Skip to main content
added 22 characters in body
Source Link
Iosif Pinelis
  • 127.8k
  • 8
  • 107
  • 229

Suppose e.g. that $X=X_1+\dots+X_n$, where the $X_i$' are independent zero-mean random vectors with $\|X_i\|\le1$ for all $i$. Then the Hoeffding--Azuma inequality (see e.g. Theorem 3.1Wikipedia) yields $$P(\|X\|>u)\ge e^{-(E\|X\|-u)^2/(2n)}$$$$P(\|X\|>u)\ge1- e^{-(E\|X\|-u)^2/(2n)}$$ for $u\le E\|X\|$.

A number of lower bounds on $P(\|X\|\ge u)$ were obtained by de Acosta A. and Samur J.D. (Infinitely divisible probability measures and the converse Kolmogorov inequality in Banach spaces. Studia Math. 1979. V. 66, 143--160). For instance, a special case, for $p=2$, of their Corollary 3.1 on p. 151 is the following: $$P(\|X\|>u)\ge \frac14\,\Big(1-\frac{(u+1)^2+u^2/2}{E\|X\|^2}\Big)$$ for $u>0$, where $X$ is just as above.

Also, in the proof of their Lemma 2.3, de Acosta and Samur showed that, if the $X_i$'s are also symmetric, then $$P(\|X\|>u)\ge\frac12\,P(\max_i\|X_i\|>u) \ge\frac12\,\Big(1-\exp\Big\{-\sum_i P(\|X_i\|>u)\Big\}\Big)$$ for $u>0$.

Suppose e.g. that $X=X_1+\dots+X_n$, where the $X_i$' are independent zero-mean random vectors with $\|X_i\|\le1$ for all $i$. Then the Hoeffding--Azuma inequality (see e.g. Theorem 3.1) yields $$P(\|X\|>u)\ge e^{-(E\|X\|-u)^2/(2n)}$$ for $u\le E\|X\|$.

A number of lower bounds on $P(\|X\|\ge u)$ were obtained by de Acosta A. and Samur J.D. (Infinitely divisible probability measures and the converse Kolmogorov inequality in Banach spaces. Studia Math. 1979. V. 66, 143--160). For instance, a special case, for $p=2$, of their Corollary 3.1 on p. 151 is the following: $$P(\|X\|>u)\ge \frac14\,\Big(1-\frac{(u+1)^2+u^2/2}{E\|X\|^2}\Big)$$ for $u>0$, where $X$ is just as above.

Also, in the proof of their Lemma 2.3, de Acosta and Samur showed that, if the $X_i$'s are also symmetric, then $$P(\|X\|>u)\ge\frac12\,P(\max_i\|X_i\|>u) \ge\frac12\,\Big(1-\exp\Big\{-\sum_i P(\|X_i\|>u)\Big\}\Big)$$ for $u>0$.

Suppose e.g. that $X=X_1+\dots+X_n$, where the $X_i$' are independent zero-mean random vectors with $\|X_i\|\le1$ for all $i$. Then the Hoeffding--Azuma inequality (see e.g. Wikipedia) yields $$P(\|X\|>u)\ge1- e^{-(E\|X\|-u)^2/(2n)}$$ for $u\le E\|X\|$.

A number of lower bounds on $P(\|X\|\ge u)$ were obtained by de Acosta A. and Samur J.D. (Infinitely divisible probability measures and the converse Kolmogorov inequality in Banach spaces. Studia Math. 1979. V. 66, 143--160). For instance, a special case, for $p=2$, of their Corollary 3.1 on p. 151 is the following: $$P(\|X\|>u)\ge \frac14\,\Big(1-\frac{(u+1)^2+u^2/2}{E\|X\|^2}\Big)$$ for $u>0$, where $X$ is just as above.

