Local time and $int_0^t operatornamesgn(B_t-x),dB_t$

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up vote
3
down vote

favorite












I would like to compute the expectation of the stopped local time $L_t^x$ with underlying Brownian Motion $B_t$ with $B_0=x$. Set



$T_a=inf s geq 0 mid B_s=a$
with $y<x<a$.



The target of interest is
$mathbbE_x[L_T_a^y]$. I would like apply optional stopping theorem. We have by Tanaka's Formula
$$|B_t-y|-|B_0-y|=int_0^t operatornamesgn(B_s-y),dB_s+L_t^y.$$



But for this I have to prove $int_0^T_a operatornamesgn(B_t-y),dB_t$ is bounded.



Nevertheless assuming this. We obtain
$|a-y|-|x-y|=mathbbE_x[L_T_a^y]$.



We have:
$|a-y|-|x-y|=a-y-|x-y|=a-x$.



So I'm not very sure about this computation. Thanks for every comment



There is some other Question which arises. If we have the case $x<y<a$ we would obtain



$mathbbE_x[L_T_a^y]=a+x-2y$. For example for $x=0$ and $y=a3/4$ we get



$mathbbE_x[L_T_a^y]=-a/2$



so there is a contradiction. But where is the mistake?



It seems the assumption fails.
So how one can handle
$mathbbE_x[L_y^T_a]$?.



It is quite strange here I used the same
expectation of stopped local time
and the result was true.



I know the Formula expectation of stopped local time
holds true nevertheless my mistake (see site 171, Diffusion process and their sample path).
Now let $a <x<y<b$ then we have



$mathbbE_x[L_y^T_a wedge T_b]=2 dfrac(x-a)(b-y)(b-a) leq mathbbE_x[L_y^T_a ] $



because the local time is increasing.
Now, we have
$mathbbE_x[L_y^T_a wedge T_b]
=2 dfrac(x-a)(b-y)(b-a)= 2 dfracb(x-a)(1-y/b)b(1-a/b)= 2 dfrac(x-a)(1-y/b)(1-a/b) rightarrow 2 (x-a)$
for $b rightarrow infty$.
Now, the continuity of the local time, dominated convergence one obtains
$mathbbE_x[L_y^T_a]=2(x-a)$.



We also have
$mathbbE_x[L_y^T_a wedge T_b]=2 dfraca(x/a-1)(b-y)a(b/a-1) =2 dfrac(x/a-1)(b-y)(b/a-1) rightarrow 2(b-y)$ for $a rightarrow -infty$.
This leads to
$mathbbE_x[L_y^ T_b]=2(b-y)$
By the way.
I hope this computation is on the right way.
I need this to compute the walk Dimension for gap diffusions.




probability stochastic-calculus brownian-motion martingales




share|cite|improve this question





















  • Instead of editing the question, it is better to post this as an answer to your own question (which is very normal). Also you should let $bto-infty$, not $a$; this would give $mathbb E_x [L_y^T_a] = 2(a-x)$, which is not unexpected at all.
    – zhoraster
    Jul 24 at 12:28










  • In my answer I change the order of $a$ and $x$ my bad.
    – Dr. NIemann
    Jul 25 at 0:25














up vote
3
down vote

favorite












I would like to compute the expectation of the stopped local time $L_t^x$ with underlying Brownian Motion $B_t$ with $B_0=x$. Set



$T_a=inf s geq 0 mid B_s=a$
with $y<x<a$.



The target of interest is
$mathbbE_x[L_T_a^y]$. I would like apply optional stopping theorem. We have by Tanaka's Formula
$$|B_t-y|-|B_0-y|=int_0^t operatornamesgn(B_s-y),dB_s+L_t^y.$$



But for this I have to prove $int_0^T_a operatornamesgn(B_t-y),dB_t$ is bounded.



Nevertheless assuming this. We obtain
$|a-y|-|x-y|=mathbbE_x[L_T_a^y]$.



