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Prove that $lim_{Lrightarrowinfty} Pleft(sup_{0leq sleq t}|B(s)|>Lright)=0$, for each $tgeq0$, where $B$ standard Brownian motion.

Proof that the stopping time for a Brownian Motion is finite for given target levelsProve that the first hitting time $tau_x:=infleft{tge 0:B_t=xright}$ of a Brownian motion is almost surely finiteconvergence in distribution of exponential of a brownian motionDistributional equalityShow that $Pleft(limsuplimits_{ttoinfty}left|B_tright|/t>cright)leq Pleft(suplimits_{tin[n,n+1]}left|B_tright|/n>c;mbox{i.o.}right)$Computing $Eleft(expleft(B_t+int_0^tB_sdB_s - frac12int_0^tB^2_sdsright)right)$ if $B$ is a standard Brownian MotionWhy is $E[ sup_{ 0leq u leq t } |X_u| ]<infty$ where $X$ is a right continuous submartingale?How to show that $P( sup_{0 leq s leq 1} |B_s| leq epsilon) > 0$ for any $epsilon > 0$?Prove the following is standard Brownian from this expression of non-standard Brownian motionProve that lim sup $|B(n)| = infty$ almost surely, where $B(n)$ is standard Brownian motion.

1

$begingroup$

Let $B(t)$, $tgeq0$, be a standard Brownian motion. I would like to prove that
$$lim_{Lrightarrowinfty} Pleft(sup_{0leq sleq t}|B(s)|>Lright)=0,$$
for each $tgeq0$.

In my class notes, I have the following proof: since $sup_{0leq sleq t}|B(s)|$ has the same probability distribution as $|B(t)|$, then
$$Pleft(sup_{0leq sleq t}|B(s)|>Lright)=P(|B(t)|>L), $$
and by Chebyshev's inequality, that probability is bounded by $L^{-1} E(|B(t)|)$, which tends to $0$ as $Lrightarrowinfty$.

I do not understand why $sup_{0leq sleq t}|B(s)|$ has the same probability distribution as $|B(t)|$. I know that $sup_{0leq sleq t}B(s)$ (with no absolute value) possesses the same probability distribution as $|B(t)|$, but this is not the same statement. Could you please elaborate on this? Thank you!

probability-theory stochastic-processes stochastic-calculus brownian-motion stochastic-analysis

share|cite|improve this question

asked Mar 18 at 12:36

user39756user39756

251313

$endgroup$

add a comment | 

1

$begingroup$

Let $B(t)$, $tgeq0$, be a standard Brownian motion. I would like to prove that
$$lim_{Lrightarrowinfty} Pleft(sup_{0leq sleq t}|B(s)|>Lright)=0,$$
for each $tgeq0$.

In my class notes, I have the following proof: since $sup_{0leq sleq t}|B(s)|$ has the same probability distribution as $|B(t)|$, then
$$Pleft(sup_{0leq sleq t}|B(s)|>Lright)=P(|B(t)|>L), $$
and by Chebyshev's inequality, that probability is bounded by $L^{-1} E(|B(t)|)$, which tends to $0$ as $Lrightarrowinfty$.

I do not understand why $sup_{0leq sleq t}|B(s)|$ has the same probability distribution as $|B(t)|$. I know that $sup_{0leq sleq t}B(s)$ (with no absolute value) possesses the same probability distribution as $|B(t)|$, but this is not the same statement. Could you please elaborate on this? Thank you!

probability-theory stochastic-processes stochastic-calculus brownian-motion stochastic-analysis

share|cite|improve this question

asked Mar 18 at 12:36

user39756user39756

251313

$endgroup$

add a comment | 

$begingroup$

Let $B(t)$, $tgeq0$, be a standard Brownian motion. I would like to prove that
$$lim_{Lrightarrowinfty} Pleft(sup_{0leq sleq t}|B(s)|>Lright)=0,$$
for each $tgeq0$.

In my class notes, I have the following proof: since $sup_{0leq sleq t}|B(s)|$ has the same probability distribution as $|B(t)|$, then
$$Pleft(sup_{0leq sleq t}|B(s)|>Lright)=P(|B(t)|>L), $$
and by Chebyshev's inequality, that probability is bounded by $L^{-1} E(|B(t)|)$, which tends to $0$ as $Lrightarrowinfty$.

