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

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.

asked yesterday

lovemath

163

New contributor

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

|
show 4 more comments

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.

asked yesterday

lovemath

163

New contributor

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

|
show 4 more comments

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.

asked yesterday

lovemath

163

New contributor

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

asked yesterday

lovemath

163

New contributor

asked yesterday

lovemath

163

New contributor

asked yesterday

lovemath

163

New contributor

asked yesterday

lovemath

163

asked yesterday

lovemath

163

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

|
show 4 more comments

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

|
show 4 more comments

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