We have:
$|a-y|-|x-y|=a-y-|x-y|=a-x$.



So I'm not very sure about this computation. Thanks for every comment



There is some other Question which arises. If we have the case $x<y<a$ we would obtain



$mathbbE_x[L_T_a^y]=a+x-2y$. For example for $x=0$ and $y=a3/4$ we get



$mathbbE_x[L_T_a^y]=-a/2$



so there is a contradiction. But where is the mistake?



It seems the assumption fails.
So how one can handle
$mathbbE_x[L_y^T_a]$?.



It is quite strange here I used the same
expectation of stopped local time
and the result was true.



I know the Formula expectation of stopped local time
holds true nevertheless my mistake (see site 171, Diffusion process and their sample path).
Now let $a <x<y<b$ then we have



$mathbbE_x[L_y^T_a wedge T_b]=2 dfrac(x-a)(b-y)(b-a) leq mathbbE_x[L_y^T_a ] $



because the local time is increasing.
Now, we have
$mathbbE_x[L_y^T_a wedge T_b]
=2 dfrac(x-a)(b-y)(b-a)= 2 dfracb(x-a)(1-y/b)b(1-a/b)= 2 dfrac(x-a)(1-y/b)(1-a/b) rightarrow 2 (x-a)$
for $b rightarrow infty$.
Now, the continuity of the local time, dominated convergence one obtains
$mathbbE_x[L_y^T_a]=2(x-a)$.



We also have
$mathbbE_x[L_y^T_a wedge T_b]=2 dfraca(x/a-1)(b-y)a(b/a-1) =2 dfrac(x/a-1)(b-y)(b/a-1) rightarrow 2(b-y)$ for $a rightarrow -infty$.
This leads to
$mathbbE_x[L_y^ T_b]=2(b-y)$
By the way.
I hope this computation is on the right way.
I need this to compute the walk Dimension for gap diffusions.




probability stochastic-calculus brownian-motion martingales




share|cite|improve this question





















  • Instead of editing the question, it is better to post this as an answer to your own question (which is very normal). Also you should let $bto-infty$, not $a$; this would give $mathbb E_x [L_y^T_a] = 2(a-x)$, which is not unexpected at all.
    – zhoraster
    Jul 24 at 12:28










  • In my answer I change the order of $a$ and $x$ my bad.
    – Dr. NIemann
    Jul 25 at 0:25












up vote
3
down vote

favorite









up vote
3
down vote

favorite











I would like to compute the expectation of the stopped local time $L_t^x$ with underlying Brownian Motion $B_t$ with $B_0=x$. Set



$T_a=inf s geq 0 mid B_s=a$
with $y<x<a$.



The target of interest is
$mathbbE_x[L_T_a^y]$. I would like apply optional stopping theorem. We have by Tanaka's Formula
$$|B_t-y|-|B_0-y|=int_0^t operatornamesgn(B_s-y),dB_s+L_t^y.$$



But for this I have to prove $int_0^T_a operatornamesgn(B_t-y),dB_t$ is bounded.



Nevertheless assuming this. We obtain
$|a-y|-|x-y|=mathbbE_x[L_T_a^y]$.



We have:
$|a-y|-|x-y|=a-y-|x-y|=a-x$.



So I'm not very sure about this computation. Thanks for every comment



There is some other Question which arises. If we have the case $x<y<a$ we would obtain



$mathbbE_x[L_T_a^y]=a+x-2y$. For example for $x=0$ and $y=a3/4$ we get



$mathbbE_x[L_T_a^y]=-a/2$



so there is a contradiction. But where is the mistake?



It seems the assumption fails.
So how one can handle
$mathbbE_x[L_y^T_a]$?.



It is quite strange here I used the same
expectation of stopped local time
and the result was true.



