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Yuval Peres
Yuval Peres
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The Question

Suppose that $X_n$ are independent random variables, $|X_n| \leq 1$, and $\mathbb{E}[X_n] = 0$. Let $S_n = \sum_{i=1}^n X_i$ and let $s_n = \sqrt{\sum_k \sigma^2(X_k)}$ be the standard deviation of $S_n$. Does $P(1 + |S_n| > \epsilon s_n)$$P(1 + |S_n| <\epsilon s_n)$ go to zero uniformly in $n$, as $\epsilon \rightarrow 0$? In other words, if you permit the abuse of notation, is $s_n = O(1 + |S_i|)$ in probability?

Some Background

I'm interested in estimating the size of $S_n$, in probability. Chebyshev's Inequality gives: $S_N = \mathcal{O}(s_n)$ in probability. In fact, if we take $T_n = \max_{1 \leq i \leq n} |S_i|$, then we also have $T_n = O(s_n)$ in probability by Kolmogorov's inequality.

In the other direction, taking advantage of $|X_i| \leq 1$, we have: $P(1 + T_n < \epsilon s_n)$ converges to zero uniformly in $n$, as $\epsilon \rightarrow 0$. A similar statement is used in a proof of Kolmogorov's 3 Series theorem, but I don't know what this theorem itself is called. Thus, by abuse of notation, $s_n = O(1 + T_n)$. Since $T_n$ and $|S_n|$ often have similar size (for example, with Ottaviani's inequality, you can prove that $P(T_n > \lambda) \leq 4 \max_{1 \leq i \leq n} P(|S_i| > \lambda/4)$), I would expect that a similar statement is true for $|S_n|$. However, I can't seem to bridge that gap.

*For reference, when I write $X_\alpha = O(Y_\alpha)$ when $X_\alpha, Y_\alpha$ are random variables and $Y_\alpha \geq 0$, I mean that $\sup_\alpha P(|X_\alpha| \leq K Y_\alpha)$ converges to 0 as $K \rightarrow \infty$.

The Question

Suppose that $X_n$ are independent random variables, $|X_n| \leq 1$, and $\mathbb{E}[X_n] = 0$. Let $S_n = \sum_{i=1}^n X_i$ and let $s_n = \sqrt{\sum_k \sigma^2(X_k)}$ be the standard deviation of $S_n$. Does $P(1 + |S_n| > \epsilon s_n)$ go to zero uniformly in $n$, as $\epsilon \rightarrow 0$? In other words, if you permit the abuse of notation, is $s_n = O(1 + |S_i|)$ in probability?

Some Background

I'm interested in estimating the size of $S_n$, in probability. Chebyshev's Inequality gives: $S_N = \mathcal{O}(s_n)$ in probability. In fact, if we take $T_n = \max_{1 \leq i \leq n} |S_i|$, then we also have $T_n = O(s_n)$ in probability by Kolmogorov's inequality.

In the other direction, taking advantage of $|X_i| \leq 1$, we have: $P(1 + T_n < \epsilon s_n)$ converges to zero uniformly in $n$, as $\epsilon \rightarrow 0$. A similar statement is used in a proof of Kolmogorov's 3 Series theorem, but I don't know what this theorem itself is called. Thus, by abuse of notation, $s_n = O(1 + T_n)$. Since $T_n$ and $|S_n|$ often have similar size (for example, with Ottaviani's inequality, you can prove that $P(T_n > \lambda) \leq 4 \max_{1 \leq i \leq n} P(|S_i| > \lambda/4)$), I would expect that a similar statement is true for $|S_n|$. However, I can't seem to bridge that gap.

*For reference, when I write $X_\alpha = O(Y_\alpha)$ when $X_\alpha, Y_\alpha$ are random variables and $Y_\alpha \geq 0$, I mean that $\sup_\alpha P(|X_\alpha| \leq K Y_\alpha)$ converges to 0 as $K \rightarrow \infty$.

The Question

Suppose that $X_n$ are independent random variables, $|X_n| \leq 1$, and $\mathbb{E}[X_n] = 0$. Let $S_n = \sum_{i=1}^n X_i$ and let $s_n = \sqrt{\sum_k \sigma^2(X_k)}$ be the standard deviation of $S_n$. Does $P(1 + |S_n| <\epsilon s_n)$ go to zero uniformly in $n$, as $\epsilon \rightarrow 0$? In other words, if you permit the abuse of notation, is $s_n = O(1 + |S_i|)$ in probability?

