Why $mathbb Eleft[supfrac{|Y_t-Y_s|}{|t-s|^alpha }right]<infty$ imply $(Y_t)_t$ continuous?











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Let $(Y_t)_t$ a stochastic process s.t. $$mathbb Eleft[sup_{s,tin [0,1], sneq t}frac{|Y_t-Y_s|}{|t-s|^alpha }right]<infty,$$ with $alpha >0$. Why does this implies that $(Y_t)_t$ is continuous a.s. ?



Does it come from the fact that if $mathbb E[X]<infty$ then $mathbb P{X<infty}=1$, and thus $$mathbb Pleft{sup_{s,tin [0,1], sneq t}frac{|Y_t-Y_s|}{|t-s|^alpha }<inftyright}=1.$$



Also
$$mathbb Pleft{sup_{s,tin [0,1], sneq t}frac{|Y_t-Y_s|}{|t-s|^alpha }<inftyright}leq mathbb Pleft{frac{|Y_t-Y_s|}{|t-s|^alpha}<inftyright}=1.$$



1) How can I continue ? Does it implies that there is $C>0$ s.t.
$$mathbb P{|Y_t-Y_s|<C|t-s|^alpha }=1,$$
or that $$mathbb P{exists C>0: |Y_t-Y_s|leq C|t-s|^alpha }=1 ?$$



2) And will it implies that $$mathbb P{lim_{tto s}|Y_t-Y_s|}=1 ?$$
If yes, why ? I don't understand why I can put the limit inside.










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  • Assuming that $$C = sup_{substack{s, t in [0, 1] \ s neq t}} frac{|Y_t - Y_s|}{|t-s|^{alpha}}$$ defines a random variable (which is far from being obvious, and in fact, reveals one of the technical subtleties that lie in the theory of stochastic processes), the hypothesis tells that $C$ is finite $mathbb{P}$-a.s. and $$ mathbb{P}left[ |Y_t - Y_s| leq C|t - s|^{alpha} : s, t in [0, 1] right] = 1. $$thus the sample path $t mapsto Y_t$ is $mathbb{P}$-.a.s. $alpha$-Holder continuous.
    – Sangchul Lee
    yesterday












  • @SangchulLee: 1) Why $C$ is not obviously a random variable ? Is there a reference I can read ? 2) So I can't prove that $mathbb P{lim_{tto s}|Y_t-Y_s|=0}=1$ ?
    – lovemath
    yesterday










  • Uncountable operations are always problematic for measurability, and as such, there is no guarantee that $C$ is measurable. Likewise, events related to regularity of sample paths, such as $$left{ lim_{tto s} |Y_t - Y_s| = 1 text{ for all } s right}=bigcap_{s}left{ lim_{tto s} |Y_t - Y_s| = 1right},$$requires uncountable operations and hence are not necessarily measurable. The usual workaround to this issue is to consider a suitable modification, for instance, by appealing to Kolmogorov-Chentsov theorem.
    – Sangchul Lee
    yesterday












  • @SangchulLee: But don't we have that $${lim_{tto s}|Y_t-Y_s|=0}=bigcap_{ninmathbb N^*}bigcup_{pinmathbb Q, p>0}bigcap_{|s-t|<p, s,tin mathbb Q}left{|Y_t-Y_s|leq frac{1}{n}right} ?$$ So the event is measurable. An other thing, an other member (but his post has been deleted) says : there is $N$ of measure $0$ s.t. for all $omega notin N$, $|Y_t(omega )-Y_s(omega )|leq C|t-s|^alpha $ and thus $lim_{tto s}|Y_t(omega )-Y_s(omega )|=0$ for all $omega notin N$. What we write "with abuse notation" by $mathbb P{lim_{tto s}|Y_t-Y_s|=0}=1$. Do you think it works ?
    – lovemath
    20 hours ago










  • The right-hand side only captures the continuity at rational points, so this cannot equal the left-hand side without further assumptions.
    – Sangchul Lee
    17 hours ago















up vote
2
down vote

favorite












Let $(Y_t)_t$ a stochastic process s.t. $$mathbb Eleft[sup_{s,tin [0,1], sneq t}frac{|Y_t-Y_s|}{|t-s|^alpha }right]<infty,$$ with $alpha >0$. Why does this implies that $(Y_t)_t$ is continuous a.s. ?



