Školjka

Očekivano rješenje ovog zadatka: Odaberimo nasumično $\varepsilon_i \in \{-1, 1\}$ s probability $\frac{1}{2}$. Svi $z_i, w_i$ su zadane konstante. Pretpostavimo da su sve spomenute slučajne varijable identitet. Tada imamo $\varepsilon_i^2 = 1$ i za $i \neq j$: $$\mathbb{E}[\varepsilon_i\varepsilon_j] = \frac{1\cdot(-1) + (-1)\cdot1 + 1\cdot1 + (-1)\cdot(-1)}{4} = 0$$ Prva nejednakost implicira: $$ \mathbb{E} \Big[\big | \sum_{i = 1}^n \varepsilon_i z_i \big|^2 - \big | \sum_{i = 1}^n \varepsilon_i w_i \big|^2 \Big] \leq 0$$ $$ \mathbb{E} \Big[\big | \sum_{i = 1}^n \varepsilon_i z_i \big|^2 \Big] \leq \mathbb{E} \Big[\big | \sum_{i = 1}^n \varepsilon_i w_i \big|^2 \Big]$$ $$ \mathbb{E} \Big[\big ( \sum_{i = 1}^n\varepsilon_i z_i \big)( \sum_{i = 1}^n\varepsilon_i \overline{z_i} \big) \Big] \leq \mathbb{E} \Big[\big ( \sum_{i = 1}^n\varepsilon_i w_i \big)( \sum_{i = 1}^n\varepsilon_i \overline{w_i} \big) \Big]$$ $$ \mathbb{E} \Big[\sum_{i=1}^n \varepsilon_i^2|z_i|^2 + \sum_{i \neq j}\varepsilon_i\varepsilon_j(z_i\overline{z_j}+\overline{z_i}z_j)\Big] \leq \mathbb{E} \Big[\sum_{i=1}^n \varepsilon_i^2|w_i|^2 + \sum_{i \neq j}\varepsilon_i\varepsilon_j(w_i\overline{w_j}+\overline{w_i}w_j)\Big]$$ $$\sum_{i=1}^n |z_i|^2 + \sum_{i \neq j} (z_i\overline{z_j}+\overline{z_i}z_j) \cdot \mathbb{E} \Big[\varepsilon_i\varepsilon_j\Big] \leq \sum_{i=1}^n |w_i|^2 + \sum_{i \neq j} (w_i\overline{w_j}+\overline{w_i}w_j) \cdot \mathbb{E} \Big[\varepsilon_i\varepsilon_j\Big] $$ $$\sum_{i=1}^n |z_i|^2 \leq \sum_{i=1}^n |w_i|^2$$ što je i trebalo dokazati.