I know the Formula expectation of stopped local time
holds true nevertheless my mistake (see site 171, Diffusion process and their sample path).
Now let $a <x<y<b$ then we have



$mathbbE_x[L_y^T_a wedge T_b]=2 dfrac(x-a)(b-y)(b-a) leq mathbbE_x[L_y^T_a ] $



because the local time is increasing.
Now, we have
$mathbbE_x[L_y^T_a wedge T_b]
=2 dfrac(x-a)(b-y)(b-a)= 2 dfracb(x-a)(1-y/b)b(1-a/b)= 2 dfrac(x-a)(1-y/b)(1-a/b) rightarrow 2 (x-a)$
for $b rightarrow infty$.
Now, the continuity of the local time, dominated convergence one obtains
$mathbbE_x[L_y^T_a]=2(x-a)$.



We also have
$mathbbE_x[L_y^T_a wedge T_b]=2 dfraca(x/a-1)(b-y)a(b/a-1) =2 dfrac(x/a-1)(b-y)(b/a-1) rightarrow 2(b-y)$ for $a rightarrow -infty$.
This leads to
$mathbbE_x[L_y^ T_b]=2(b-y)$
By the way.
I hope this computation is on the right way.
I need this to compute the walk Dimension for gap diffusions.




probability stochastic-calculus brownian-motion martingales




share|cite|improve this question













I would like to compute the expectation of the stopped local time $L_t^x$ with underlying Brownian Motion $B_t$ with $B_0=x$. Set



$T_a=inf s geq 0 mid B_s=a$
with $y<x<a$.



The target of interest is
$mathbbE_x[L_T_a^y]$. I would like apply optional stopping theorem. We have by Tanaka's Formula
$$|B_t-y|-|B_0-y|=int_0^t operatornamesgn(B_s-y),dB_s+L_t^y.$$



But for this I have to prove $int_0^T_a operatornamesgn(B_t-y),dB_t$ is bounded.



Nevertheless assuming this. We obtain
$|a-y|-|x-y|=mathbbE_x[L_T_a^y]$.



We have:
$|a-y|-|x-y|=a-y-|x-y|=a-x$.



So I'm not very sure about this computation. Thanks for every comment



There is some other Question which arises. If we have the case $x<y<a$ we would obtain



$mathbbE_x[L_T_a^y]=a+x-2y$. For example for $x=0$ and $y=a3/4$ we get



$mathbbE_x[L_T_a^y]=-a/2$



so there is a contradiction. But where is the mistake?



It seems the assumption fails.
So how one can handle
$mathbbE_x[L_y^T_a]$?.



It is quite strange here I used the same
expectation of stopped local time
and the result was true.



I know the Formula expectation of stopped local time
holds true nevertheless my mistake (see site 171, Diffusion process and their sample path).
Now let $a <x<y<b$ then we have



$mathbbE_x[L_y^T_a wedge T_b]=2 dfrac(x-a)(b-y)(b-a) leq mathbbE_x[L_y^T_a ] $



because the local time is increasing.
Now, we have
$mathbbE_x[L_y^T_a wedge T_b]
=2 dfrac(x-a)(b-y)(b-a)= 2 dfracb(x-a)(1-y/b)b(1-a/b)= 2 dfrac(x-a)(1-y/b)(1-a/b) rightarrow 2 (x-a)$
for $b rightarrow infty$.
Now, the continuity of the local time, dominated convergence one obtains
$mathbbE_x[L_y^T_a]=2(x-a)$.



We also have
$mathbbE_x[L_y^T_a wedge T_b]=2 dfraca(x/a-1)(b-y)a(b/a-1) =2 dfrac(x/a-1)(b-y)(b/a-1) rightarrow 2(b-y)$ for $a rightarrow -infty$.
This leads to
$mathbbE_x[L_y^ T_b]=2(b-y)$
By the way.
I hope this computation is on the right way.
I need this to compute the walk Dimension for gap diffusions.





