Consider a family of densitites $f(x,\theta)=\frac{\exp(-\sqrt{x})}{\theta}$. Let $X_1$ be a single observation from this family. I have shown that $\sqrt{X_1}/2$ is an unbiased estimator. Now consider $n$ observations $X_1,\ldots,X_n$. I have shown that $T(X)=\sqrt{X_1}+\cdots+\sqrt{X_n}$ is a sufficient statistic for $\theta$. Now use the Rao Blackwell theorem to find an improved estimator.
The improved estimator is $E(\sqrt{x}/2\mid T)$. For this I need the conditional density. How do I find this?
Thanks
Your function $x\mapsto\dfrac{\exp(-\sqrt{x})}{\theta}$ is not a probability density on $(0,\infty)$ unless $\theta=2$. Here's a guess: Maybe you meant $\dfrac{\exp\left(-\sqrt{x/\theta\, {}}\right)}{2\theta}$.
At any rate, suppose you have i.i.d. observations, and you want $$ \mathbb E\left(\frac{\sqrt{X_1}}{2} \mid \sqrt{X_1}+\cdots+\sqrt{X_n}\right). $$
You know \begin{align} \frac {\sqrt{X_1}+\cdots+\sqrt{X_n}} 2 & = \mathbb E \left(\frac{ \sqrt{X_1}+\cdots+\sqrt{X_n}}{2} \mid \sqrt{X_1}+\cdots+\sqrt{X_n}\right) \\[15pt] & = \mathbb E\left(\frac{\sqrt{X_1}}{2} \mid \sqrt{X_1}+\cdots+\sqrt{X_n}\right) + \cdots\cdots\cdots \\[10pt] & {}\qquad \cdots\cdots + \mathbb E\left(\frac{\sqrt{X_n}}{2} \mid \sqrt{X_1}+\cdots+\sqrt{X_n}\right). \end{align} By symmetry, all of the terms on the right must be equal to each other, so each must be equal to $1/n$ times the term on the left.