Kolmogorov's maximal inequality with shifted starting point

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probability  

$\newcommand{\argmin}{\mathop{\mathrm{argmin}}\limits}$ $\newcommand{\argmax}{\mathop{\mathrm{argmax}}\limits}$

Kolmogorov’s maximal inequality provides result similar to that of Chebyshev’s inequality to maximum of partial sum of random variables.

If $X_i$'s are independent, $EX_i=0$, $\mathrm{Var}(X_i) < \infty$, $S_n := \sum\limits_{k=1}^n X_k$, then the following inequality holds. $$P(\max_{1 \leq k \leq n} |S_k| > x) \leq \frac{1}{x^2} \mathrm{Var}(S_n)$$
Let $A_k = \{ S_k > x,\: S_j \leq x,\: 1\leq j \leq n \}$. $A_k$'s are disjoint and $\bigcup\limits_{k=1}^n A_k = \{ \max_{1 \leq k \leq n} \|S_k\| > x \}$. $$\begin{align} \mathrm{Var}(S_n) &= \mathrm{E}(S_n^2) \geq \mathrm{E}(S_n^2; \cup_{k=1}^n A_k)\\ &= \sum\limits_{k=1}^n \int_{A_k} S_n^2 \: dP \\ &= \sum\limits_{k=1}^n \int_{A_k} (S_n - S_k + S_k)^2 \: dP \\ &= \sum\limits_{k=1}^n \int_{A_k} (S_n - S_k)^2 + S_k^2 + 2S_k (S_n - S_k) \: dP \\ &\geq \sum\limits_{k=1}^n \big\{ \int_{A_k} S_k^2 + 2\int_{\Omega}S_k\boldsymbol{1}_{A_k}\: dP \int_{\Omega}(S_n - S_k) \: dP \big\}\\ &= \sum\limits_{k=1}^n \int_{A_k} S_k^2 \: dP \\ &\geq \sum\limits_{k=1}^n \int_{A_k} x^2 \: dP \\ &= x^2P(\cup_{k=1}^n A_k) \\ &= x^2 P(\max_{1 \leq k \leq n} |S_k| > x) \end{align}$$


Let $Y_n = X_{n+m}$ and $T_n= \sum\limits_{k=1}^n Y_k$, then $Y_i$’s are also independent with expectaion 0 and finite variance. Hence, the following also holds,

\[P(\max_{0 \leq k \leq n-m} |T_k| > x) \leq \frac{1}{x^2} \mathrm{Var}(T_{n-m})\]

which means

\[P(\max_{m \leq k \leq n} |S_k - S_m| > x) \leq \frac{1}{x^2} \mathrm{Var}(S_n - S_m)\]

Thus, even with different starting point of summation, and even if $X_i$’s are not identically distributed, the same form of inequality can be used. This result is useful when proving Kolmogorov’s three-series theorem.