 Solution

Use ${\mathcal{F}}_m=\sigma ({\xi }_1,\dots ,{\xi }_m)$ and $S_0=0$. Independence and zero means imply $\mathbb{E}[S_{m+1}\mid {\mathcal{F}}_m]=S_m$, so the partial sums form a square-integrable martingale. More explicitly,Thus $S_m^2$ is a nonnegative submartingale. The Doob maximal inequality at threshold $x^2$ now proves the Kolmogorov maximal inequality:Finally, expanding the terminal square and using independence and the zero means gives $\mathbb{E}{S}_n^2={\sum }_{j=1}^n\mathbb{E}{\xi }_j^2$. Identical distributions are not required.