Sum of variables of a martingale
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I have the sequence $X_1, X_2,...X_n$ as a martingale, each of which is bounded. Now I want to explore some upper bound for the sum $S_n=X_1+X_2+...+X_n$, e.g., the format like Hoeffding inequality or Berstein inequality. Any method or conclusions suggestions for this? I have no idea at all and can't find any related conclusions.
probability-theory martingales concentration-of-measure large-deviation-theory
asked Dec 16 '18 at 14:58
JorisJoris
11
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add a comment |
$begingroup$
I have the sequence $X_1, X_2,...X_n$ as a martingale, each of which is bounded. Now I want to explore some upper bound for the sum $S_n=X_1+X_2+...+X_n$, e.g., the format like Hoeffding inequality or Berstein inequality. Any method or conclusions suggestions for this? I have no idea at all and can't find any related conclusions.
probability-theory martingales concentration-of-measure large-deviation-theory
asked Dec 16 '18 at 14:58
JorisJoris
11
$endgroup$
$begingroup$
Maybe this can be what you are looking for en.m.wikipedia.org/wiki/Azuma%27s_inequality. But the assumption should be that $X_n$ is a martingale difference sequence.
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– Song
Dec 16 '18 at 15:10
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Thank you. I have considered that but the assumption is different as you pointed out, which makes it difficult to apply it to my problem.
$endgroup$
– Joris
Dec 16 '18 at 15:13
add a comment |
$begingroup$
I have the sequence $X_1, X_2,...X_n$ as a martingale, each of which is bounded. Now I want to explore some upper bound for the sum $S_n=X_1+X_2+...+X_n$, e.g., the format like Hoeffding inequality or Berstein inequality. Any method or conclusions suggestions for this? I have no idea at all and can't find any related conclusions.
probability-theory martingales concentration-of-measure large-deviation-theory
asked Dec 16 '18 at 14:58
JorisJoris
11
$endgroup$
I have the sequence $X_1, X_2,...X_n$ as a martingale, each of which is bounded. Now I want to explore some upper bound for the sum $S_n=X_1+X_2+...+X_n$, e.g., the format like Hoeffding inequality or Berstein inequality. Any method or conclusions suggestions for this? I have no idea at all and can't find any related conclusions.
probability-theory martingales concentration-of-measure large-deviation-theory
probability-theory martingales concentration-of-measure large-deviation-theory
asked Dec 16 '18 at 14:58
JorisJoris
11
asked Dec 16 '18 at 14:58
JorisJoris
11
edited Dec 16 '18 at 15:05
Joris
asked Dec 16 '18 at 14:58
JorisJoris
11
asked Dec 16 '18 at 14:58
JorisJoris
11
asked Dec 16 '18 at 14:58
JorisJoris
11
11
$begingroup$
Maybe this can be what you are looking for en.m.wikipedia.org/wiki/Azuma%27s_inequality. But the assumption should be that $X_n$ is a martingale difference sequence.
$endgroup$
– Song
Dec 16 '18 at 15:10
$begingroup$
Thank you. I have considered that but the assumption is different as you pointed out, which makes it difficult to apply it to my problem.
$endgroup$
– Joris
Dec 16 '18 at 15:13
add a comment |
$begingroup$
Maybe this can be what you are looking for en.m.wikipedia.org/wiki/Azuma%27s_inequality. But the assumption should be that $X_n$ is a martingale difference sequence.
$endgroup$
– Song
Dec 16 '18 at 15:10
$begingroup$
Thank you. I have considered that but the assumption is different as you pointed out, which makes it difficult to apply it to my problem.
$endgroup$
– Joris
Dec 16 '18 at 15:13
$begingroup$
Maybe this can be what you are looking for en.m.wikipedia.org/wiki/Azuma%27s_inequality. But the assumption should be that $X_n$ is a martingale difference sequence.
$endgroup$
– Song
Dec 16 '18 at 15:10
$begingroup$
Maybe this can be what you are looking for en.m.wikipedia.org/wiki/Azuma%27s_inequality. But the assumption should be that $X_n$ is a martingale difference sequence.
$endgroup$
– Song
Dec 16 '18 at 15:10
$begingroup$
Thank you. I have considered that but the assumption is different as you pointed out, which makes it difficult to apply it to my problem.
$endgroup$
– Joris
Dec 16 '18 at 15:13
$begingroup$
Thank you. I have considered that but the assumption is different as you pointed out, which makes it difficult to apply it to my problem.
$endgroup$
– Joris
Dec 16 '18 at 15:13
add a comment |
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$begingroup$
Maybe this can be what you are looking for en.m.wikipedia.org/wiki/Azuma%27s_inequality. But the assumption should be that $X_n$ is a martingale difference sequence.
$endgroup$
– Song
Dec 16 '18 at 15:10
$begingroup$
Thank you. I have considered that but the assumption is different as you pointed out, which makes it difficult to apply it to my problem.
$endgroup$
– Joris
Dec 16 '18 at 15:13