x}\,dx
\end{align}
Don't think this can be simplified by hand though.
