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I work with i.i.d. variables $X_1, \dots, X_{N}$ such that $0 \le X_i \le 1$, $E[X_i] = \mu$, $Var[X_i] = \sigma^2$.

I am gradually sampling $X_1, X_2, \dots$ and want to ensure that the natural sample variance estimate stays within reasonable bounds. More formally, define $A_n = \sum_{i = 1}^n (X_i - \bar X_n)^2 - (n-1)\sigma^2$ where $\bar X_n = \sum_{i = 1}^{n} X_i/n$. We have $E[A_n] = 0$ for every $n$. Also, using that $X_i$s are bounded, we can crudely upper bound the variance as $Var[A_n] \le n\sigma^2$.

Using Chebyshev's inequality, we can conclude that $P[|A_{N}| > 100\sqrt{N\sigma^2}] < 0.5$. However, I would like to have a stronger result $P[\max_{n = 1}^{N} |A_{n}| > 100\sqrt{N\sigma^2}] < 0.5$. My question is: How do we achieve this bound? Is there some well-known inequality that proves this?

Note that if $A_n$ was a martingale, we could use Kolmogorov's inequality to arrive at this conclusion. My intuition why the inequality holds is that $A_n$ behaves very similar to a martingale.

I work with i.i.d. variables $X_1, \dots, X_{N}$ such that $0 \le X_i \le 1$, $E[X_i] = \mu$, $\operatorname{Var}[X_i] = \sigma^2$.

I am gradually sampling $X_1, X_2, \dots$ and want to ensure that the natural sample variance estimate stays within reasonable bounds. More formally, define $A_n = \sum_{i = 1}^n (X_i - \bar X_n)^2 - (n-1)\sigma^2$ where $\bar X_n = \sum_{i = 1}^{n} X_i/n$. We have $E[A_n] = 0$ for every $n$. Also, using that $X_i$s are bounded, we can crudely upper bound the variance as $\operatorname{Var}[A_n] \le n\sigma^2$.

Using Chebyshev's inequality, we can conclude that $P[|A_{N}| > 100\sqrt{N\sigma^2}] < 0.5$. However, I would like to have a stronger result $P[\max_{n = 1}^{N} |A_{n}| > 100\sqrt{N\sigma^2}] < 0.5$. My question is: How do we achieve this bound? Is there some well-known inequality that proves this?

Note that if $A_n$ was a martingale, we could use Kolmogorov's inequality to arrive at this conclusion. My intuition why the inequality holds is that $A_n$ behaves very similar to a martingale.

I work with i.i.d. variables $X_1, \dots, X_{n_0}$ such that $0 \le X_i \le 1$, $E[X_i] = \mu$, $Var[X_i] = \sigma^2$.

I am gradually sampling $X_1, X_2, \dots$ and want to ensure that the natural sample variance estimate stays within reasonable bounds. More formally, define $A_n = \sum_{i = 1}^n (X_i - \bar X_n)^2 - (n-1)\sigma^2$ where $\bar X_n = \sum_{i = 1}^{n} X_i/n$. We have $E[A_n] = 0$ for every $n$. Also, using that $X_i$s are bounded, we can crudely upper bound the variance as $Var[A_n] \le n\sigma^2$.

Using Chebyshev's inequality, we can conclude that $P[|A_{n_0}| > 100\sqrt{n_0\sigma^2}] < 0.5$. However, I would like to have a stronger result $P[\max_{n = 1}^{n_0} |A_{n}| > 100\sqrt{n_0\sigma^2}] < 0.5$. My question is: How do we achieve this bound? Is there some well-known inequality that proves this?

Note that if $A_n$ was a martingale, we could use Kolmogorov's inequality to arrive at this conclusion. My intuition why the inequality holds is that $A_n$ behaves very similar to a martingale.

I work with i.i.d. variables $X_1, \dots, X_{N}$ such that $0 \le X_i \le 1$, $E[X_i] = \mu$, $Var[X_i] = \sigma^2$.

I am gradually sampling $X_1, X_2, \dots$ and want to ensure that the natural sample variance estimate stays within reasonable bounds. More formally, define $A_n = \sum_{i = 1}^n (X_i - \bar X_n)^2 - (n-1)\sigma^2$ where $\bar X_n = \sum_{i = 1}^{n} X_i/n$. We have $E[A_n] = 0$ for every $n$. Also, using that $X_i$s are bounded, we can crudely upper bound the variance as $Var[A_n] \le n\sigma^2$.

Using Chebyshev's inequality, we can conclude that $P[|A_{N}| > 100\sqrt{N\sigma^2}] < 0.5$. However, I would like to have a stronger result $P[\max_{n = 1}^{N} |A_{n}| > 100\sqrt{N\sigma^2}] < 0.5$. My question is: How do we achieve this bound? Is there some well-known inequality that proves this?

Note that if $A_n$ was a martingale, we could use Kolmogorov's inequality to arrive at this conclusion. My intuition why the inequality holds is that $A_n$ behaves very similar to a martingale.

I work with i.i.d. variables $X_1, \dots, X_{n_0}$ such that $0 \le X_i \le 1$, $E[X_i] = \mu$, $Var[X_i] = \sigma^2$.

I am gradually sampling $X_1, X_2, \dots$ and want to ensure that the natural sample variance estimate stays within reasonable bounds. More formally, define $A_n = \sum_{i = 1}^n (x_i - \hat{\mu}_n)^2 - (n-1)\sigma^2$ where $\hat{\mu_n} = \sum_{i = 1}^{n} x_i/n$. We have $E[A_n] = 0$ for every $n$. Also, using that $X_i$s are bounded, we can crudely upper bound the variance as $Var[A_n] \le n\sigma^2$.

Using Chebyshev's inequality, we can conclude that $P[|A_{n_0}| > 100\sqrt{n_0\sigma^2}] < 0.5$. However, I would like to have a stronger result $P[\max_{n = 1}^{n_0} |A_{n}| > 100\sqrt{n_0\sigma^2}] < 0.5$. My question is: How do we achieve this bound? Is there some well-known inequality that proves this?

Note that if $A_n$ was a martingale, we could use Kolmogorov's inequality to arrive at this conclusion. My intuition why the inequality holds is that $A_n$ behaves very similar to a martingale.

I work with i.i.d. variables $X_1, \dots, X_{n_0}$ such that $0 \le X_i \le 1$, $E[X_i] = \mu$, $Var[X_i] = \sigma^2$.

I am gradually sampling $X_1, X_2, \dots$ and want to ensure that the natural sample variance estimate stays within reasonable bounds. More formally, define $A_n = \sum_{i = 1}^n (X_i - \bar X_n)^2 - (n-1)\sigma^2$ where $\bar X_n = \sum_{i = 1}^{n} X_i/n$. We have $E[A_n] = 0$ for every $n$. Also, using that $X_i$s are bounded, we can crudely upper bound the variance as $Var[A_n] \le n\sigma^2$.

Using Chebyshev's inequality, we can conclude that $P[|A_{n_0}| > 100\sqrt{n_0\sigma^2}] < 0.5$. However, I would like to have a stronger result $P[\max_{n = 1}^{n_0} |A_{n}| > 100\sqrt{n_0\sigma^2}] < 0.5$. My question is: How do we achieve this bound? Is there some well-known inequality that proves this?

Note that if $A_n$ was a martingale, we could use Kolmogorov's inequality to arrive at this conclusion. My intuition why the inequality holds is that $A_n$ behaves very similar to a martingale.