Why is $P^b(B_{T-T_b} in (-infty,b))=1/2$












1












$begingroup$


Why is $P^b(B_{T-T_b} in (-infty,b))=1/2$ on the set ${T_b<t} $ where $T_b=inf{t ge 0 :B_t=b}$ and $T=t 1_{{T_b<t}}+infty 1_{{T_b ge 0}}$.



I am trying to understand Proposition 2.6.19 in Shreve
and trying to figure out why $$(U_{T(omega)-T_b(omega)}1_Gamma)(B_{S(omega)}(omega))=1/2$$
My attempt
I write $$(U_{T(omega)-T_b(omega)}1_Gamma)(B_{S(omega)}(omega))=P^b(B_{t-T_b)} in (-infty,b))=P^0(B_{t-T_b)} in (-infty,0))$$



But then can I somehow argue that the distribution of $B_{t-T_b}$ is normal with mean $0$? I cant see why this should be true? Or am I totally off track



enter image description here






stochastic-processes brownian-motion markov-process stochastic-analysis







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  • $begingroup$
    What is $U$....?
    $endgroup$
    – saz
    Dec 10 '18 at 17:52










  • $begingroup$
    $(U_tf)(x)=E^x(f(X_t)$
    $endgroup$
    – user3503589
    Dec 10 '18 at 17:53
















1












$begingroup$


Why is $P^b(B_{T-T_b} in (-infty,b))=1/2$ on the set ${T_b<t} $ where $T_b=inf{t ge 0 :B_t=b}$ and $T=t 1_{{T_b<t}}+infty 1_{{T_b ge 0}}$.



I am trying to understand Proposition 2.6.19 in Shreve
and trying to figure out why $$(U_{T(omega)-T_b(omega)}1_Gamma)(B_{S(omega)}(omega))=1/2$$
My attempt
I write $$(U_{T(omega)-T_b(omega)}1_Gamma)(B_{S(omega)}(omega))=P^b(B_{t-T_b)} in (-infty,b))=P^0(B_{t-T_b)} in (-infty,0))$$



But then can I somehow argue that the distribution of $B_{t-T_b}$ is normal with mean $0$? I cant see why this should be true? Or am I totally off track



enter image description here






stochastic-processes brownian-motion markov-process stochastic-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    What is $U$....?
    $endgroup$
    – saz
    Dec 10 '18 at 17:52










  • $begingroup$
    $(U_tf)(x)=E^x(f(X_t)$
    $endgroup$
    – user3503589
    Dec 10 '18 at 17:53














1












1








1





$begingroup$


Why is $P^b(B_{T-T_b} in (-infty,b))=1/2$ on the set ${T_b<t} $ where $T_b=inf{t ge 0 :B_t=b}$ and $T=t 1_{{T_b<t}}+infty 1_{{T_b ge 0}}$.



I am trying to understand Proposition 2.6.19 in Shreve
and trying to figure out why $$(U_{T(omega)-T_b(omega)}1_Gamma)(B_{S(omega)}(omega))=1/2$$
My attempt
I write $$(U_{T(omega)-T_b(omega)}1_Gamma)(B_{S(omega)}(omega))=P^b(B_{t-T_b)} in (-infty,b))=P^0(B_{t-T_b)} in (-infty,0))$$



But then can I somehow argue that the distribution of $B_{t-T_b}$ is normal with mean $0$? I cant see why this should be true? Or am I totally off track



enter image description here






stochastic-processes brownian-motion markov-process stochastic-analysis







share|cite|improve this question











$endgroup$




Why is $P^b(B_{T-T_b} in (-infty,b))=1/2$ on the set ${T_b<t} $ where $T_b=inf{t ge 0 :B_t=b}$ and $T=t 1_{{T_b<t}}+infty 1_{{T_b ge 0}}$.



