Why is $P^b(B_{T-T_b} in (-infty,b))=1/2$
$begingroup$
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
stochastic-processes brownian-motion markov-process stochastic-analysis
edited Dec 11 '18 at 14:11
saz
81.8k862131
asked Dec 10 '18 at 16:02
user3503589user3503589
1,3051821
$endgroup$
add a comment |
$begingroup$
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
stochastic-processes brownian-motion markov-process stochastic-analysis
edited Dec 11 '18 at 14:11
saz
81.8k862131
asked Dec 10 '18 at 16:02
user3503589user3503589
1,3051821
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
stochastic-processes brownian-motion markov-process stochastic-analysis
edited Dec 11 '18 at 14:11
saz
81.8k862131
asked Dec 10 '18 at 16:02
user3503589user3503589
1,3051821
$endgroup$
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
stochastic-processes brownian-motion markov-process stochastic-analysis
stochastic-processes brownian-motion markov-process stochastic-analysis
edited Dec 11 '18 at 14:11
saz
81.8k862131
asked Dec 10 '18 at 16:02
user3503589user3503589
1,3051821
edited Dec 11 '18 at 14:11
saz
81.8k862131
asked Dec 10 '18 at 16:02
user3503589user3503589
1,3051821
edited Dec 11 '18 at 14:11
saz
81.8k862131
edited Dec 11 '18 at 14:11
saz
81.8k862131
edited Dec 11 '18 at 14:11
saz
81.8k862131
81.8k862131
asked Dec 10 '18 at 16:02
user3503589user3503589
1,3051821
asked Dec 10 '18 at 16:02
user3503589user3503589
1,3051821
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 $omega in {T<infty} subseteq {S<T}$.
answered Dec 10 '18 at 18:01
sazsaz
81.8k862131
$endgroup$
$begingroup$
. 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
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
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 $omega in {T<infty} subseteq {S<T}$.
answered Dec 10 '18 at 18:01
sazsaz
81.8k862131
$endgroup$
$begingroup$
. 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
add a comment |
$begingroup$
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 $omega in {T<infty} subseteq {S<T}$.
answered Dec 10 '18 at 18:01
sazsaz
81.8k862131
$endgroup$
$begingroup$
. 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
add a comment |
$begingroup$
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 $omega in {T<infty} subseteq {S<T}$.
answered Dec 10 '18 at 18:01
sazsaz
81.8k862131
$endgroup$
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 $omega in {T<infty} subseteq {S<T}$.
answered Dec 10 '18 at 18:01
sazsaz
81.8k862131
answered Dec 10 '18 at 18:01
sazsaz
81.8k862131
answered Dec 10 '18 at 18:01
sazsaz
81.8k862131
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
add a comment |
$begingroup$
. 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
add a comment |
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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