probability stochastic-calculus brownian-motion martingales





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 23 at 14:44
























asked Jul 22 at 16:00









Dr. NIemann

605




605











  • Instead of editing the question, it is better to post this as an answer to your own question (which is very normal). Also you should let $bto-infty$, not $a$; this would give $mathbb E_x [L_y^T_a] = 2(a-x)$, which is not unexpected at all.
    – zhoraster
    Jul 24 at 12:28










  • In my answer I change the order of $a$ and $x$ my bad.
    – Dr. NIemann
    Jul 25 at 0:25
















  • Instead of editing the question, it is better to post this as an answer to your own question (which is very normal). Also you should let $bto-infty$, not $a$; this would give $mathbb E_x [L_y^T_a] = 2(a-x)$, which is not unexpected at all.
    – zhoraster
    Jul 24 at 12:28










  • In my answer I change the order of $a$ and $x$ my bad.
    – Dr. NIemann
    Jul 25 at 0:25















Instead of editing the question, it is better to post this as an answer to your own question (which is very normal). Also you should let $bto-infty$, not $a$; this would give $mathbb E_x [L_y^T_a] = 2(a-x)$, which is not unexpected at all.
– zhoraster
Jul 24 at 12:28




Instead of editing the question, it is better to post this as an answer to your own question (which is very normal). Also you should let $bto-infty$, not $a$; this would give $mathbb E_x [L_y^T_a] = 2(a-x)$, which is not unexpected at all.
– zhoraster
Jul 24 at 12:28












In my answer I change the order of $a$ and $x$ my bad.
– Dr. NIemann
Jul 25 at 0:25




In my answer I change the order of $a$ and $x$ my bad.
– Dr. NIemann
Jul 25 at 0:25










2 Answers
2






active

oldest

votes

















up vote
0
down vote



accepted










Because the integrand is bounded, the stochastic intergal is uniformly integrable. So $int_0^T_aoperatornamesgn(B_s-y),dB_s$ has mean zero. The rest of your computation is fine.






share|cite|improve this answer

















  • 2




    Hmmm... The same reasoning would "prove" that $a=E_y(B_T_a)=y$, which is obviously untrue.
    – Did
    Jul 23 at 10:54










  • So It is not a uniform integrable martingal?
    – Dr. NIemann
    Jul 23 at 11:03










  • Not, it is not.
    – zhoraster
    Jul 23 at 11:06










  • Ok, this will explain the contradiction. Is there an other way to handle $mathbbE_x[L_y^T_a]$?
    – Dr. NIemann
    Jul 23 at 11:08







  • 1




    Replace $T_a$ by $R_n:=min( T_a, T_z,n)$, where $z$ is far to the left of $y$. and take expectations: $$ Bbb E^x[|B_R_n-y|] -|x-y| = Bbb E^x[L^y_R_n]. $$ The right side of this converges, as $ntoinfty$, to $Bbb E^x[L^y_min(T_a, T_z)]$ by monotone convergence. The left side converges, as $ntoinfty$, to $|a-y|cdot Bbb P^x[T_a<T_z]+|z-y|cdotBbb P^x[T_z<T_a]-|x-y|$ by dominated convergence. From the well-known formula for hitting probabilities, this is equal to $$ |a-y|cdotx-zover a-z+|z-y|cdota-xover a-z-|x-y|. $$ Finally, let $zto-infty$.
    – John Dawkins
    Jul 24 at 17:31

















up vote
0
down vote













Let $a <x<y<b$ then we have



$mathbbE_x[L_y^T_a wedge T_b]=2 dfrac(x-a)(b-y)(b-a) leq mathbbE_x[L_y^T_a ] $



because the local time is increasing.
Now, we have
$mathbbE_x[L_y^T_a wedge T_b]
=2 dfrac(x-a)(b-y)(b-a)= 2 dfracb(x-a)(1-y/b)b(1-a/b)= 2 dfrac(x-a)(1-y/b)(1-a/b) rightarrow 2 (x-a)$
for $b rightarrow infty$.
Now, the continuity of the local time, monotone convergence one obtains
$mathbbE_x[L_y^T_a]=2(x-a)$.



