How to strengthen the maximal inquality in stochastic process? [closed]
Let ${X_n}$ be a submartingale. Strengthen the maximal inequality to
$lambda Pr{max_{0leq k leq n} X_k>lambda}leq E[X_n I{max_{0leq kleq n} X_k>lambda}]leq E[X_n^{+}]leq E[|X_n|]$, $lambda>0$
Note, the maximal inequality requires $X_kgeq 0$ for all $k$. The above does not.
Consequently, for a martingale ${X_n}$, we have,
$lambda Pr{max_{0leq k leq n} |X_k|>lambda}leq E[X_n I{max_{0leq kleq n} |X_k|>lambda}]$, $lambda>0$
stochastic-processes
asked Dec 27 '18 at 8:45
gloria
42
closed as unclear what you're asking by Saad, saz, Eric Wofsey, KReiser, Paul Frost Dec 28 '18 at 12:19
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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Let ${X_n}$ be a submartingale. Strengthen the maximal inequality to
$lambda Pr{max_{0leq k leq n} X_k>lambda}leq E[X_n I{max_{0leq kleq n} X_k>lambda}]leq E[X_n^{+}]leq E[|X_n|]$, $lambda>0$
Note, the maximal inequality requires $X_kgeq 0$ for all $k$. The above does not.
Consequently, for a martingale ${X_n}$, we have,
$lambda Pr{max_{0leq k leq n} |X_k|>lambda}leq E[X_n I{max_{0leq kleq n} |X_k|>lambda}]$, $lambda>0$
stochastic-processes
asked Dec 27 '18 at 8:45
gloria
42
closed as unclear what you're asking by Saad, saz, Eric Wofsey, KReiser, Paul Frost Dec 28 '18 at 12:19
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
Let ${X_n}$ be a submartingale. Strengthen the maximal inequality to
$lambda Pr{max_{0leq k leq n} X_k>lambda}leq E[X_n I{max_{0leq kleq n} X_k>lambda}]leq E[X_n^{+}]leq E[|X_n|]$, $lambda>0$
Note, the maximal inequality requires $X_kgeq 0$ for all $k$. The above does not.
Consequently, for a martingale ${X_n}$, we have,
$lambda Pr{max_{0leq k leq n} |X_k|>lambda}leq E[X_n I{max_{0leq kleq n} |X_k|>lambda}]$, $lambda>0$
stochastic-processes
asked Dec 27 '18 at 8:45
gloria
42
Let ${X_n}$ be a submartingale. Strengthen the maximal inequality to
$lambda Pr{max_{0leq k leq n} X_k>lambda}leq E[X_n I{max_{0leq kleq n} X_k>lambda}]leq E[X_n^{+}]leq E[|X_n|]$, $lambda>0$
Note, the maximal inequality requires $X_kgeq 0$ for all $k$. The above does not.
Consequently, for a martingale ${X_n}$, we have,
$lambda Pr{max_{0leq k leq n} |X_k|>lambda}leq E[X_n I{max_{0leq kleq n} |X_k|>lambda}]$, $lambda>0$
stochastic-processes
stochastic-processes
asked Dec 27 '18 at 8:45
gloria
42
asked Dec 27 '18 at 8:45
gloria
42
asked Dec 27 '18 at 8:45
gloria
42
asked Dec 27 '18 at 8:45
gloria
42
asked Dec 27 '18 at 8:45
gloria
42
42
closed as unclear what you're asking by Saad, saz, Eric Wofsey, KReiser, Paul Frost Dec 28 '18 at 12:19
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Saad, saz, Eric Wofsey, KReiser, Paul Frost Dec 28 '18 at 12:19
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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