On the complexity of stochastic integration (Q2701558): Difference between revisions

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Fix \(T,L,K>0\), and consider functions \(f\) of \([0,T]\times \mathbb{R}\) into \(\mathbb{R}\) such that \(\partial_1f\) and \(\partial^2_2f\) are continuous, and bounded by \(L,K\), respectively. Set \(I(f):= \int^T_0 f(t,B_t)dB_t\), \(B_t\) being a Brownian motion, and for \(n\in\mathbb{N}^*\): NEWLINE\[NEWLINE\begin{multlined} A_n(f):= \sum^n_{j=1} f\bigl(((j-1)/n) T,B_{((j-1)/n)T} \bigr)\times \bigl(B_{(j/n)T}- B_{((j-1)/n)T}\bigr) \\ +{1\over 2}\sum^n_{j=1} \partial_2f\bigl(((j-1)/n)T,B_{((j-1)/n)T}\bigr) \times \Bigl[(B_{(j/n)T}-B_{((j-1)/n)T})^2 -{T\over n}\Bigr]. \end{multlined}NEWLINE\]NEWLINE Then NEWLINE\[NEWLINE\bigl \|I(f)-A_n(f) \bigr\|_2\leq {T^{3/2}\over n}\times \sqrt{{2\over 3} L^2+K^2},NEWLINE\]NEWLINE and this estimate is sharp, up to some multiplicative constant.
Property / review text: Fix \(T,L,K>0\), and consider functions \(f\) of \([0,T]\times \mathbb{R}\) into \(\mathbb{R}\) such that \(\partial_1f\) and \(\partial^2_2f\) are continuous, and bounded by \(L,K\), respectively. Set \(I(f):= \int^T_0 f(t,B_t)dB_t\), \(B_t\) being a Brownian motion, and for \(n\in\mathbb{N}^*\): NEWLINE\[NEWLINE\begin{multlined} A_n(f):= \sum^n_{j=1} f\bigl(((j-1)/n) T,B_{((j-1)/n)T} \bigr)\times \bigl(B_{(j/n)T}- B_{((j-1)/n)T}\bigr) \\ +{1\over 2}\sum^n_{j=1} \partial_2f\bigl(((j-1)/n)T,B_{((j-1)/n)T}\bigr) \times \Bigl[(B_{(j/n)T}-B_{((j-1)/n)T})^2 -{T\over n}\Bigr]. \end{multlined}NEWLINE\]NEWLINE Then NEWLINE\[NEWLINE\bigl \|I(f)-A_n(f) \bigr\|_2\leq {T^{3/2}\over n}\times \sqrt{{2\over 3} L^2+K^2},NEWLINE\]NEWLINE and this estimate is sharp, up to some multiplicative constant. / rank
 
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Property / reviewed by: Jacques Franchi / rank
 
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Latest revision as of 14:36, 10 April 2025

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On the complexity of stochastic integration
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    On the complexity of stochastic integration (English)
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    19 February 2001
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    approximation of Itô integral
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    complexity
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    optimal algorithm
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    Fix \(T,L,K>0\), and consider functions \(f\) of \([0,T]\times \mathbb{R}\) into \(\mathbb{R}\) such that \(\partial_1f\) and \(\partial^2_2f\) are continuous, and bounded by \(L,K\), respectively. Set \(I(f):= \int^T_0 f(t,B_t)dB_t\), \(B_t\) being a Brownian motion, and for \(n\in\mathbb{N}^*\): NEWLINE\[NEWLINE\begin{multlined} A_n(f):= \sum^n_{j=1} f\bigl(((j-1)/n) T,B_{((j-1)/n)T} \bigr)\times \bigl(B_{(j/n)T}- B_{((j-1)/n)T}\bigr) \\ +{1\over 2}\sum^n_{j=1} \partial_2f\bigl(((j-1)/n)T,B_{((j-1)/n)T}\bigr) \times \Bigl[(B_{(j/n)T}-B_{((j-1)/n)T})^2 -{T\over n}\Bigr]. \end{multlined}NEWLINE\]NEWLINE Then NEWLINE\[NEWLINE\bigl \|I(f)-A_n(f) \bigr\|_2\leq {T^{3/2}\over n}\times \sqrt{{2\over 3} L^2+K^2},NEWLINE\]NEWLINE and this estimate is sharp, up to some multiplicative constant.
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