We also have
$mathbbE_x[L_y^T_a wedge T_b]=2 dfraca(x/a-1)(b-y)a(b/a-1) =2 dfrac(x/a-1)(b-y)(b/a-1) rightarrow 2(b-y)$ for $a rightarrow -infty$.
This leads to
$mathbbE_x[L_y^ T_b]=2(b-y)$






share|cite|improve this answer























  • Yes, but in OP, you have $y<x$. With this, the numerator would be $(y-a)(b-x)$, and the limit would become $2(b-x)$.
    – zhoraster
    Jul 27 at 4:13










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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
0
down vote



accepted










Because the integrand is bounded, the stochastic intergal is uniformly integrable. So $int_0^T_aoperatornamesgn(B_s-y),dB_s$ has mean zero. The rest of your computation is fine.






share|cite|improve this answer

















  • 2




    Hmmm... The same reasoning would "prove" that $a=E_y(B_T_a)=y$, which is obviously untrue.
    – Did
    Jul 23 at 10:54










  • So It is not a uniform integrable martingal?
    – Dr. NIemann
    Jul 23 at 11:03










  • Not, it is not.
    – zhoraster
    Jul 23 at 11:06










  • Ok, this will explain the contradiction. Is there an other way to handle $mathbbE_x[L_y^T_a]$?
    – Dr. NIemann
    Jul 23 at 11:08







  • 1




    Replace $T_a$ by $R_n:=min( T_a, T_z,n)$, where $z$ is far to the left of $y$. and take expectations: $$ Bbb E^x[|B_R_n-y|] -|x-y| = Bbb E^x[L^y_R_n]. $$ The right side of this converges, as $ntoinfty$, to $Bbb E^x[L^y_min(T_a, T_z)]$ by monotone convergence. The left side converges, as $ntoinfty$, to $|a-y|cdot Bbb P^x[T_a<T_z]+|z-y|cdotBbb P^x[T_z<T_a]-|x-y|$ by dominated convergence. From the well-known formula for hitting probabilities, this is equal to $$ |a-y|cdotx-zover a-z+|z-y|cdota-xover a-z-|x-y|. $$ Finally, let $zto-infty$.
    – John Dawkins
    Jul 24 at 17:31














up vote
0
down vote



accepted










Because the integrand is bounded, the stochastic intergal is uniformly integrable. So $int_0^T_aoperatornamesgn(B_s-y),dB_s$ has mean zero. The rest of your computation is fine.






share|cite|improve this answer

















  • 2




    Hmmm... The same reasoning would "prove" that $a=E_y(B_T_a)=y$, which is obviously untrue.
    – Did
    Jul 23 at 10:54










  • So It is not a uniform integrable martingal?
    – Dr. NIemann
    Jul 23 at 11:03










  • Not, it is not.
    – zhoraster
    Jul 23 at 11:06










  • Ok, this will explain the contradiction. Is there an other way to handle $mathbbE_x[L_y^T_a]$?
    – Dr. NIemann
    Jul 23 at 11:08







  • 1




    Replace $T_a$ by $R_n:=min( T_a, T_z,n)$, where $z$ is far to the left of $y$. and take expectations: $$ Bbb E^x[|B_R_n-y|] -|x-y| = Bbb E^x[L^y_R_n]. $$ The right side of this converges, as $ntoinfty$, to $Bbb E^x[L^y_min(T_a, T_z)]$ by monotone convergence. The left side converges, as $ntoinfty$, to $|a-y|cdot Bbb P^x[T_a<T_z]+|z-y|cdotBbb P^x[T_z<T_a]-|x-y|$ by dominated convergence. From the well-known formula for hitting probabilities, this is equal to $$ |a-y|cdotx-zover a-z+|z-y|cdota-xover a-z-|x-y|. $$ Finally, let $zto-infty$.
    – John Dawkins
    Jul 24 at 17:31












up vote
0
down vote



accepted







up vote
0
down vote



accepted






Because the integrand is bounded, the stochastic intergal is uniformly integrable. So $int_0^T_aoperatornamesgn(B_s-y),dB_s$ has mean zero. The rest of your computation is fine.






share|cite|improve this answer













Because the integrand is bounded, the stochastic intergal is uniformly integrable. So $int_0^T_aoperatornamesgn(B_s-y),dB_s$ has mean zero. The rest of your computation is fine.







