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Good day to All.

Let $S_{1,n} = \sum_{i=1}^{n}\xi_{i}$, where $(\xi_{i})_{i \in \mathbb{N}}$ be independent RV with values in some Banach space.

On pages 79-80 in this book author provides an example that illustrates the fact that in the infinite dimensional spaces it may be not enough to have assumptions on the distribution of the individual summands $\xi_{i}$ in order to control the deviations of $\lVert {S_{1,n}}\rVert $, where $\lVert \cdot \rVert $ is the norm of corresponding Banach space. Author says:"The following example serves to show the relative independence of restrictions on concentration of the sum and those concerning the distributions of individual summands ".

In his example 3.0.1 let the space $c_0$ (with the norm $\|x\|_{\infty} = \sup_{n \in \mathbb{N}} |x^{(n)}|$) be the space of real-valued convergent to zero sequences and we consider an array of i.i.d. real-valued $\xi_{j}^{i} = \frac{1}{2}\delta_{-1}+\frac{1}{2}\delta_{1}$.

Consider the sequence $\xi_{j} = ( \frac{\xi_{j}^{(i)}}{\ln\ln(10 i)}, i \in \mathbb{N} )$ which are centered, in $c_{0}$ and a.s. bounded, i.e. $\|\xi_j\|_{\infty} = \frac{1}{\ln \ln 10}$. Consider RV's $\eta_{n}^{(i)} = \sum_{j=1}^{n}\xi_{j}^{(i)}$ and notice that they are independent. One easily sees that $\mathbb{P}(\eta_{n}^{i}=n) = 2^{-n}$ and therefore it implies: \begin{align*} \mathbb{P}(\max_{i \leq 2^n}\eta_{n}^{(i)}<n) = \prod_{k=1}^{2^n}(1-\mathbb{P}(\eta_{n}^{(i)}=n)) \leq \exp({-2^{n} 2^{-n})} = e^{-1}, \end{align*} where we used classical $1-x \leq \exp(-x)$ for $x\geq 0$. Thefore, for all $n$ we have: \begin{align*} P\left(\|S_{1,n}\| \geq \frac{n}{\ln \ln (10 * 2^n)}\right) \geq P\left(\max_{i \leq 2^n}\eta_{n}^{(i)} \geq n\right) \geq 1-e^{-1}. \end{align*}

Thus, the norm of sum scales (with constant probability) like $\frac{n}{\ln (n)}$, whereas $\|\xi_{j}\|_{\infty} = \frac{1}{\ln \ln (10)}$.

My question: can't we also consider $l^{\infty} := \{x=(x^{(n)}) : \sup_{n\in \mathbb N}| x^{(n)}| \lt \infty\}$ (which is Banach space with respect to the norm $\|x\|_{\infty} = \sup_{n \in \mathbb N} |x^{(n)}|$) and just the sequences $\xi_{j} = (\xi_{j}^{j}, j \in \mathbb{N})$ (without additional log weight ) and $\xi_{j}^{i} = \frac{1}{2}\delta_{-1}+\frac{1}{2}\delta_{1}$ i.i.d in $i,j$ as in example 3.0.1? Then, it seems the same reasoning as in the book applies for this case.

Good day to All.

Let $S_{1,n} = \sum_{i=1}^{n}\xi_{i}$, where $(\xi_{i})_{i \in \mathbb{N}}$ be independent RV with values in some Banach space.

On pages 79-80 in this book author provides an example that illustrates the fact that in the infinite dimensional spaces it may be not enough to have assumptions on the distribution of the individual summands $\xi_{i}$ in order to control the deviations of $\lVert {S_{1,n}}\rVert $, where $\lVert \cdot \rVert $ is the norm of corresponding Banach space. 

Author says:"In infinite dimensions, "individual" conditions may not suffice: it is sometimes necessary to include restrictions on the distribution of the sum in order to obtain stronger bounds for its "tails".The following example serves to show the relative independence of restrictions on concentration of the sum and those concerning the distributions of individual summands ".

In his example 3.0.1 let the space $c_0$ (with the norm $\|x\|_{\infty} = \sup_{n \in \mathbb{N}} |x^{(n)}|$) be the space of real-valued convergent to zero sequences and we consider an array of i.i.d. real-valued $\xi_{j}^{i} = \frac{1}{2}\delta_{-1}+\frac{1}{2}\delta_{1}$.