I do not understand why $sup_{0leq sleq t}|B(s)|$ has the same probability distribution as $|B(t)|$. I know that $sup_{0leq sleq t}B(s)$ (with no absolute value) possesses the same probability distribution as $|B(t)|$, but this is not the same statement. Could you please elaborate on this? Thank you!

probability-theory stochastic-processes stochastic-calculus brownian-motion stochastic-analysis

share|cite|improve this question

asked Mar 18 at 12:36

user39756user39756

251313

$endgroup$

share|cite|improve this question

asked Mar 18 at 12:36

user39756user39756

251313

share|cite|improve this question

asked Mar 18 at 12:36

user39756user39756

251313

asked Mar 18 at 12:36

user39756user39756

251313

asked Mar 18 at 12:36

user39756user39756

251313

1 Answer
1

active

oldest

votes

2

$begingroup$

You are correct in your doubts. $sup_{s<t}|B_s|$ does not have the same distribution as $|B_t|$, so your notes have a mistake.

One way around this; you can write
$$
sup_{0<s<t}|B_s|=maxleft(sup_{0<s<t}B_s,sup_{0<s<t}-B_tright)
$$
and then the event the LHS is more than $L$ is the union of the events that both arguments of the RHS are more than $L$. Therefore, $P(sup_{0<s<t}|B_s|>L)le 2P(|B_t|>L)$.

However, there is an even easier proof. For any (finite) random variable $X$, we have $lim_{Ltoinfty}P(X>L)=0$. This follows from continuity of probability, since the events ${X>L}$ decrease to the empty set as $Ltoinfty$.

share|cite|improve this answer

edited Mar 19 at 15:28

answered Mar 18 at 15:38

Mike EarnestMike Earnest

26.7k22151

$endgroup$

add a comment | 

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2

$begingroup$

You are correct in your doubts. $sup_{s<t}|B_s|$ does not have the same distribution as $|B_t|$, so your notes have a mistake.

One way around this; you can write
$$
sup_{0<s<t}|B_s|=maxleft(sup_{0<s<t}B_s,sup_{0<s<t}-B_tright)
$$
and then the event the LHS is more than $L$ is the union of the events that both arguments of the RHS are more than $L$. Therefore, $P(sup_{0<s<t}|B_s|>L)le 2P(|B_t|>L)$.

However, there is an even easier proof. For any (finite) random variable $X$, we have $lim_{Ltoinfty}P(X>L)=0$. This follows from continuity of probability, since the events ${X>L}$ decrease to the empty set as $Ltoinfty$.

share|cite|improve this answer

edited Mar 19 at 15:28

answered Mar 18 at 15:38

Mike EarnestMike Earnest

26.7k22151

$endgroup$

add a comment | 

2

$begingroup$

You are correct in your doubts. $sup_{s<t}|B_s|$ does not have the same distribution as $|B_t|$, so your notes have a mistake.

One way around this; you can write
$$
sup_{0<s<t}|B_s|=maxleft(sup_{0<s<t}B_s,sup_{0<s<t}-B_tright)
$$
and then the event the LHS is more than $L$ is the union of the events that both arguments of the RHS are more than $L$. Therefore, $P(sup_{0<s<t}|B_s|>L)le 2P(|B_t|>L)$.

However, there is an even easier proof. For any (finite) random variable $X$, we have $lim_{Ltoinfty}P(X>L)=0$. This follows from continuity of probability, since the events ${X>L}$ decrease to the empty set as $Ltoinfty$.

share|cite|improve this answer

edited Mar 19 at 15:28

answered Mar 18 at 15:38

Mike EarnestMike Earnest

26.7k22151

$endgroup$

add a comment | 

$begingroup$

You are correct in your doubts. $sup_{s<t}|B_s|$ does not have the same distribution as $|B_t|$, so your notes have a mistake.

One way around this; you can write
$$
sup_{0<s<t}|B_s|=maxleft(sup_{0<s<t}B_s,sup_{0<s<t}-B_tright)
$$
and then the event the LHS is more than $L$ is the union of the events that both arguments of the RHS are more than $L$. Therefore, $P(sup_{0<s<t}|B_s|>L)le 2P(|B_t|>L)$.

However, there is an even easier proof. For any (finite) random variable $X$, we have $lim_{Ltoinfty}P(X>L)=0$. This follows from continuity of probability, since the events ${X>L}$ decrease to the empty set as $Ltoinfty$.

share|cite|improve this answer

edited Mar 19 at 15:28

answered Mar 18 at 15:38

Mike EarnestMike Earnest

26.7k22151

$endgroup$

share|cite|improve this answer

edited Mar 19 at 15:28

answered Mar 18 at 15:38

Mike EarnestMike Earnest

26.7k22151

answered Mar 18 at 15:38

Mike EarnestMike Earnest

26.7k22151

answered Mar 18 at 15:38

Mike EarnestMike Earnest

26.7k22151