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 22 at 17:19









John Dawkins

12.5k1917




12.5k1917







  • 2




    Hmmm... The same reasoning would "prove" that $a=E_y(B_T_a)=y$, which is obviously untrue.
    – Did
    Jul 23 at 10:54










  • So It is not a uniform integrable martingal?
    – Dr. NIemann
    Jul 23 at 11:03










  • Not, it is not.
    – zhoraster
    Jul 23 at 11:06










  • Ok, this will explain the contradiction. Is there an other way to handle $mathbbE_x[L_y^T_a]$?
    – Dr. NIemann
    Jul 23 at 11:08







  • 1




    Replace $T_a$ by $R_n:=min( T_a, T_z,n)$, where $z$ is far to the left of $y$. and take expectations: $$ Bbb E^x[|B_R_n-y|] -|x-y| = Bbb E^x[L^y_R_n]. $$ The right side of this converges, as $ntoinfty$, to $Bbb E^x[L^y_min(T_a, T_z)]$ by monotone convergence. The left side converges, as $ntoinfty$, to $|a-y|cdot Bbb P^x[T_a<T_z]+|z-y|cdotBbb P^x[T_z<T_a]-|x-y|$ by dominated convergence. From the well-known formula for hitting probabilities, this is equal to $$ |a-y|cdotx-zover a-z+|z-y|cdota-xover a-z-|x-y|. $$ Finally, let $zto-infty$.
    – John Dawkins
    Jul 24 at 17:31












  • 2




    Hmmm... The same reasoning would "prove" that $a=E_y(B_T_a)=y$, which is obviously untrue.
    – Did
    Jul 23 at 10:54










  • So It is not a uniform integrable martingal?
    – Dr. NIemann
    Jul 23 at 11:03










  • Not, it is not.
    – zhoraster
    Jul 23 at 11:06










  • Ok, this will explain the contradiction. Is there an other way to handle $mathbbE_x[L_y^T_a]$?
    – Dr. NIemann
    Jul 23 at 11:08







  • 1




    Replace $T_a$ by $R_n:=min( T_a, T_z,n)$, where $z$ is far to the left of $y$. and take expectations: $$ Bbb E^x[|B_R_n-y|] -|x-y| = Bbb E^x[L^y_R_n]. $$ The right side of this converges, as $ntoinfty$, to $Bbb E^x[L^y_min(T_a, T_z)]$ by monotone convergence. The left side converges, as $ntoinfty$, to $|a-y|cdot Bbb P^x[T_a<T_z]+|z-y|cdotBbb P^x[T_z<T_a]-|x-y|$ by dominated convergence. From the well-known formula for hitting probabilities, this is equal to $$ |a-y|cdotx-zover a-z+|z-y|cdota-xover a-z-|x-y|. $$ Finally, let $zto-infty$.
    – John Dawkins
    Jul 24 at 17:31







2




2




Hmmm... The same reasoning would "prove" that $a=E_y(B_T_a)=y$, which is obviously untrue.
– Did
Jul 23 at 10:54




Hmmm... The same reasoning would "prove" that $a=E_y(B_T_a)=y$, which is obviously untrue.
– Did
Jul 23 at 10:54












