HDS

Exercise 15.9: Achievable Rates for Uniform Shift Family

chapter 15

Since $\min \lbrace Y_1, \ldots , Y_n \rbrace - \theta = \min \lbrace Y_1 - \theta, \ldots , Y_n - \theta \rbrace$, we can w.l.o.g. take $\theta = 0$ and $Y_i \sim \Unif[0, 1]$ i.i.d.

Let $Z = \min_i Y_i$ and observe $\P(Z \geq t) = \P(Y_1 > t)^n = (1 - t)^n$. Since $Z$ is non-negative \begin{align} \E [Z^2] = \int_0^1 \! \P(Z^2 > t) \, \sd t = \int_0^1 (1 - \sqrt{t})^n \, \sd t = \frac{2}{n^2 + 3n +2} \, . \end{align}

Published on 29 January 2022.