Also, in the proof of their Lemma 2.3, de Acosta and Samur showed that, if the $X_i$'s are also symmetric, then $$P(\|X\|>u)\ge\frac12\,P(\max_i\|X_i\|>u) \ge\frac12\,\Big(1-\exp\Big\{-\sum_i P(\|X_i\|>u)\Big\}\Big)$$ for $u>0$.

deleted 3 characters in body
Source Link
Iosif Pinelis
  • 127.8k
  • 8
  • 107
  • 229

Suppose e.g. that $X=X_1+\dots+X_n$, where the $X_i$' are independent zero-mean random vectors with $\|X_i\|\le1$ for all $i$. Then the Hoeffding--Azuma inequality (see e.g. Theorem 3.1) yields $$P(\|X\|>u)\ge e^{-(E\|X\|-u)^2/(2n)}$$ for $u\le E\|X\|$.

A number of lower bounds on $P(\|X\|\ge u)$ were obtained by de Acosta A. and Samur J.D. (Infinitely divisible probability measures and the converse Kolmogorov inequality in Banach spaces. Studia Math. 1979. V. 66, 143--160). For instance, a special case, for $p=2$, of their Corollary 3.1 on p. 151 is the following: $$P(\|X\|>u)\ge \frac14\,\Big(1-\frac{(u+1)^2+u^2/2}{E\|X\|^2}\Big)$$ for $u>0$, where $X$ is as just as above.

Also, in the proof of their Lemma 2.3, de Acosta and Samur showed that, if the $X_i$'s are also symmetric, then $$P(\|X\|>u)\ge\frac12\,P(\max_i\|X_i\|>u) \ge\frac12\,\Big(1-\exp\Big\{-\sum_i P(\|X_i\|>u)\Big\}\Big)$$ for $u>0$.

Suppose e.g. that $X=X_1+\dots+X_n$, where the $X_i$' are independent zero-mean random vectors with $\|X_i\|\le1$ for all $i$. Then the Hoeffding--Azuma inequality (see e.g. Theorem 3.1) yields $$P(\|X\|>u)\ge e^{-(E\|X\|-u)^2/(2n)}$$ for $u\le E\|X\|$.

A number of lower bounds on $P(\|X\|\ge u)$ were obtained by de Acosta A. and Samur J.D. (Infinitely divisible probability measures and the converse Kolmogorov inequality in Banach spaces. Studia Math. 1979. V. 66, 143--160). For instance, a special case, for $p=2$, of their Corollary 3.1 on p. 151 is the following: $$P(\|X\|>u)\ge \frac14\,\Big(1-\frac{(u+1)^2+u^2/2}{E\|X\|^2}\Big)$$ for $u>0$, where $X$ is as just as above.

Also, in the proof of their Lemma 2.3, de Acosta and Samur showed that, if the $X_i$'s are also symmetric, then $$P(\|X\|>u)\ge\frac12\,P(\max_i\|X_i\|>u) \ge\frac12\,\Big(1-\exp\Big\{-\sum_i P(\|X_i\|>u)\Big\}\Big)$$ for $u>0$.

Suppose e.g. that $X=X_1+\dots+X_n$, where the $X_i$' are independent zero-mean random vectors with $\|X_i\|\le1$ for all $i$. Then the Hoeffding--Azuma inequality (see e.g. Theorem 3.1) yields $$P(\|X\|>u)\ge e^{-(E\|X\|-u)^2/(2n)}$$ for $u\le E\|X\|$.

A number of lower bounds on $P(\|X\|\ge u)$ were obtained by de Acosta A. and Samur J.D. (Infinitely divisible probability measures and the converse Kolmogorov inequality in Banach spaces. Studia Math. 1979. V. 66, 143--160). For instance, a special case, for $p=2$, of their Corollary 3.1 on p. 151 is the following: $$P(\|X\|>u)\ge \frac14\,\Big(1-\frac{(u+1)^2+u^2/2}{E\|X\|^2}\Big)$$ for $u>0$, where $X$ is just as above.

