Risk function of binomial random variable

Question

Suppose $X\sim Binomial(100,\theta)$, True estimator($\delta$) $= X/100$;

$$R(\theta ,\delta) = E_{\theta}\left[\left(\theta - \frac{X}{100}\right)^2\right] = \theta(1-\theta)/100$$

I am copying this same from text book but I trying to prove the above ?

Answer 1

$R(\theta ,\delta) = E_{\theta}\left[\left(\theta - \frac{X}{100}\right)^2\right]= E_{\theta}\left[\theta^2-2\frac{\theta X}{100} + \left(\frac{X}{100}\right)^2\right]=E_{\theta}[\theta^2]-\frac{1}{50}E_{\theta}[\theta X]+\frac{1}{100^2}E_{\theta}[X^2]=\theta^2-2\theta^2+\frac{1}{100^2}\left[100\theta(1-\theta)+(100\theta)^2\right]=\theta(1-\theta)/100$

This is because, $E_{\theta}[X]=n\theta,Var_{\theta}=n\theta(1-\theta)$ for $X\sim binomial(n,\theta)$