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JakobTejs
JakobTejs
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Let $X_1, \ldots, X_n$ be independent symmetric variables. Now I would like to know whether there exists a constant $C$, such that $$ \Pr[\max_{k \in [n]} |\sum_{i \in [n] \setminus \{k\}} X_i| \ge t] \le C\Pr[|\sum_{i \in [n]} X_i| \ge t] $$ Now similar inequalities are Levy's inequalities \begin{align} \label{eq:levy-sum} \Pr[\max_{1 \le l \le n} |\sum_{i = 1}^l X_i| \ge t] &\le 2 \Pr[|\sum_{i = 1}^n X_i] \\ \label{eq:levy-individual} \Pr[\max_{1 \le l \le n} |X_l| \ge t] &\le 2 \Pr[|\sum_{i = 1}^n X_i] \end{align}\begin{align} \label{eq:levy-sum} \Pr[\max_{1 \le l \le n} |\sum_{i = 1}^l X_i| \ge t] &\le 2 \Pr[|\sum_{i = 1}^n X_i| \ge t] \\ \label{eq:levy-individual} \Pr[\max_{1 \le l \le n} |X_l| \ge t] &\le 2 \Pr[|\sum_{i = 1}^n X_i| \ge t] \end{align} Unfortunately, I am unable to use the techniques used to prove Levy's inequality in my case.

The best result I have been able to prove is the following \begin{align*} \Pr[\max_{k \in [n]} |\sum_{i \in [n] \setminus \{k\}} X_i| \ge t] &\le \Pr[\max_{k \in [n]} |X_k| + |\sum_{i \in [n]} X_i| \ge t] \\&\le \Pr[\max_{k \in [n]} |X_k| \ge t/2] + \Pr[|\sum_{i \in [n]} X_i| \ge t/2] \\&\le 3\Pr[|\sum_{i \in [n]} X_i| \ge t/2] \end{align*} The last inequality follows from using the second of Levy's inequalities. But here we lose a factor of 2 on $t$, which I would like to avoid.

Let $X_1, \ldots, X_n$ be independent symmetric variables. Now I would like to know whether there exists a constant $C$, such that $$ \Pr[\max_{k \in [n]} |\sum_{i \in [n] \setminus \{k\}} X_i| \ge t] \le C\Pr[|\sum_{i \in [n]} X_i| \ge t] $$ Now similar inequalities are Levy's inequalities \begin{align} \label{eq:levy-sum} \Pr[\max_{1 \le l \le n} |\sum_{i = 1}^l X_i| \ge t] &\le 2 \Pr[|\sum_{i = 1}^n X_i] \\ \label{eq:levy-individual} \Pr[\max_{1 \le l \le n} |X_l| \ge t] &\le 2 \Pr[|\sum_{i = 1}^n X_i] \end{align} Unfortunately, I am unable to use the techniques used to prove Levy's inequality in my case.

The best result I have been able to prove is the following \begin{align*} \Pr[\max_{k \in [n]} |\sum_{i \in [n] \setminus \{k\}} X_i| \ge t] &\le \Pr[\max_{k \in [n]} |X_k| + |\sum_{i \in [n]} X_i| \ge t] \\&\le \Pr[\max_{k \in [n]} |X_k| \ge t/2] + \Pr[|\sum_{i \in [n]} X_i| \ge t/2] \\&\le 3\Pr[|\sum_{i \in [n]} X_i| \ge t/2] \end{align*} The last inequality follows from using the second of Levy's inequalities. But here we lose a factor of 2 on $t$, which I would like to avoid.

Let $X_1, \ldots, X_n$ be independent symmetric variables. Now I would like to know whether there exists a constant $C$, such that $$ \Pr[\max_{k \in [n]} |\sum_{i \in [n] \setminus \{k\}} X_i| \ge t] \le C\Pr[|\sum_{i \in [n]} X_i| \ge t] $$ Now similar inequalities are Levy's inequalities \begin{align} \label{eq:levy-sum} \Pr[\max_{1 \le l \le n} |\sum_{i = 1}^l X_i| \ge t] &\le 2 \Pr[|\sum_{i = 1}^n X_i| \ge t] \\ \label{eq:levy-individual} \Pr[\max_{1 \le l \le n} |X_l| \ge t] &\le 2 \Pr[|\sum_{i = 1}^n X_i| \ge t] \end{align} Unfortunately, I am unable to use the techniques used to prove Levy's inequality in my case.

The best result I have been able to prove is the following \begin{align*} \Pr[\max_{k \in [n]} |\sum_{i \in [n] \setminus \{k\}} X_i| \ge t] &\le \Pr[\max_{k \in [n]} |X_k| + |\sum_{i \in [n]} X_i| \ge t] \\&\le \Pr[\max_{k \in [n]} |X_k| \ge t/2] + \Pr[|\sum_{i \in [n]} X_i| \ge t/2] \\&\le 3\Pr[|\sum_{i \in [n]} X_i| \ge t/2] \end{align*} The last inequality follows from using the second of Levy's inequalities. But here we lose a factor of 2 on $t$, which I would like to avoid.

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JakobTejs
JakobTejs
  • 175
  • 6

Does there exist a constant $C$, such that, $\Pr[\max_k|\sum_{i\neq k}X_i|\ge t]\le C\Pr[|\sum_i X_i|\ge t]$ for all independent symmetric variables?

Let $X_1, \ldots, X_n$ be independent symmetric variables. Now I would like to know whether there exists a constant $C$, such that $$ \Pr[\max_{k \in [n]} |\sum_{i \in [n] \setminus \{k\}} X_i| \ge t] \le C\Pr[|\sum_{i \in [n]} X_i| \ge t] $$ Now similar inequalities are Levy's inequalities \begin{align} \label{eq:levy-sum} \Pr[\max_{1 \le l \le n} |\sum_{i = 1}^l X_i| \ge t] &\le 2 \Pr[|\sum_{i = 1}^n X_i] \\ \label{eq:levy-individual} \Pr[\max_{1 \le l \le n} |X_l| \ge t] &\le 2 \Pr[|\sum_{i = 1}^n X_i] \end{align} Unfortunately, I am unable to use the techniques used to prove Levy's inequality in my case.

The best result I have been able to prove is the following \begin{align*} \Pr[\max_{k \in [n]} |\sum_{i \in [n] \setminus \{k\}} X_i| \ge t] &\le \Pr[\max_{k \in [n]} |X_k| + |\sum_{i \in [n]} X_i| \ge t] \\&\le \Pr[\max_{k \in [n]} |X_k| \ge t/2] + \Pr[|\sum_{i \in [n]} X_i| \ge t/2] \\&\le 3\Pr[|\sum_{i \in [n]} X_i| \ge t/2] \end{align*} The last inequality follows from using the second of Levy's inequalities. But here we lose a factor of 2 on $t$, which I would like to avoid.