Mixing
Conditional expectation for Brownian motion
$begingroup$
Consider two Brownian motions $(W_t)_{tge 0}$ with starting point $x$ and $(W'_t)_{tge 0}$ with starting point $y$. Define $T:=inf{tge 0:W_t=0}$, the first time when $W_t$ is equal to $0$. Show that for every $x,y,t>0$ it holds:
$$P(W_{twedge T}le y mid {W_0=x})=P(|W'_t|ge x mid {W'_0=y}).$$
My attempt:
Rewrite $W'_t=B'_t+y$ for a standard Brownian motion $(B'_t)_{tge 0}$. Then $P(W'_0=y)=1$ and $1_{{W'_0=y}}=1_Omega$, such that
begin{align}
P(|W'_t|ge x mid {W'_0=y})&=frac{1}{P(W'_0=y)}EBig[1_{{|W'_t|ge x}}1_{{W'_0=y}}Big]=Pbig(|B_t'+y|ge xbig)
end{align}
Rewrite $W_t=B_t+x$ for a standard Brownian motion $(B_t)_{tge 0}$. Then $P(W_0=x)=1$ and $1_{{W_0=x}}=1_Omega$, such that
begin{align}
P(W_{twedge T}le y mid {W_0=x})&=frac{1}{P(W_0=x)}EBig[1_{{W_{twedge T}le y}}1_{{W_0=x}}Big]=Pbig(B_{twedge T}+xle ybig)\
end{align}
Here I get stuck. Maybe I have to use $0le B_{twedge T}+x$ and the reflection principle? I would really appreciate some help! Thanks in advance!
probability-theory stochastic-processes stochastic-calculus brownian-motion stopping-times
edited Jan 26 at 19:13
saz
81.5k861127
asked Jan 21 at 20:53
user408858user408858
482213
$endgroup$
add a comment |
$begingroup$
Consider two Brownian motions $(W_t)_{tge 0}$ with starting point $x$ and $(W'_t)_{tge 0}$ with starting point $y$. Define $T:=inf{tge 0:W_t=0}$, the first time when $W_t$ is equal to $0$. Show that for every $x,y,t>0$ it holds:
$$P(W_{twedge T}le y mid {W_0=x})=P(|W'_t|ge x mid {W'_0=y}).$$
My attempt:
Rewrite $W'_t=B'_t+y$ for a standard Brownian motion $(B'_t)_{tge 0}$. Then $P(W'_0=y)=1$ and $1_{{W'_0=y}}=1_Omega$, such that
begin{align}
P(|W'_t|ge x mid {W'_0=y})&=frac{1}{P(W'_0=y)}EBig[1_{{|W'_t|ge x}}1_{{W'_0=y}}Big]=Pbig(|B_t'+y|ge xbig)
end{align}
Rewrite $W_t=B_t+x$ for a standard Brownian motion $(B_t)_{tge 0}$. Then $P(W_0=x)=1$ and $1_{{W_0=x}}=1_Omega$, such that
begin{align}
P(W_{twedge T}le y mid {W_0=x})&=frac{1}{P(W_0=x)}EBig[1_{{W_{twedge T}le y}}1_{{W_0=x}}Big]=Pbig(B_{twedge T}+xle ybig)\
end{align}
Here I get stuck. Maybe I have to use $0le B_{twedge T}+x$ and the reflection principle? I would really appreciate some help! Thanks in advance!
probability-theory stochastic-processes stochastic-calculus brownian-motion stopping-times
edited Jan 26 at 19:13
saz
81.5k861127
asked Jan 21 at 20:53
user408858user408858
482213
$endgroup$
$begingroup$
What does $t wedge T$ mean? For some $t<T$ ? I guess you assume that $x>y$...
$endgroup$
– Diger
Jan 24 at 1:48
$begingroup$
$twedge T=min(t,T)$
$endgroup$
– user408858
Jan 24 at 1:49
$begingroup$
Never seen it, thanks.
$endgroup$
– Diger
Jan 24 at 1:50
$begingroup$
Where did you encounter this problem?
$endgroup$
– saz
Jan 24 at 7:54
$begingroup$
Probably you are telling me that you need rigorous proof, but from a heuristic perspective isn't it fairly clear that this is so, since the Normal-distribution is symmetric i.e. it doesn't matter whether you go left (from x to y) or right (from y to x); they appear with equal probability?
