Exercise 5.9 - Posterior median is optimal estimate under L1 loss

Answers

The posterior loss expectation is (where we have omitted $𝒟$ w.l.o.g.):

\begin{aligned} ρ (a) & = \int | y - a | p (y) d y = \int_{- \infty}^{a} (a - y) p (y) d y + \int_{a}^{+ \infty} (y - a) p (y) d y \\ = a {\int_{- \infty}^{a} p (y) d y - \int_{a}^{+ \infty} p (y) d y} - \int_{- \infty}^{a} 𝑦𝑝 (y) d y + \int_{a}^{+ \infty} 𝑦𝑝 (y) d y . \end{aligned}

Differentiating the loss w.r.t. $a$ ends up with:

\frac{\partial}{∂𝑎} ρ (a) = {\int_{- \infty}^{a} p (y) 𝑑𝑦 - \int_{a}^{+ \infty} p (y)} + a \cdot 2 p (a) - a \cdot 2 p (a) .

Set it to zero so:

\int_{- \infty}^{a} p (y) 𝑑𝑦 = \int_{a}^{+ \infty} p (y),

whose summation is unity, hence $a$ satisfies:

\int_{- \infty}^{a} p (y) 𝑑𝑦 = \int_{a}^{+ \infty} p (y) = \frac{1}{2},

which identifies the median.

solour_lfq

2021-03-24 13:42