Also, in the proof of their Lemma 2.3, de Acosta and Samur showed that, if the $X_i$'s are also symmetric, then $$P(\|X\|>u)\ge\frac12\,P(\max_i\|X_i\|>u) \ge\frac12\,\Big(1-\exp\Big\{-\sum_i P(\|X_i\|>u)\Big\}\Big)$$ for $u>0$.

added 242 characters in body
Source Link
Iosif Pinelis
  • 127.8k
  • 8
  • 107
  • 229

Suppose e.g. that $X=X_1+\dots+X_n$, where the $X_i$' are independent zero-mean random vectors with $\|X_i\|\le1$ for all $i$. Then the Hoeffding--Azuma inequality (see e.g. Theorem 3.1) yields $$P(\|X\|>u)\ge e^{-(E\|X\|-u)^2/(2n)}$$ for $u\le E\|X\|$.

A number of lower bounds on $P(\|X\|\ge u)$ were obtained inby de Acosta A., and Samur J.D. Infinitely(Infinitely divisible probability measures and the converse Kolmogorov inequality in Banach spaces. Studia Math. 1979. V. 66, 143--160). For instance, a special case, for $p=2$, of their Corollary 3.1 on p. 151 is the following: $$P(\|X\|>u)\ge \frac14\,\Big(1-\frac{(u+1)^2+u^2/2}{E\|X\|^2}\Big)$$ for $u>0$, where $X$ is as just as above.

Also, in the proof of their Lemma 2.3, de Acosta and Samur showed that, if the $X_i$'s are also symmetric, then $$P(\|X\|>u)\ge\frac12\,P(\max_i\|X_i\|>u) \ge\frac12\,\Big(1-\exp\Big\{-\sum_i P(\|X_i\|>u)\Big\}\Big)$$ for $u>0$.

Suppose e.g. that $X=X_1+\dots+X_n$, where the $X_i$' are independent zero-mean random vectors with $\|X_i\|\le1$ for all $i$. Then the Hoeffding--Azuma inequality (see e.g. Theorem 3.1) yields $$P(\|X\|>u)\ge e^{-(E\|X\|-u)^2/(2n)}$$ for $u\le E\|X\|$.

A number of lower bounds on $P(\|X\|\ge u)$ were obtained in de Acosta A., Samur J.D. Infinitely divisible probability measures and the converse Kolmogorov inequality in Banach spaces. Studia Math. 1979. V. 66, 143--160. For instance, a special case, for $p=2$, of their Corollary 3.1 on p. 151 is the following: $$P(\|X\|>u)\ge \frac14\,\Big(1-\frac{(u+1)^2+u^2/2}{E\|X\|^2}\Big)$$ for $u>0$, where $X$ is as just as above.

Suppose e.g. that $X=X_1+\dots+X_n$, where the $X_i$' are independent zero-mean random vectors with $\|X_i\|\le1$ for all $i$. Then the Hoeffding--Azuma inequality (see e.g. Theorem 3.1) yields $$P(\|X\|>u)\ge e^{-(E\|X\|-u)^2/(2n)}$$ for $u\le E\|X\|$.

A number of lower bounds on $P(\|X\|\ge u)$ were obtained by de Acosta A. and Samur J.D. (Infinitely divisible probability measures and the converse Kolmogorov inequality in Banach spaces. Studia Math. 1979. V. 66, 143--160). For instance, a special case, for $p=2$, of their Corollary 3.1 on p. 151 is the following: $$P(\|X\|>u)\ge \frac14\,\Big(1-\frac{(u+1)^2+u^2/2}{E\|X\|^2}\Big)$$ for $u>0$, where $X$ is as just as above.

Also, in the proof of their Lemma 2.3, de Acosta and Samur showed that, if the $X_i$'s are also symmetric, then $$P(\|X\|>u)\ge\frac12\,P(\max_i\|X_i\|>u) \ge\frac12\,\Big(1-\exp\Big\{-\sum_i P(\|X_i\|>u)\Big\}\Big)$$ for $u>0$.

Source Link
Iosif Pinelis
  • 127.8k
  • 8
  • 107
  • 229
Loading