Consider the sequence $\xi_{j} = ( \frac{\xi_{j}^{(i)}}{\ln\ln(10 i)}, i \in \mathbb{N} )$ which is centered, in $c_{0}$ and a.s. bounded, i.e. $\|\xi_j\|_{\infty} = \frac{1}{\ln \ln 10}$. Consider RV's $\eta_{n}^{(i)} = \sum_{j=1}^{n}\xi_{j}^{(i)}$ and notice that they are independent. One easily sees that $\mathbb{P}(\eta_{n}^{i}=n) = 2^{-n}$ and therefore it implies: \begin{align*} \mathbb{P}(\max_{i \leq 2^n}\eta_{n}^{(i)}<n) = \prod_{k=1}^{2^n}(1-\mathbb{P}(\eta_{n}^{(i)}=n)) \leq \exp({-2^{n} 2^{-n})} = e^{-1}, \end{align*} where we used classical $1-x \leq \exp(-x)$ for $x\geq 0$. Thefore, for all $n$ we have: \begin{align*} P\left(\|S_{1,n}\|_{\infty} \geq \frac{n}{\ln \ln (10 * 2^n)}\right) \geq P\left(\max_{i \leq 2^n}\eta_{n}^{(i)} \geq n\right) \geq 1-e^{-1}. \end{align*}

Thus, the norm of sum scales (with constant probability separated from zero) like $\frac{n}{\ln (n)}$, even though its coordinates have variance of order $\sqrt{n}$.

I don't really understand why author gives the final remark about the variance of the coordinates, although (as it seemed to me, maybe I am wrong) the qoal was to show that even though we have for each $i \leq n$, $\|\xi_{i}\|_{\infty} \leq 1$ a.s., the typical deviations of $\|S_{1,n}\|_{\infty}$ is of order $n/\ln(n)$ (and not for example $\sqrt{n}$ as it would be in some Hilbert space).

Also my question: can't we also consider $l^{\infty} := \{x=(x^{(n)}) : \sup_{n\in \mathbb N}| x^{(n)}| \lt \infty\}$ (which is Banach space with respect to the norm $\|x\|_{\infty} = \sup_{n \in \mathbb N} |x^{(n)}|$) and just the sequences $\xi_{j} = (\xi_{j}^{j}, j \in \mathbb{N})$ (without additional log weight ) and $\xi_{j}^{i} = \frac{1}{2}\delta_{-1}+\frac{1}{2}\delta_{1}$ i.i.d in $i,j$ as in example 3.0.1? Then, it seems the same reasoning as in the book applies for this case.

Good day to All.

Let $S_{1,n} = \sum_{i=1}^{n}\xi_{i}$, where $(\xi_{i})_{i \in \mathbb{N}}$ be independent RV with values in some Banach space.

On pages 79-80 in this book author provides an example that illustrates the fact that in the infinite dimensional spaces it may be not enough to have assumptions on the distribution of the individual summands $\xi_{i}$ in order to control the deviations of $\lVert {S_{1,n}}\rVert $, where $\lVert \cdot \rVert $ is the norm of corresponding Banach space. Author says:

\begin{quote} The following example serves to show the relative independence of restrictions on concentration of the sum and those concerning the distributions of individual summands. \end{quote}

In his example 3.0.1 let the space $c_0$ (with the norm $\|x\|_{\infty} = \sup_{n \in \mathbb{N}} |x^{(n)}|$) be the space of real-valued convergent to zero sequences and we consider an array of i.i.d. real-valued $\xi_{j}^{i} = \frac{1}{2}\delta_{-1}+\frac{1}{2}\delta_{1}$.

Consider the sequence $\xi_{j} = (\xi_{j}^{i} = \frac{\xi_{j}^{i}}{\ln\ln(10 i)}, i \in \mathbb{N} )$ which are centered, in $c_{0}$ and a.s. bounded, i.e. $\|\xi_j\|_{\infty} = \frac{1}{\ln \ln 10}$. Consider RV's $\eta_{n}^{i} = \sum_{j=1}^{n}\xi_{j}^{i}$ and notice that they are independent. One easily sees that $\mathbb{P}(\eta_{n}^{i}=n) = 2^{-n}$ and therefore it implies: \begin{align*} \mathbb{P}(\max_{i \leq 2^n}\eta_{n}^{i}<n) = \prod_{k=1}^{2^n}(1-\mathbb{P}(\eta_{n}^{i}=n)) \leq \exp\paren{-2^{n} 2^{-n}} = e^{-1}, \end{align*} where we used classical $1-x \leq \exp(-x)$ for $x\geq 0$. Thefore, for all $n$ we have: \begin{align*} P\left(\|S_{1,n}\| \geq \frac{n}{\ln \ln (10 2^n)}\right) \geq P\left(\max_{i \leq 2^n}\eta_{n}^{i} \geq n\right) \geq 1-e^{-1}. \end{align*}

Thus, the norm of sum scales (with constant probability) like $\frac{n}{\ln (n)}$, whereas $\|\xi\|_{\infty} = \frac{1}{\ln \ln (10)}$.