Some Background

I'm interested in estimating the size of $S_n$, in probability. Chebyshev's Inequality gives: $S_N = \mathcal{O}(s_n)$ in probability. In fact, if we take $T_n = \max_{1 \leq i \leq n} |S_i|$, then we also have $T_n = O(s_n)$ in probability by Kolmogorov's inequality.

In the other direction, taking advantage of $|X_i| \leq 1$, we have: $P(1 + T_n < \epsilon s_n)$ converges to zero uniformly in $n$, as $\epsilon \rightarrow 0$. A similar statement is used in a proof of Kolmogorov's 3 Series theorem, but I don't know what this theorem itself is called. Thus, by abuse of notation, $s_n = O(1 + T_n)$. Since $T_n$ and $|S_n|$ often have similar size (for example, with Ottaviani's inequality, you can prove that $P(T_n > \lambda) \leq 4 \max_{1 \leq i \leq n} P(|S_i| > \lambda/4)$), I would expect that a similar statement is true for $|S_n|$. However, I can't seem to bridge that gap.

*For reference, when I write $X_\alpha = O(Y_\alpha)$ when $X_\alpha, Y_\alpha$ are random variables and $Y_\alpha \geq 0$, I mean that $\sup_\alpha P(|X_\alpha| \leq K Y_\alpha)$ converges to 0 as $K \rightarrow \infty$.

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Theone
Theone
  • 143
  • 5

The Question

Suppose that $X_n$ are independent random variables, $|X_n| \leq 1$, and $\mathbb{E}[X_n] = 0$. Let $S_n = \sum_{i=1}^n X_i$ and let $s_n = \sqrt{\sum_k \sigma^2(X_k)}$ be the standard deviation of $S_n$. Does $P(1 + |S_n| > \epsilon s_n)$ go to zero uniformly in $n$, as $\epsilon \rightarrow 0$? In other words, if you permit the abuse of notation, is $s_n = O(1 + |S_i|)$ in probability?

Some Background

I'm interested in estimating the size of $S_n$, in probability. Chebyshev's Inequality gives: $S_N = \mathcal{O}(s_n)$ in probability. In fact, if we take $T_n = \max_{i=1}^n |S_i|$$T_n = \max_{1 \leq i \leq n} |S_i|$, then we also have $T_n = O(s_n)$ in probability by Kolmogorov's inequality.

In the other direction, taking advantage of $|X_i| \leq 1$, we have: $P(1 + T_n < \epsilon s_n)$ converges to zero uniformly in $n$, as $\epsilon \rightarrow 0$. A similar statement is used in a proof of Kolmogorov's 3 Series theorem, but I don't know what this theorem itself is called. Thus, by abuse of notation, $s_n = O(1 + T_n)$. Since $T_n$ and $|S_n|$ often have similar size (for example, with Ottaviani's inequality, you can prove that $P(T_n > \lambda) \leq 2 P(|S_n| > \lambda)$ by Levy's Inequality$P(T_n > \lambda) \leq 4 \max_{1 \leq i \leq n} P(|S_i| > \lambda/4)$), I would expect that a similar statement is true for $|S_n|$. However, I can't seem to bridge that gap.

*For reference, when I write $X_\alpha = O(Y_\alpha)$ when $X_\alpha, Y_\alpha$ are random variables and $Y_\alpha \geq 0$, I mean that $\sup_\alpha P(|X_\alpha| \leq K Y_\alpha)$ converges to 0 as $K \rightarrow \infty$.

The Question

Suppose that $X_n$ are independent random variables, $|X_n| \leq 1$, and $\mathbb{E}[X_n] = 0$. Let $S_n = \sum_{i=1}^n X_i$ and let $s_n = \sqrt{\sum_k \sigma^2(X_k)}$ be the standard deviation of $S_n$. Does $P(1 + |S_n| > \epsilon s_n)$ go to zero uniformly in $n$, as $\epsilon \rightarrow 0$? In other words, if you permit the abuse of notation, is $s_n = O(1 + |S_i|)$ in probability?

Some Background

I'm interested in estimating the size of $S_n$, in probability. Chebyshev's Inequality gives: $S_N = \mathcal{O}(s_n)$ in probability. In fact, if we take $T_n = \max_{i=1}^n |S_i|$, then we also have $T_n = O(s_n)$ in probability by Kolmogorov's inequality.