I am trying to understand Proposition 2.6.19 in Shreve
and trying to figure out why $$(U_{T(omega)-T_b(omega)}1_Gamma)(B_{S(omega)}(omega))=1/2$$
My attempt
I write $$(U_{T(omega)-T_b(omega)}1_Gamma)(B_{S(omega)}(omega))=P^b(B_{t-T_b)} in (-infty,b))=P^0(B_{t-T_b)} in (-infty,0))$$



But then can I somehow argue that the distribution of $B_{t-T_b}$ is normal with mean $0$? I cant see why this should be true? Or am I totally off track



enter image description here






stochastic-processes brownian-motion markov-process stochastic-analysis




stochastic-processes brownian-motion markov-process stochastic-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 11 '18 at 14:11









saz

81.8k862131




81.8k862131










asked Dec 10 '18 at 16:02









user3503589user3503589

1,3051821




1,3051821












  • $begingroup$
    What is $U$....?
    $endgroup$
    – saz
    Dec 10 '18 at 17:52










  • $begingroup$
    $(U_tf)(x)=E^x(f(X_t)$
    $endgroup$
    – user3503589
    Dec 10 '18 at 17:53


















  • $begingroup$
    What is $U$....?
    $endgroup$
    – saz
    Dec 10 '18 at 17:52










  • $begingroup$
    $(U_tf)(x)=E^x(f(X_t)$
    $endgroup$
    – user3503589
    Dec 10 '18 at 17:53
















$begingroup$
What is $U$....?
$endgroup$
– saz
Dec 10 '18 at 17:52




$begingroup$
What is $U$....?
$endgroup$
– saz
Dec 10 '18 at 17:52












$begingroup$
$(U_tf)(x)=E^x(f(X_t)$
$endgroup$
– user3503589
Dec 10 '18 at 17:53




$begingroup$
$(U_tf)(x)=E^x(f(X_t)$
$endgroup$
– user3503589
Dec 10 '18 at 17:53










1 Answer
1






active

oldest

votes


















2












$begingroup$

Fix $t>0$. By the very definition of $S=T_b$ we have $B_{S(omega)}(omega)=b$, i.e.



$$(U_{t} 1_{(-infty,b)})(B_{S(omega)}(omega)) = (U_{t} 1_{(-infty,b)})(b).$$



Since $$ (U_t 1_{(-infty,b)})(x) = mathbb{P}^x(B_t < b)$$



for any $x in mathbb{R}$, we find



$$(U_{t} 1_{(-infty,b)})(B_{S(omega)}(omega)) = mathbb{P}^b(B_t < b).$$



As $mathbb{P}^b(B_t < b) = mathbb{P}^0(b+B_t < b)$ this implies



$$(U_{t} 1_{(-infty,b)})(B_{S(omega)}(omega)) = mathbb{P}^0(B_t < 0) = frac{1}{2}.$$



Since $t>0$ is arbitrary, we can choose in particular $t:=T(omega)-S(omega)$ for $omega in {T<infty} subseteq {S<T}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    . Thank you . Now it’s all clear .
    $endgroup$
    – user3503589
    Dec 10 '18 at 19:44






  • 1




    $begingroup$
    @user3503589 You are welcome; glad I could help you.
    $endgroup$
    – saz
    Dec 10 '18 at 20:26











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1 Answer
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1 Answer
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2












$begingroup$

Fix $t>0$. By the very definition of $S=T_b$ we have $B_{S(omega)}(omega)=b$, i.e.



$$(U_{t} 1_{(-infty,b)})(B_{S(omega)}(omega)) = (U_{t} 1_{(-infty,b)})(b).$$



Since $$ (U_t 1_{(-infty,b)})(x) = mathbb{P}^x(B_t < b)$$



for any $x in mathbb{R}$, we find



$$(U_{t} 1_{(-infty,b)})(B_{S(omega)}(omega)) = mathbb{P}^b(B_t < b).$$



As $mathbb{P}^b(B_t < b) = mathbb{P}^0(b+B_t < b)$ this implies



$$(U_{t} 1_{(-infty,b)})(B_{S(omega)}(omega)) = mathbb{P}^0(B_t < 0) = frac{1}{2}.$$



Since $t>0$ is arbitrary, we can choose in particular $t:=T(omega)-S(omega)$ for $omega in {T<infty} subseteq {S<T}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    . Thank you . Now it’s all clear .
    $endgroup$
    – user3503589
    Dec 10 '18 at 19:44






  • 1




    $begingroup$
    @user3503589 You are welcome; glad I could help you.
    $endgroup$
    – saz
    Dec 10 '18 at 20:26
















2












$begingroup$

Fix $t>0$. By the very definition of $S=T_b$ we have $B_{S(omega)}(omega)=b$, i.e.