My question: can't we also consider $l^{\infty} := \{x=(x^{(n)}) : \sup_{n\in \mathbb N}| x^{(n)}| \lt \infty\}$ (which is Banach space with respect to the norm $\|x\|_{\infty} = \sup_{n \in \mathbb N} |x^{(n)}|$) and just the sequences $\xi_{j} = (\xi_{j}^{j}, j \in \mathbb{N})$ (without additional log weight ) and $\xi_{j}^{i} = \frac{1}{2}\delta_{-1}+\frac{1}{2}\delta_{1}$ i.i.d in $i,j$ as in example 3.0.1 to illustrate the same fact?

Good day to All.

Let $S_{1,n} = \sum_{i=1}^{n}\xi_{i}$, where $(\xi_{i})_{i \in \mathbb{N}}$ be independent RV with values in some Banach space.

On pages 79-80 in this book author provides an example that illustrates the fact that in the infinite dimensional spaces it may be not enough to have assumptions on the distribution of the individual summands $\xi_{i}$ in order to control the deviations of $\lVert {S_{1,n}}\rVert $, where $\lVert \cdot \rVert $ is the norm of corresponding Banach space. Author says:"The following example serves to show the relative independence of restrictions on concentration of the sum and those concerning the distributions of individual summands ".

In his example 3.0.1 let the space $c_0$ (with the norm $\|x\|_{\infty} = \sup_{n \in \mathbb{N}} |x^{(n)}|$) be the space of real-valued convergent to zero sequences and we consider an array of i.i.d. real-valued $\xi_{j}^{i} = \frac{1}{2}\delta_{-1}+\frac{1}{2}\delta_{1}$.

Consider the sequence $\xi_{j} = ( \frac{\xi_{j}^{(i)}}{\ln\ln(10 i)}, i \in \mathbb{N} )$ which are centered, in $c_{0}$ and a.s. bounded, i.e. $\|\xi_j\|_{\infty} = \frac{1}{\ln \ln 10}$. Consider RV's $\eta_{n}^{(i)} = \sum_{j=1}^{n}\xi_{j}^{(i)}$ and notice that they are independent. One easily sees that $\mathbb{P}(\eta_{n}^{i}=n) = 2^{-n}$ and therefore it implies: \begin{align*} \mathbb{P}(\max_{i \leq 2^n}\eta_{n}^{(i)}<n) = \prod_{k=1}^{2^n}(1-\mathbb{P}(\eta_{n}^{(i)}=n)) \leq \exp({-2^{n} 2^{-n})} = e^{-1}, \end{align*} where we used classical $1-x \leq \exp(-x)$ for $x\geq 0$. Thefore, for all $n$ we have: \begin{align*} P\left(\|S_{1,n}\| \geq \frac{n}{\ln \ln (10 * 2^n)}\right) \geq P\left(\max_{i \leq 2^n}\eta_{n}^{(i)} \geq n\right) \geq 1-e^{-1}. \end{align*}

Thus, the norm of sum scales (with constant probability) like $\frac{n}{\ln (n)}$, whereas $\|\xi_{j}\|_{\infty} = \frac{1}{\ln \ln (10)}$.

My question: can't we also consider $l^{\infty} := \{x=(x^{(n)}) : \sup_{n\in \mathbb N}| x^{(n)}| \lt \infty\}$ (which is Banach space with respect to the norm $\|x\|_{\infty} = \sup_{n \in \mathbb N} |x^{(n)}|$) and just the sequences $\xi_{j} = (\xi_{j}^{j}, j \in \mathbb{N})$ (without additional log weight ) and $\xi_{j}^{i} = \frac{1}{2}\delta_{-1}+\frac{1}{2}\delta_{1}$ i.i.d in $i,j$ as in example 3.0.1? Then, it seems the same reasoning as in the book applies for this case.

Good day to All.