Does it come from the fact that if $mathbb E[X]<infty$ then $mathbb P{X<infty}=1$, and thus $$mathbb Pleft{sup_{s,tin [0,1], sneq t}frac{|Y_t-Y_s|}{|t-s|^alpha }<inftyright}=1.$$



Also
$$mathbb Pleft{sup_{s,tin [0,1], sneq t}frac{|Y_t-Y_s|}{|t-s|^alpha }<inftyright}leq mathbb Pleft{frac{|Y_t-Y_s|}{|t-s|^alpha}<inftyright}=1.$$



1) How can I continue ? Does it implies that there is $C>0$ s.t.
$$mathbb P{|Y_t-Y_s|<C|t-s|^alpha }=1,$$
or that $$mathbb P{exists C>0: |Y_t-Y_s|leq C|t-s|^alpha }=1 ?$$



2) And will it implies that $$mathbb P{lim_{tto s}|Y_t-Y_s|}=1 ?$$
If yes, why ? I don't understand why I can put the limit inside.










share|cite|improve this question







New contributor




lovemath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • Assuming that $$C = sup_{substack{s, t in [0, 1] \ s neq t}} frac{|Y_t - Y_s|}{|t-s|^{alpha}}$$ defines a random variable (which is far from being obvious, and in fact, reveals one of the technical subtleties that lie in the theory of stochastic processes), the hypothesis tells that $C$ is finite $mathbb{P}$-a.s. and $$ mathbb{P}left[ |Y_t - Y_s| leq C|t - s|^{alpha} : s, t in [0, 1] right] = 1. $$thus the sample path $t mapsto Y_t$ is $mathbb{P}$-.a.s. $alpha$-Holder continuous.
    – Sangchul Lee
    yesterday












  • @SangchulLee: 1) Why $C$ is not obviously a random variable ? Is there a reference I can read ? 2) So I can't prove that $mathbb P{lim_{tto s}|Y_t-Y_s|=0}=1$ ?
    – lovemath
    yesterday










  • Uncountable operations are always problematic for measurability, and as such, there is no guarantee that $C$ is measurable. Likewise, events related to regularity of sample paths, such as $$left{ lim_{tto s} |Y_t - Y_s| = 1 text{ for all } s right}=bigcap_{s}left{ lim_{tto s} |Y_t - Y_s| = 1right},$$requires uncountable operations and hence are not necessarily measurable. The usual workaround to this issue is to consider a suitable modification, for instance, by appealing to Kolmogorov-Chentsov theorem.
    – Sangchul Lee
    yesterday












  • @SangchulLee: But don't we have that $${lim_{tto s}|Y_t-Y_s|=0}=bigcap_{ninmathbb N^*}bigcup_{pinmathbb Q, p>0}bigcap_{|s-t|<p, s,tin mathbb Q}left{|Y_t-Y_s|leq frac{1}{n}right} ?$$ So the event is measurable. An other thing, an other member (but his post has been deleted) says : there is $N$ of measure $0$ s.t. for all $omega notin N$, $|Y_t(omega )-Y_s(omega )|leq C|t-s|^alpha $ and thus $lim_{tto s}|Y_t(omega )-Y_s(omega )|=0$ for all $omega notin N$. What we write "with abuse notation" by $mathbb P{lim_{tto s}|Y_t-Y_s|=0}=1$. Do you think it works ?
    – lovemath
    20 hours ago










  • The right-hand side only captures the continuity at rational points, so this cannot equal the left-hand side without further assumptions.
    – Sangchul Lee
    17 hours ago













up vote
2
down vote

favorite









up vote
2
down vote

favorite











Let $(Y_t)_t$ a stochastic process s.t. $$mathbb Eleft[sup_{s,tin [0,1], sneq t}frac{|Y_t-Y_s|}{|t-s|^alpha }right]<infty,$$ with $alpha >0$. Why does this implies that $(Y_t)_t$ is continuous a.s. ?



