Random variable for variance Given a binomial distributed measure $P_\theta$, with parameter $\theta\in (0,1)$ on $\Omega:=\{0,1,\ldots, n\}$. I want to find a random variable
$$R: \Omega\rightarrow \mathbb{R},\quad R(x):=?$$
such that
$$E_\theta[R]=\theta(1-\theta).$$
So my first clue was some kind of using the idea that
$$Var_\theta[x]=E_\theta[x^2]-E_\theta[x]^2=n\theta(1-\theta).$$
So I would like to use something like
$$R(x):=\frac{1}{n}(x^2-x)$$
but this only gives 
$$E_\theta[R]=\frac{1}{n}(E_\theta[x^2]-E_\theta[x]),$$
such that I am missing the square.
Any hints for this problem?
Thanks in advance!
Edit: Of course $R$ shall be independent of $\theta$. So you cannot just take
$$R(x):=\frac{1-\theta}{n}x$$
for each $\theta$ or something similiar depending on $\theta$.
 A: You were actually very close.
The requested item is something which expectation is $\theta(1 - \theta) = \theta - \theta^2$. One option is to find an estimator that is unbiased. Without further requirements, the method of moments suffices.
Since it is know $E_{\theta}[x] = n\theta$ such that $G(x) \equiv \frac{x}n$ is an unbiased estimator for $\theta$, as in $E_{\theta}[G] = \theta$ , we can try the simple-minded
$$\theta - \theta^2 \mapsto G - G^2 \qquad \implies \text{ansatz:} \quad R(x) \propto A(x) := \frac{x}n - \left( \frac{x}n \right)^2$$ 
Now let's examine the expectation of $A(x)$.
\begin{align}
E_{\theta}[ A ] &= E_{\theta}\left[ \frac{x}n - \left( \frac{x}n \right)^2 \right] \\
&= \frac1n E_{\theta}[x] - \frac1{ n^2} E_{\theta}[x^2] \\
&= \frac1n \cdot n \theta - \frac1{ n^2} \left( Var_{\theta}[x] + E_{\theta}[x]^2 \right) \\
&= \theta - \frac1{ n^2} \left( n\theta(1-\theta) + n^2\theta^2 \right) \\
&= \theta - \frac1{ n^2} \left( n\theta + n(n-1)\theta^2 \right) \\
&= \frac{n-1}n \left( \theta  - \theta^2\right)
\end{align}
That is, $A(x)$ is a biased estimator for $\theta(1 - \theta)$, we can simply correct the bias by scaling.
$$R(x) := \frac{n}{n - 1} A(x) = \frac1{n - 1} x \left( 1 - \frac{x}n \right)$$
is the desired random variable such that $E_{\theta}[R] = \frac{n}{n-1} E_{\theta}[ A ] = \theta( 1 - \theta)$ as requested.
