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.



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?
Alas, no such inequality can hold. Suppose that the symmetric $X\_i$ take values $\pm 1$ and $t=n-1$. Then
$$
 \Pr[\max\_{k \in [n]} |\sum\_{i \in [n] \setminus \{k\}} X\_i| \ge t] =(n+1)2^{1-n}
$$
but
$$\Pr[|\sum\_{i \in [n]} X\_i| \ge t]=2^{1-n} \,.
$$