Does it come from the fact that if $mathbb E[X]<infty$ then $mathbb P{X<infty}=1$, and thus $$mathbb Pleft{sup_{s,tin [0,1], sneq t}frac{|Y_t-Y_s|}{|t-s|^alpha }<inftyright}=1.$$



Also
$$mathbb Pleft{sup_{s,tin [0,1], sneq t}frac{|Y_t-Y_s|}{|t-s|^alpha }<inftyright}leq mathbb Pleft{frac{|Y_t-Y_s|}{|t-s|^alpha}<inftyright}=1.$$



1) How can I continue ? Does it implies that there is $C>0$ s.t.
$$mathbb P{|Y_t-Y_s|<C|t-s|^alpha }=1,$$
or that $$mathbb P{exists C>0: |Y_t-Y_s|leq C|t-s|^alpha }=1 ?$$



2) And will it implies that $$mathbb P{lim_{tto s}|Y_t-Y_s|}=1 ?$$
If yes, why ? I don't understand why I can put the limit inside.










share|cite|improve this question







New contributor




lovemath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











Let $(Y_t)_t$ a stochastic process s.t. $$mathbb Eleft[sup_{s,tin [0,1], sneq t}frac{|Y_t-Y_s|}{|t-s|^alpha }right]<infty,$$ with $alpha >0$. Why does this implies that $(Y_t)_t$ is continuous a.s. ?



Does it come from the fact that if $mathbb E[X]<infty$ then $mathbb P{X<infty}=1$, and thus $$mathbb Pleft{sup_{s,tin [0,1], sneq t}frac{|Y_t-Y_s|}{|t-s|^alpha }<inftyright}=1.$$



Also
$$mathbb Pleft{sup_{s,tin [0,1], sneq t}frac{|Y_t-Y_s|}{|t-s|^alpha }<inftyright}leq mathbb Pleft{frac{|Y_t-Y_s|}{|t-s|^alpha}<inftyright}=1.$$



1) How can I continue ? Does it implies that there is $C>0$ s.t.
$$mathbb P{|Y_t-Y_s|<C|t-s|^alpha }=1,$$
or that $$mathbb P{exists C>0: |Y_t-Y_s|leq C|t-s|^alpha }=1 ?$$



2) And will it implies that $$mathbb P{lim_{tto s}|Y_t-Y_s|}=1 ?$$
If yes, why ? I don't understand why I can put the limit inside.







probability stochastic-processes






share|cite|improve this question







New contributor




lovemath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question







New contributor




lovemath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question






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Check out our Code of Conduct.









asked yesterday









lovemath

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Check out our Code of Conduct.





New contributor





lovemath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






lovemath is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • Assuming that $$C = sup_{substack{s, t in [0, 1] \ s neq t}} frac{|Y_t - Y_s|}{|t-s|^{alpha}}$$ defines a random variable (which is far from being obvious, and in fact, reveals one of the technical subtleties that lie in the theory of stochastic processes), the hypothesis tells that $C$ is finite $mathbb{P}$-a.s. and $$ mathbb{P}left[ |Y_t - Y_s| leq C|t - s|^{alpha} : s, t in [0, 1] right] = 1. $$thus the sample path $t mapsto Y_t$ is $mathbb{P}$-.a.s. $alpha$-Holder continuous.
    – Sangchul Lee
    yesterday












  • @SangchulLee: 1) Why $C$ is not obviously a random variable ? Is there a reference I can read ? 2) So I can't prove that $mathbb P{lim_{tto s}|Y_t-Y_s|=0}=1$ ?
    – lovemath
    yesterday










  • Uncountable operations are always problematic for measurability, and as such, there is no guarantee that $C$ is measurable. Likewise, events related to regularity of sample paths, such as $$left{ lim_{tto s} |Y_t - Y_s| = 1 text{ for all } s right}=bigcap_{s}left{ lim_{tto s} |Y_t - Y_s| = 1right},$$requires uncountable operations and hence are not necessarily measurable. The usual workaround to this issue is to consider a suitable modification, for instance, by appealing to Kolmogorov-Chentsov theorem.
    – Sangchul Lee
    yesterday












