question about Loss and Risk functions

Question

Let X1, . . . , X25 be a random sample from a distribution that is N(θ, 1), for −∞ < θ < ∞. Let Y = X¯, the mean of the random sample. Compare the two decision functions given by d1(y) = y and d2(y) = 0 for −∞ < y < ∞.

(i)Let the loss function be given by L[θ, d(y)] = (θ − d(y))2 . Show that R(θ, d1) = 1/25 and R(θ, d2) = θ 2 . Which of the decision functions yields the smaller maximum risk?

(ii)Let the loss function be given by L[θ, d(y)] = |θ − d(y)|. Show that R(θ, d1) = 1/5sqrt(2/π) 1(norm) and R(θ, d2) = |θ|. Which of the decision functions yields the smaller maximum risk?

I have been able to solve (i) using the definition of variance of X bar but I am lost as how to prove that R(θ,d1) = (1/5)sqrt(2/pi), any help would be greatly appreciated.

Answer 1

If I understand correctly, you need to take the expectation of $|\theta - Y|$, which using this is

$$ \mathbb{E}[|\theta - Y|] = \int_{-\infty}^{\infty} p_Y(y) |\theta-y| dy = \int_{-\infty}^{\infty} p_Y(y-\theta) |y| dy $$

where $Y$ is normally distributed with mean $\theta$ and variance $\sigma^2 = 1/25$. Since this integral is symmetric around 0 we can rewrite as

$$ \mathbb{E}[|\theta - Y|] = 2\int_{0}^{\infty} y p_Y(y-\theta) dy = \frac{2}{\sqrt{2\pi \sigma^2}} \int_{0}^{\infty} y e^{-y^2/2\sigma^2} dy. $$

The integral comes out to $\sigma^2$, so combined with the prefactor the solution is

$$ \mathbb{E}[|\theta - Y|] = \sqrt{\frac{2\sigma^2}{\pi}} = \frac{1}{5} \sqrt{\frac{2}{\pi}} $$