So It is not a uniform integrable martingal?
– Dr. NIemann
Jul 23 at 11:03




So It is not a uniform integrable martingal?
– Dr. NIemann
Jul 23 at 11:03












Not, it is not.
– zhoraster
Jul 23 at 11:06




Not, it is not.
– zhoraster
Jul 23 at 11:06












Ok, this will explain the contradiction. Is there an other way to handle $mathbbE_x[L_y^T_a]$?
– Dr. NIemann
Jul 23 at 11:08





Ok, this will explain the contradiction. Is there an other way to handle $mathbbE_x[L_y^T_a]$?
– Dr. NIemann
Jul 23 at 11:08





1




1




Replace $T_a$ by $R_n:=min( T_a, T_z,n)$, where $z$ is far to the left of $y$. and take expectations: $$ Bbb E^x[|B_R_n-y|] -|x-y| = Bbb E^x[L^y_R_n]. $$ The right side of this converges, as $ntoinfty$, to $Bbb E^x[L^y_min(T_a, T_z)]$ by monotone convergence. The left side converges, as $ntoinfty$, to $|a-y|cdot Bbb P^x[T_a<T_z]+|z-y|cdotBbb P^x[T_z<T_a]-|x-y|$ by dominated convergence. From the well-known formula for hitting probabilities, this is equal to $$ |a-y|cdotx-zover a-z+|z-y|cdota-xover a-z-|x-y|. $$ Finally, let $zto-infty$.
– John Dawkins
Jul 24 at 17:31




Replace $T_a$ by $R_n:=min( T_a, T_z,n)$, where $z$ is far to the left of $y$. and take expectations: $$ Bbb E^x[|B_R_n-y|] -|x-y| = Bbb E^x[L^y_R_n]. $$ The right side of this converges, as $ntoinfty$, to $Bbb E^x[L^y_min(T_a, T_z)]$ by monotone convergence. The left side converges, as $ntoinfty$, to $|a-y|cdot Bbb P^x[T_a<T_z]+|z-y|cdotBbb P^x[T_z<T_a]-|x-y|$ by dominated convergence. From the well-known formula for hitting probabilities, this is equal to $$ |a-y|cdotx-zover a-z+|z-y|cdota-xover a-z-|x-y|. $$ Finally, let $zto-infty$.
– John Dawkins
Jul 24 at 17:31










up vote
0
down vote













Let $a <x<y<b$ then we have



$mathbbE_x[L_y^T_a wedge T_b]=2 dfrac(x-a)(b-y)(b-a) leq mathbbE_x[L_y^T_a ] $



because the local time is increasing.
Now, we have
$mathbbE_x[L_y^T_a wedge T_b]
=2 dfrac(x-a)(b-y)(b-a)= 2 dfracb(x-a)(1-y/b)b(1-a/b)= 2 dfrac(x-a)(1-y/b)(1-a/b) rightarrow 2 (x-a)$
for $b rightarrow infty$.
Now, the continuity of the local time, monotone convergence one obtains
$mathbbE_x[L_y^T_a]=2(x-a)$.



We also have
$mathbbE_x[L_y^T_a wedge T_b]=2 dfraca(x/a-1)(b-y)a(b/a-1) =2 dfrac(x/a-1)(b-y)(b/a-1) rightarrow 2(b-y)$ for $a rightarrow -infty$.
This leads to
$mathbbE_x[L_y^ T_b]=2(b-y)$






share|cite|improve this answer























  • Yes, but in OP, you have $y<x$. With this, the numerator would be $(y-a)(b-x)$, and the limit would become $2(b-x)$.
    – zhoraster
    Jul 27 at 4:13














up vote
0
down vote













Let $a <x<y<b$ then we have



$mathbbE_x[L_y^T_a wedge T_b]=2 dfrac(x-a)(b-y)(b-a) leq mathbbE_x[L_y^T_a ] $



because the local time is increasing.
Now, we have
$mathbbE_x[L_y^T_a wedge T_b]
=2 dfrac(x-a)(b-y)(b-a)= 2 dfracb(x-a)(1-y/b)b(1-a/b)= 2 dfrac(x-a)(1-y/b)(1-a/b) rightarrow 2 (x-a)$
for $b rightarrow infty$.
Now, the continuity of the local time, monotone convergence one obtains
$mathbbE_x[L_y^T_a]=2(x-a)$.