  • @SangchulLee: But don't we have that $${lim_{tto s}|Y_t-Y_s|=0}=bigcap_{ninmathbb N^*}bigcup_{pinmathbb Q, p>0}bigcap_{|s-t|<p, s,tin mathbb Q}left{|Y_t-Y_s|leq frac{1}{n}right} ?$$ So the event is measurable. An other thing, an other member (but his post has been deleted) says : there is $N$ of measure $0$ s.t. for all $omega notin N$, $|Y_t(omega )-Y_s(omega )|leq C|t-s|^alpha $ and thus $lim_{tto s}|Y_t(omega )-Y_s(omega )|=0$ for all $omega notin N$. What we write "with abuse notation" by $mathbb P{lim_{tto s}|Y_t-Y_s|=0}=1$. Do you think it works ?
    – lovemath
    20 hours ago










  • The right-hand side only captures the continuity at rational points, so this cannot equal the left-hand side without further assumptions.
    – Sangchul Lee
    17 hours ago


















  • Assuming that $$C = sup_{substack{s, t in [0, 1] \ s neq t}} frac{|Y_t - Y_s|}{|t-s|^{alpha}}$$ defines a random variable (which is far from being obvious, and in fact, reveals one of the technical subtleties that lie in the theory of stochastic processes), the hypothesis tells that $C$ is finite $mathbb{P}$-a.s. and $$ mathbb{P}left[ |Y_t - Y_s| leq C|t - s|^{alpha} : s, t in [0, 1] right] = 1. $$thus the sample path $t mapsto Y_t$ is $mathbb{P}$-.a.s. $alpha$-Holder continuous.
    – Sangchul Lee
    yesterday












  • @SangchulLee: 1) Why $C$ is not obviously a random variable ? Is there a reference I can read ? 2) So I can't prove that $mathbb P{lim_{tto s}|Y_t-Y_s|=0}=1$ ?
    – lovemath
    yesterday










  • Uncountable operations are always problematic for measurability, and as such, there is no guarantee that $C$ is measurable. Likewise, events related to regularity of sample paths, such as $$left{ lim_{tto s} |Y_t - Y_s| = 1 text{ for all } s right}=bigcap_{s}left{ lim_{tto s} |Y_t - Y_s| = 1right},$$requires uncountable operations and hence are not necessarily measurable. The usual workaround to this issue is to consider a suitable modification, for instance, by appealing to Kolmogorov-Chentsov theorem.
    – Sangchul Lee
    yesterday












  • @SangchulLee: But don't we have that $${lim_{tto s}|Y_t-Y_s|=0}=bigcap_{ninmathbb N^*}bigcup_{pinmathbb Q, p>0}bigcap_{|s-t|<p, s,tin mathbb Q}left{|Y_t-Y_s|leq frac{1}{n}right} ?$$ So the event is measurable. An other thing, an other member (but his post has been deleted) says : there is $N$ of measure $0$ s.t. for all $omega notin N$, $|Y_t(omega )-Y_s(omega )|leq C|t-s|^alpha $ and thus $lim_{tto s}|Y_t(omega )-Y_s(omega )|=0$ for all $omega notin N$. What we write "with abuse notation" by $mathbb P{lim_{tto s}|Y_t-Y_s|=0}=1$. Do you think it works ?
    – lovemath
    20 hours ago










  • The right-hand side only captures the continuity at rational points, so this cannot equal the left-hand side without further assumptions.
    – Sangchul Lee
    17 hours ago
















Assuming that $$C = sup_{substack{s, t in [0, 1] \ s neq t}} frac{|Y_t - Y_s|}{|t-s|^{alpha}}$$ defines a random variable (which is far from being obvious, and in fact, reveals one of the technical subtleties that lie in the theory of stochastic processes), the hypothesis tells that $C$ is finite $mathbb{P}$-a.s. and $$ mathbb{P}left[ |Y_t - Y_s| leq C|t - s|^{alpha} : s, t in [0, 1] right] = 1. $$thus the sample path $t mapsto Y_t$ is $mathbb{P}$-.a.s. $alpha$-Holder continuous.
– Sangchul Lee
yesterday






