#math/linear-algebra #math/probability #TODO %%--%% The Cauchy-Schwarz Inequality states %%?%% $ \begin{gather*} a_i, b_i \in \mathbb{R} \quad \text{for } i = 1, \dots, n \\ (a_{1}b_{1} + \dots + a_{n}b_{n})^2 \leq (a_{1}^2 + \dots + a_{n}^2)(b_{1}^2 + \dots + b_{n}^2)\\ \\ \text{or} \\ \\ \left( \sum_{i=1}^{n} a_i b_i \right)^2 \leq \left( \sum_{i=1}^{n} a_i^2 \right) \left( \sum_{i=1}^{n} b_i^2 \right) \\ \\ \text{or}\\ \\ u, v \in \mathrm{R}^n \\ \lvert u \cdot v \rvert^2 \leq (u \cdot u)(v \cdot v) \\ \end{gather*} $ %%--%% Square-rooting both sides gives you $ \begin{gather*} u, v \in \mathrm{R}^n \\ \lvert u \cdot v \rvert \leq \lvert \lvert u \rvert \rvert \text{ } \lvert \lvert v \rvert \rvert \end{gather*} $ Note that the two sides are equal if and only if $u$ and $v$ are [[Linearly Dependent]]. ## Applied to Probability %%--%% The trick to use Cauchy-Schwarz in a probability setting is to %%?%%assign one vector, $p$, to the probabilities of the outcome space and another vector, $b$, to a ones vector. $ \begin{align} (p_{1} + \dots + p_{n})^{2} &\leq (p_{1}^2 + \dots + p_{n}^2)(1^2 + \dots + 1^2) \\ 1 &\leq (p_{1}^2 + \dots + p_{n}^2)(n)\\ \frac{1}{n} &\leq (p_{1}^2 + \dots + p_{n}^2) \end{align} $ Using a feature from Cauchy-Schwarz, when the two vectors, $p$ and $b$, are not scalar multiples of one another, i.e. when $p$ is non-uniform, we can drop the equality. $ \begin{gather*} \exists{} p_{i}, p_{j} \mid p_{i} \neq p_{j} \\ \frac{1}{n} < (p_{1}^2 + \dots + p_{n}^2) \end{gather*} $ %%--%%