Let $S_{1,n} = \sum_{i=1}^{n}\xi_{i}$, where $(\xi_{i})_{i \in \mathbb{N}}$ be independent RV with values in some Banach space.

On page 80 in this book author provides an example that illustrates the fact that in the infinite dimensional spaces it may be not enough to have assumptions on the distribution of the individual summands $\xi_{i}$ in order to control the deviations of $\lVert {S_{1,n}}\rVert $, where $\lVert \cdot \rVert $ is the norm of corresponding Banach space. Namely, in his example 3.0.1 the space $c_0$ (with the norm $\|x\|_{\infty} = \sup_{n \in \mathbb N} |x^{(n)}|$) of real-valued convergent to zero sequences is considered.

My question: can't we also consider $l^{\infty} := \{x=(x^{(n)}) : \sup_{n\in \mathbb N}| x^{(n)}| \lt \infty\}$ (which is Banach space with respect to the norm $\|x\|_{\infty} = \sup_{n \in \mathbb N} |x^{(n)}|$) and just the sequences $\xi_{j} = (\xi_{j}^{j}, j \in \mathbb{N})$ (without additional log weight ) and $\xi_{j}^{i} = \frac{1}{2}\delta_{-1}+\frac{1}{2}\delta_{1}$ i.i.d in $i,j$ as in example 3.0.1 to illustrate the same fact?

Good day to All.

Let $S_{1,n} = \sum_{i=1}^{n}\xi_{i}$, where $(\xi_{i})_{i \in \mathbb{N}}$ be independent RV with values in some Banach space.

On pages 79-80 in this book author provides an example that illustrates the fact that in the infinite dimensional spaces it may be not enough to have assumptions on the distribution of the individual summands $\xi_{i}$ in order to control the deviations of $\lVert {S_{1,n}}\rVert $, where $\lVert \cdot \rVert $ is the norm of corresponding Banach space. Author says:

\begin{quote} The following example serves to show the relative independence of restrictions on concentration of the sum and those concerning the distributions of individual summands. \end{quote}

In his example 3.0.1 let the space $c_0$ (with the norm $\|x\|_{\infty} = \sup_{n \in \mathbb{N}} |x^{(n)}|$) be the space of real-valued convergent to zero sequences and we consider an array of i.i.d. real-valued $\xi_{j}^{i} = \frac{1}{2}\delta_{-1}+\frac{1}{2}\delta_{1}$.

Consider the sequence $\xi_{j} = (\xi_{j}^{i} = \frac{\xi_{j}^{i}}{\ln\ln(10 i)}, i \in \mathbb{N} )$ which are centered, in $c_{0}$ and a.s. bounded, i.e. $\|\xi_j\|_{\infty} = \frac{1}{\ln \ln 10}$. Consider RV's $\eta_{n}^{i} = \sum_{j=1}^{n}\xi_{j}^{i}$ and notice that they are independent. One easily sees that $\mathbb{P}(\eta_{n}^{i}=n) = 2^{-n}$ and therefore it implies: \begin{align*} \mathbb{P}(\max_{i \leq 2^n}\eta_{n}^{i}<n) = \prod_{k=1}^{2^n}(1-\mathbb{P}(\eta_{n}^{i}=n)) \leq \exp\paren{-2^{n} 2^{-n}} = e^{-1}, \end{align*} where we used classical $1-x \leq \exp(-x)$ for $x\geq 0$. Thefore, for all $n$ we have: \begin{align*} P\left(\|S_{1,n}\| \geq \frac{n}{\ln \ln (10 2^n)}\right) \geq P\left(\max_{i \leq 2^n}\eta_{n}^{i} \geq n\right) \geq 1-e^{-1}. \end{align*}

Thus, the norm of sum scales (with constant probability) like $\frac{n}{\ln (n)}$, whereas $\|\xi\|_{\infty} = \frac{1}{\ln \ln (10)}$.

My question: can't we also consider $l^{\infty} := \{x=(x^{(n)}) : \sup_{n\in \mathbb N}| x^{(n)}| \lt \infty\}$ (which is Banach space with respect to the norm $\|x\|_{\infty} = \sup_{n \in \mathbb N} |x^{(n)}|$) and just the sequences $\xi_{j} = (\xi_{j}^{j}, j \in \mathbb{N})$ (without additional log weight ) and $\xi_{j}^{i} = \frac{1}{2}\delta_{-1}+\frac{1}{2}\delta_{1}$ i.i.d in $i,j$ as in example 3.0.1 to illustrate the same fact?