$endgroup$
– Diger
Jan 24 at 12:55
add a comment |
$begingroup$
Consider two Brownian motions $(W_t)_{tge 0}$ with starting point $x$ and $(W'_t)_{tge 0}$ with starting point $y$. Define $T:=inf{tge 0:W_t=0}$, the first time when $W_t$ is equal to $0$. Show that for every $x,y,t>0$ it holds:
$$P(W_{twedge T}le y mid {W_0=x})=P(|W'_t|ge x mid {W'_0=y}).$$
My attempt:
Rewrite $W'_t=B'_t+y$ for a standard Brownian motion $(B'_t)_{tge 0}$. Then $P(W'_0=y)=1$ and $1_{{W'_0=y}}=1_Omega$, such that
begin{align}
P(|W'_t|ge x mid {W'_0=y})&=frac{1}{P(W'_0=y)}EBig[1_{{|W'_t|ge x}}1_{{W'_0=y}}Big]=Pbig(|B_t'+y|ge xbig)
end{align}
Rewrite $W_t=B_t+x$ for a standard Brownian motion $(B_t)_{tge 0}$. Then $P(W_0=x)=1$ and $1_{{W_0=x}}=1_Omega$, such that
begin{align}
P(W_{twedge T}le y mid {W_0=x})&=frac{1}{P(W_0=x)}EBig[1_{{W_{twedge T}le y}}1_{{W_0=x}}Big]=Pbig(B_{twedge T}+xle ybig)\
end{align}
Here I get stuck. Maybe I have to use $0le B_{twedge T}+x$ and the reflection principle? I would really appreciate some help! Thanks in advance!
probability-theory stochastic-processes stochastic-calculus brownian-motion stopping-times
edited Jan 26 at 19:13
saz
81.5k861127
asked Jan 21 at 20:53
user408858user408858
482213
$endgroup$
Consider two Brownian motions $(W_t)_{tge 0}$ with starting point $x$ and $(W'_t)_{tge 0}$ with starting point $y$. Define $T:=inf{tge 0:W_t=0}$, the first time when $W_t$ is equal to $0$. Show that for every $x,y,t>0$ it holds:
$$P(W_{twedge T}le y mid {W_0=x})=P(|W'_t|ge x mid {W'_0=y}).$$
My attempt:
Rewrite $W'_t=B'_t+y$ for a standard Brownian motion $(B'_t)_{tge 0}$. Then $P(W'_0=y)=1$ and $1_{{W'_0=y}}=1_Omega$, such that
begin{align}
P(|W'_t|ge x mid {W'_0=y})&=frac{1}{P(W'_0=y)}EBig[1_{{|W'_t|ge x}}1_{{W'_0=y}}Big]=Pbig(|B_t'+y|ge xbig)
end{align}
Rewrite $W_t=B_t+x$ for a standard Brownian motion $(B_t)_{tge 0}$. Then $P(W_0=x)=1$ and $1_{{W_0=x}}=1_Omega$, such that
begin{align}
P(W_{twedge T}le y mid {W_0=x})&=frac{1}{P(W_0=x)}EBig[1_{{W_{twedge T}le y}}1_{{W_0=x}}Big]=Pbig(B_{twedge T}+xle ybig)\
end{align}
Here I get stuck. Maybe I have to use $0le B_{twedge T}+x$ and the reflection principle? I would really appreciate some help! Thanks in advance!
probability-theory stochastic-processes stochastic-calculus brownian-motion stopping-times
probability-theory stochastic-processes stochastic-calculus brownian-motion stopping-times
edited Jan 26 at 19:13
saz
81.5k861127
asked Jan 21 at 20:53
user408858user408858
482213
edited Jan 26 at 19:13
saz
81.5k861127
asked Jan 21 at 20:53
user408858user408858
482213
edited Jan 26 at 19:13
saz
81.5k861127
edited Jan 26 at 19:13
saz
81.5k861127
edited Jan 26 at 19:13
saz
81.5k861127
81.5k861127
asked Jan 21 at 20:53
user408858user408858
482213
asked Jan 21 at 20:53
user408858user408858
482213
asked Jan 21 at 20:53
user408858user408858
482213
482213
$begingroup$
What does $t wedge T$ mean? For some $t<T$ ? I guess you assume that $x>y$...
$endgroup$
– Diger
Jan 24 at 1:48
$begingroup$
$twedge T=min(t,T)$
$endgroup$
– user408858
Jan 24 at 1:49
$begingroup$
Never seen it, thanks.
$endgroup$
– Diger
Jan 24 at 1:50
$begingroup$
Where did you encounter this problem?
