How does the Cauchy-Schwarz inequality imply this

cauchy-schwarz-inequalityprobability theoryproof-explanation

Let $(X_i)_{1\leq i \leq n}$ be a sequence of orthogonal random variables (pairwise independent random variables) and let $S_i = \sum_{k=1}^{i}{X_k}$.

in Almost sure convergence by William F. Stout, Page 18, it is said :

Note that $S_{i}{ }^{2} \leq i \sum_{j=1}^{i} X_{j}^{2}$ follows from the Cauchy-Schwarz inequality.

I'm a bit confused as to how the CS inequality is used here, the usual inner product is the expectation but I fail to see which inner product is being considered here, any clarifications will be greatly appreciated, thanks.

Best Answer

It's just the vector form of Cauchy-Schwarz applied to vectors $(1,1,\ldots,1)$ and $(x_1,\ldots,x_i)$. One gets $$(x_1+\cdots+x_i)^2\le(1^2+\cdots+1^2)(x_1^2+\cdots+x_i^2)= i(x_1^2+\cdots+x_i^2).$$

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