We also have
$mathbbE_x[L_y^T_a wedge T_b]=2 dfraca(x/a-1)(b-y)a(b/a-1) =2 dfrac(x/a-1)(b-y)(b/a-1) rightarrow 2(b-y)$ for $a rightarrow -infty$.
This leads to
$mathbbE_x[L_y^ T_b]=2(b-y)$






share|cite|improve this answer























  • Yes, but in OP, you have $y<x$. With this, the numerator would be $(y-a)(b-x)$, and the limit would become $2(b-x)$.
    – zhoraster
    Jul 27 at 4:13












up vote
0
down vote










up vote
0
down vote









Let $a <x<y<b$ then we have



$mathbbE_x[L_y^T_a wedge T_b]=2 dfrac(x-a)(b-y)(b-a) leq mathbbE_x[L_y^T_a ] $



because the local time is increasing.
Now, we have
$mathbbE_x[L_y^T_a wedge T_b]
=2 dfrac(x-a)(b-y)(b-a)= 2 dfracb(x-a)(1-y/b)b(1-a/b)= 2 dfrac(x-a)(1-y/b)(1-a/b) rightarrow 2 (x-a)$
for $b rightarrow infty$.
Now, the continuity of the local time, monotone convergence one obtains
$mathbbE_x[L_y^T_a]=2(x-a)$.



We also have
$mathbbE_x[L_y^T_a wedge T_b]=2 dfraca(x/a-1)(b-y)a(b/a-1) =2 dfrac(x/a-1)(b-y)(b/a-1) rightarrow 2(b-y)$ for $a rightarrow -infty$.
This leads to
$mathbbE_x[L_y^ T_b]=2(b-y)$






share|cite|improve this answer















Let $a <x<y<b$ then we have



$mathbbE_x[L_y^T_a wedge T_b]=2 dfrac(x-a)(b-y)(b-a) leq mathbbE_x[L_y^T_a ] $



because the local time is increasing.
Now, we have
$mathbbE_x[L_y^T_a wedge T_b]
=2 dfrac(x-a)(b-y)(b-a)= 2 dfracb(x-a)(1-y/b)b(1-a/b)= 2 dfrac(x-a)(1-y/b)(1-a/b) rightarrow 2 (x-a)$
for $b rightarrow infty$.
Now, the continuity of the local time, monotone convergence one obtains
$mathbbE_x[L_y^T_a]=2(x-a)$.



We also have
$mathbbE_x[L_y^T_a wedge T_b]=2 dfraca(x/a-1)(b-y)a(b/a-1) =2 dfrac(x/a-1)(b-y)(b/a-1) rightarrow 2(b-y)$ for $a rightarrow -infty$.
This leads to
$mathbbE_x[L_y^ T_b]=2(b-y)$







share|cite|improve this answer















share|cite|improve this answer



share|cite|improve this answer








edited Jul 25 at 0:10


























answered Jul 25 at 0:02









Dr. NIemann

605




605











  • Yes, but in OP, you have $y<x$. With this, the numerator would be $(y-a)(b-x)$, and the limit would become $2(b-x)$.
    – zhoraster
    Jul 27 at 4:13
















  • Yes, but in OP, you have $y<x$. With this, the numerator would be $(y-a)(b-x)$, and the limit would become $2(b-x)$.
    – zhoraster
    Jul 27 at 4:13















Yes, but in OP, you have $y<x$. With this, the numerator would be $(y-a)(b-x)$, and the limit would become $2(b-x)$.
– zhoraster
Jul 27 at 4:13




Yes, but in OP, you have $y<x$. With this, the numerator would be $(y-a)(b-x)$, and the limit would become $2(b-x)$.
– zhoraster
Jul 27 at 4:13












 

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