Assuming that $$C = sup_{substack{s, t in [0, 1] \ s neq t}} frac{|Y_t - Y_s|}{|t-s|^{alpha}}$$ defines a random variable (which is far from being obvious, and in fact, reveals one of the technical subtleties that lie in the theory of stochastic processes), the hypothesis tells that $C$ is finite $mathbb{P}$-a.s. and $$ mathbb{P}left[ |Y_t - Y_s| leq C|t - s|^{alpha} : s, t in [0, 1] right] = 1. $$thus the sample path $t mapsto Y_t$ is $mathbb{P}$-.a.s. $alpha$-Holder continuous.
– Sangchul Lee
yesterday














@SangchulLee: 1) Why $C$ is not obviously a random variable ? Is there a reference I can read ? 2) So I can't prove that $mathbb P{lim_{tto s}|Y_t-Y_s|=0}=1$ ?
– lovemath
yesterday




@SangchulLee: 1) Why $C$ is not obviously a random variable ? Is there a reference I can read ? 2) So I can't prove that $mathbb P{lim_{tto s}|Y_t-Y_s|=0}=1$ ?
– lovemath
yesterday












Uncountable operations are always problematic for measurability, and as such, there is no guarantee that $C$ is measurable. Likewise, events related to regularity of sample paths, such as $$left{ lim_{tto s} |Y_t - Y_s| = 1 text{ for all } s right}=bigcap_{s}left{ lim_{tto s} |Y_t - Y_s| = 1right},$$requires uncountable operations and hence are not necessarily measurable. The usual workaround to this issue is to consider a suitable modification, for instance, by appealing to Kolmogorov-Chentsov theorem.
– Sangchul Lee
yesterday






Uncountable operations are always problematic for measurability, and as such, there is no guarantee that $C$ is measurable. Likewise, events related to regularity of sample paths, such as $$left{ lim_{tto s} |Y_t - Y_s| = 1 text{ for all } s right}=bigcap_{s}left{ lim_{tto s} |Y_t - Y_s| = 1right},$$requires uncountable operations and hence are not necessarily measurable. The usual workaround to this issue is to consider a suitable modification, for instance, by appealing to Kolmogorov-Chentsov theorem.
– Sangchul Lee
yesterday














@SangchulLee: But don't we have that $${lim_{tto s}|Y_t-Y_s|=0}=bigcap_{ninmathbb N^*}bigcup_{pinmathbb Q, p>0}bigcap_{|s-t|<p, s,tin mathbb Q}left{|Y_t-Y_s|leq frac{1}{n}right} ?$$ So the event is measurable. An other thing, an other member (but his post has been deleted) says : there is $N$ of measure $0$ s.t. for all $omega notin N$, $|Y_t(omega )-Y_s(omega )|leq C|t-s|^alpha $ and thus $lim_{tto s}|Y_t(omega )-Y_s(omega )|=0$ for all $omega notin N$. What we write "with abuse notation" by $mathbb P{lim_{tto s}|Y_t-Y_s|=0}=1$. Do you think it works ?
– lovemath
20 hours ago




@SangchulLee: But don't we have that $${lim_{tto s}|Y_t-Y_s|=0}=bigcap_{ninmathbb N^*}bigcup_{pinmathbb Q, p>0}bigcap_{|s-t|<p, s,tin mathbb Q}left{|Y_t-Y_s|leq frac{1}{n}right} ?$$ So the event is measurable. An other thing, an other member (but his post has been deleted) says : there is $N$ of measure $0$ s.t. for all $omega notin N$, $|Y_t(omega )-Y_s(omega )|leq C|t-s|^alpha $ and thus $lim_{tto s}|Y_t(omega )-Y_s(omega )|=0$ for all $omega notin N$. What we write "with abuse notation" by $mathbb P{lim_{tto s}|Y_t-Y_s|=0}=1$. Do you think it works ?
– lovemath
20 hours ago












The right-hand side only captures the continuity at rational points, so this cannot equal the left-hand side without further assumptions.
– Sangchul Lee
17 hours ago




The right-hand side only captures the continuity at rational points, so this cannot equal the left-hand side without further assumptions.
– Sangchul Lee
17 hours ago















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