Prove that $E_θ(θ-\frac{X}{100})^2 = \frac {θ(1-θ)}{100}$

Question

$X \text {~} Binomial(100,θ)$, $δ(X)=\frac {X}{100}$, $g(θ)=θ$ and the loss function is given by $L(θ,d)=(θ-d)^2$.

The risk function for δ is $R(θ,δ) = E_θ(θ-\frac{X}{100})^2 = \frac {θ(1-θ)}{100}$

I want to know why $E_θ(θ-\frac{X}{100})^2 = \frac {θ(1-θ)}{100}$

Answer 1

Observe $$\begin{align} \mathbb{E}[\delta(X)] &= \dfrac{\mathbb{E}[X]}{100} = \dfrac{100\theta}{100} = \theta\\ \text{Var}(\delta(X)) &= \dfrac{1}{100^2}\text{Var}(X) = \dfrac{1}{100^2}(100)\theta(1-\theta) = \dfrac{\theta(1-\theta)}{100} \end{align}$$ and then lastly, $$\text{Var}(\delta(X)) = \mathbb{E}\left[\left(\delta(X) - \mathbb{E}[\delta(X)] \right)^2 \right] = \mathbb{E}\left[\left(\mathbb{E}[\delta(X)] - \delta(X) \right)^2 \right]\text{.}$$