Uniformly minimum variance unbiased estimator of theta. Let $X_1,X_2,....X_n$ be a random sample from a population with probability density function 
$$f(x|\theta)=\dfrac{\theta}{2}e^{-\theta|x|} ,-\infty < x < \infty,\theta>0$$ 
Then a UMVE of $\theta$ is ?
Can someone tell me if i did everything right or not in the following steps. I dont have answer written for this problem in my solution manual. 
Steps : 
1.Its an even function that's why pdf changes as $$f(x|\theta)=\theta e^{-\theta x} ,0 < x < \infty,\theta>0$$
2.Using Rao blackwell theorem i found MVUE $\dfrac{n-1}{\sum_{i=1}^{n} X_i}$
Now MVUE is $\dfrac{n-1}{\sum_{i=1}^{n} X_i}$ or $\dfrac{n-1}{\sum_{i=1}^{n} |X_i|}$ ? Please give your thoughts and tell me where did i follow wrong track? 

Write 
$$f(x)=\dfrac{\theta}{2}e^{-\theta|x|}=h(x)g(\theta)\exp\left(\eta(\theta)\cdot T(x)\right)$$
with $h(x) = 1$, $g(\theta) = \theta/2$, $\eta(\theta)=-\theta$, and $T(x) = |x|$. 
It follows that $f$ is of the exponential family.
Furthermore, note that the parameter space
$$\Theta=(0, \infty)\supset(0, 1)$$
so $\Theta$ contains an open set. Therefore, it follows that $T(\mathbf{X}) = \sum_{i=1}^{n}|X_i|$ is sufficient and complete.
Next, we need to find the CDF of $|X_1|=|X|$. Observe that due to that $f$ is even,
$$F_{|X|}(x)=\mathbb{P}(|X| \leq x)=\mathbb{P}(-x\leq X \leq x)=\int_{-x}^{x}f(t \mid \theta)\text{ d}t=2\int_{0}^{x}f(t \mid \theta)\text{ d}t$$
and we can see that
$$\int_{0}^{x}\dfrac{\theta}{2}e^{-\theta|t|}\text{ d}t=\dfrac{\theta}{2}\int_{0}^{x}e^{-\theta t}\text{ d}t=\dfrac{\theta}{2}\cdot\dfrac{1}{-\theta}(e^{-\theta x}-1)=\dfrac{1-e^{-\theta x}}{2}$$
so it follows that $$F_{|X|}(x)=1-e^{-\theta x}$$
for $x > 0$, with probability density function
$$f_{|X|}(x) = \theta e^{-\theta x}$$
for $x > 0$ and $f_{|X|}(x) = 0$ otherwise, 
 hence $|X|$ follows an exponential distribution with mean $1/\theta$. It follows that $T(\mathbf{X})$ has a Gamma distribution with $\alpha = n$ and $\beta = \theta$. 
With some work, it can be shown that $\dfrac{1}{T(\mathbf{X})}$ follows an Inverse-Gamma distribution with mean $\dfrac{\theta}{n-1}$ as long as $n > 1$; hence, $$\mathbb{E}\left[\dfrac{n-1}{T(\mathbf{X})}\right]=\mathbb{E}\left[\dfrac{n-1}{\sum_{i=1}^{n}|X_i|}\right]=\theta$$
so by Lehmann-Scheffe, $\dfrac{n-1}{\sum_{i=1}^{n}|X_i|}$ is the UMVUE of $\theta$.