In the other direction, taking advantage of $|X_i| \leq 1$, we have: $P(1 + T_n < \epsilon s_n)$ converges to zero uniformly in $n$, as $\epsilon \rightarrow 0$. A similar statement is used in a proof of Kolmogorov's 3 Series theorem, but I don't know what this theorem itself is called. Thus, by abuse of notation, $s_n = O(1 + T_n)$. Since $T_n$ and $|S_n|$ often have similar size (for example, $P(T_n > \lambda) \leq 2 P(|S_n| > \lambda)$ by Levy's Inequality), I would expect that a similar statement is true for $|S_n|$. However, I can't seem to bridge that gap.

*For reference, when I write $X_\alpha = O(Y_\alpha)$ when $X_\alpha, Y_\alpha$ are random variables and $Y_\alpha \geq 0$, I mean that $\sup_\alpha P(|X_\alpha| \leq K Y_\alpha)$ converges to 0 as $K \rightarrow \infty$.

The Question

Suppose that $X_n$ are independent random variables, $|X_n| \leq 1$, and $\mathbb{E}[X_n] = 0$. Let $S_n = \sum_{i=1}^n X_i$ and let $s_n = \sqrt{\sum_k \sigma^2(X_k)}$ be the standard deviation of $S_n$. Does $P(1 + |S_n| > \epsilon s_n)$ go to zero uniformly in $n$, as $\epsilon \rightarrow 0$? In other words, if you permit the abuse of notation, is $s_n = O(1 + |S_i|)$ in probability?

Some Background

I'm interested in estimating the size of $S_n$, in probability. Chebyshev's Inequality gives: $S_N = \mathcal{O}(s_n)$ in probability. In fact, if we take $T_n = \max_{1 \leq i \leq n} |S_i|$, then we also have $T_n = O(s_n)$ in probability by Kolmogorov's inequality.

In the other direction, taking advantage of $|X_i| \leq 1$, we have: $P(1 + T_n < \epsilon s_n)$ converges to zero uniformly in $n$, as $\epsilon \rightarrow 0$. A similar statement is used in a proof of Kolmogorov's 3 Series theorem, but I don't know what this theorem itself is called. Thus, by abuse of notation, $s_n = O(1 + T_n)$. Since $T_n$ and $|S_n|$ often have similar size (for example, with Ottaviani's inequality, you can prove that $P(T_n > \lambda) \leq 4 \max_{1 \leq i \leq n} P(|S_i| > \lambda/4)$), I would expect that a similar statement is true for $|S_n|$. However, I can't seem to bridge that gap.

*For reference, when I write $X_\alpha = O(Y_\alpha)$ when $X_\alpha, Y_\alpha$ are random variables and $Y_\alpha \geq 0$, I mean that $\sup_\alpha P(|X_\alpha| \leq K Y_\alpha)$ converges to 0 as $K \rightarrow \infty$.

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Theone
Theone
  • 143
  • 5

Is a sum of a bounded random variables the same order as its standard deviation?

The Question

Suppose that $X_n$ are independent random variables, $|X_n| \leq 1$, and $\mathbb{E}[X_n] = 0$. Let $S_n = \sum_{i=1}^n X_i$ and let $s_n = \sqrt{\sum_k \sigma^2(X_k)}$ be the standard deviation of $S_n$. Does $P(1 + |S_n| > \epsilon s_n)$ go to zero uniformly in $n$, as $\epsilon \rightarrow 0$? In other words, if you permit the abuse of notation, is $s_n = O(1 + |S_i|)$ in probability?

Some Background

I'm interested in estimating the size of $S_n$, in probability. Chebyshev's Inequality gives: $S_N = \mathcal{O}(s_n)$ in probability. In fact, if we take $T_n = \max_{i=1}^n |S_i|$, then we also have $T_n = O(s_n)$ in probability by Kolmogorov's inequality.

In the other direction, taking advantage of $|X_i| \leq 1$, we have: $P(1 + T_n < \epsilon s_n)$ converges to zero uniformly in $n$, as $\epsilon \rightarrow 0$. A similar statement is used in a proof of Kolmogorov's 3 Series theorem, but I don't know what this theorem itself is called. Thus, by abuse of notation, $s_n = O(1 + T_n)$. Since $T_n$ and $|S_n|$ often have similar size (for example, $P(T_n > \lambda) \leq 2 P(|S_n| > \lambda)$ by Levy's Inequality), I would expect that a similar statement is true for $|S_n|$. However, I can't seem to bridge that gap.

*For reference, when I write $X_\alpha = O(Y_\alpha)$ when $X_\alpha, Y_\alpha$ are random variables and $Y_\alpha \geq 0$, I mean that $\sup_\alpha P(|X_\alpha| \leq K Y_\alpha)$ converges to 0 as $K \rightarrow \infty$.