$$(U_{t} 1_{(-infty,b)})(B_{S(omega)}(omega)) = (U_{t} 1_{(-infty,b)})(b).$$



Since $$ (U_t 1_{(-infty,b)})(x) = mathbb{P}^x(B_t < b)$$



for any $x in mathbb{R}$, we find



$$(U_{t} 1_{(-infty,b)})(B_{S(omega)}(omega)) = mathbb{P}^b(B_t < b).$$



As $mathbb{P}^b(B_t < b) = mathbb{P}^0(b+B_t < b)$ this implies



$$(U_{t} 1_{(-infty,b)})(B_{S(omega)}(omega)) = mathbb{P}^0(B_t < 0) = frac{1}{2}.$$



Since $t>0$ is arbitrary, we can choose in particular $t:=T(omega)-S(omega)$ for $omega in {T<infty} subseteq {S<T}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    . Thank you . Now it’s all clear .
    $endgroup$
    – user3503589
    Dec 10 '18 at 19:44






  • 1




    $begingroup$
    @user3503589 You are welcome; glad I could help you.
    $endgroup$
    – saz
    Dec 10 '18 at 20:26














2












2








2





$begingroup$

Fix $t>0$. By the very definition of $S=T_b$ we have $B_{S(omega)}(omega)=b$, i.e.



$$(U_{t} 1_{(-infty,b)})(B_{S(omega)}(omega)) = (U_{t} 1_{(-infty,b)})(b).$$



Since $$ (U_t 1_{(-infty,b)})(x) = mathbb{P}^x(B_t < b)$$



for any $x in mathbb{R}$, we find



$$(U_{t} 1_{(-infty,b)})(B_{S(omega)}(omega)) = mathbb{P}^b(B_t < b).$$



As $mathbb{P}^b(B_t < b) = mathbb{P}^0(b+B_t < b)$ this implies



$$(U_{t} 1_{(-infty,b)})(B_{S(omega)}(omega)) = mathbb{P}^0(B_t < 0) = frac{1}{2}.$$



Since $t>0$ is arbitrary, we can choose in particular $t:=T(omega)-S(omega)$ for $omega in {T<infty} subseteq {S<T}$.






share|cite|improve this answer









$endgroup$



Fix $t>0$. By the very definition of $S=T_b$ we have $B_{S(omega)}(omega)=b$, i.e.



$$(U_{t} 1_{(-infty,b)})(B_{S(omega)}(omega)) = (U_{t} 1_{(-infty,b)})(b).$$



Since $$ (U_t 1_{(-infty,b)})(x) = mathbb{P}^x(B_t < b)$$



for any $x in mathbb{R}$, we find



$$(U_{t} 1_{(-infty,b)})(B_{S(omega)}(omega)) = mathbb{P}^b(B_t < b).$$



As $mathbb{P}^b(B_t < b) = mathbb{P}^0(b+B_t < b)$ this implies



$$(U_{t} 1_{(-infty,b)})(B_{S(omega)}(omega)) = mathbb{P}^0(B_t < 0) = frac{1}{2}.$$



Since $t>0$ is arbitrary, we can choose in particular $t:=T(omega)-S(omega)$ for $omega in {T<infty} subseteq {S<T}$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 10 '18 at 18:01









sazsaz

81.8k862131




81.8k862131












  • $begingroup$
    . Thank you . Now it’s all clear .
    $endgroup$
    – user3503589
    Dec 10 '18 at 19:44






  • 1




    $begingroup$
    @user3503589 You are welcome; glad I could help you.
    $endgroup$
    – saz
    Dec 10 '18 at 20:26


















  • $begingroup$
    . Thank you . Now it’s all clear .
    $endgroup$
    – user3503589
    Dec 10 '18 at 19:44






  • 1




    $begingroup$
    @user3503589 You are welcome; glad I could help you.
    $endgroup$
    – saz
    Dec 10 '18 at 20:26
















$begingroup$
. Thank you . Now it’s all clear .
$endgroup$
– user3503589
Dec 10 '18 at 19:44




$begingroup$
. Thank you . Now it’s all clear .
$endgroup$
– user3503589
Dec 10 '18 at 19:44




1




1




$begingroup$
@user3503589 You are welcome; glad I could help you.
$endgroup$
– saz
Dec 10 '18 at 20:26




$begingroup$
@user3503589 You are welcome; glad I could help you.
$endgroup$
– saz
Dec 10 '18 at 20:26


















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