$endgroup$
– saz
Jan 24 at 7:54
$begingroup$
Probably you are telling me that you need rigorous proof, but from a heuristic perspective isn't it fairly clear that this is so, since the Normal-distribution is symmetric i.e. it doesn't matter whether you go left (from x to y) or right (from y to x); they appear with equal probability?
$endgroup$
– Diger
Jan 24 at 12:55
add a comment |
$begingroup$
What does $t wedge T$ mean? For some $t<T$ ? I guess you assume that $x>y$...
$endgroup$
– Diger
Jan 24 at 1:48
$begingroup$
$twedge T=min(t,T)$
$endgroup$
– user408858
Jan 24 at 1:49
$begingroup$
Never seen it, thanks.
$endgroup$
– Diger
Jan 24 at 1:50
$begingroup$
Where did you encounter this problem?
$endgroup$
– saz
Jan 24 at 7:54
$begingroup$
Probably you are telling me that you need rigorous proof, but from a heuristic perspective isn't it fairly clear that this is so, since the Normal-distribution is symmetric i.e. it doesn't matter whether you go left (from x to y) or right (from y to x); they appear with equal probability?
$endgroup$
– Diger
Jan 24 at 12:55
$begingroup$
What does $t wedge T$ mean? For some $t<T$ ? I guess you assume that $x>y$...
$endgroup$
– Diger
Jan 24 at 1:48
$begingroup$
What does $t wedge T$ mean? For some $t<T$ ? I guess you assume that $x>y$...
$endgroup$
– Diger
Jan 24 at 1:48
$begingroup$
$twedge T=min(t,T)$
$endgroup$
– user408858
Jan 24 at 1:49
$begingroup$
$twedge T=min(t,T)$
$endgroup$
– user408858
Jan 24 at 1:49
$begingroup$
Never seen it, thanks.
$endgroup$
– Diger
Jan 24 at 1:50
$begingroup$
Never seen it, thanks.
$endgroup$
– Diger
Jan 24 at 1:50
$begingroup$
Where did you encounter this problem?
$endgroup$
– saz
Jan 24 at 7:54
$begingroup$
Where did you encounter this problem?
$endgroup$
– saz
Jan 24 at 7:54
$begingroup$
Probably you are telling me that you need rigorous proof, but from a heuristic perspective isn't it fairly clear that this is so, since the Normal-distribution is symmetric i.e. it doesn't matter whether you go left (from x to y) or right (from y to x); they appear with equal probability?
$endgroup$
– Diger
Jan 24 at 12:55
$begingroup$
Probably you are telling me that you need rigorous proof, but from a heuristic perspective isn't it fairly clear that this is so, since the Normal-distribution is symmetric i.e. it doesn't matter whether you go left (from x to y) or right (from y to x); they appear with equal probability?
$endgroup$
– Diger
Jan 24 at 12:55
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let's write the assertion in term of a standard Brownian motion $(B_t)_{t geq 0}$.
If we set $T_{x} = inf{t geq 0; B_t = -x}$ then we need to show that
$$mathbb{P}(x+ B_{t wedge T_{x}} leq y) = mathbb{P}(|B_t+y| geq x) tag{1}$$
for any $x,y,t>0$.
Fix $x,y,t>0$. Since
$$B_{t wedge T_x} = begin{cases} -x & T_x leq t, \ B_t, & T_x>t end{cases}$$
we have
$$mathbb{P}(x+B_{t wedge T_x} leq y) = mathbb{P}(T_x leq t) + mathbb{P}(x+B_t leq y, T_x>t). tag{2}$$
Since the reflection principle allows us to calculate explicitly the joint distribution $(B_t,inf_{s leq t} B_s)$ (see below), we get
begin{align*}mathbb{P}(x+B_t leq y, T_x>t) &= mathbb{P} left( B_t leq y-x, inf_{s leq t} B_s>-x right) \ &= - frac{2}{sqrt{2pi t^3}} int_{-x}^0 int_{u leq z leq y-x} (2u-z) exp left(- frac{(2u-z)^2}{2t} right) , dz , du \ &= frac{2}{sqrt{2pi t}} int_{-x}^0 left[-exp left(- frac{(2u-y+x)^2}{2t} right) + exp left( -frac{u^2}{2t} right)right] , du.end{align*}
Using the linearity of the integral to split the integral on the right-hand side into two parts and performing a simple change of variables we get
$$mathbb{P}(x+B_t leq y, T_x>t) = sqrt{frac{2}{pi t}} int_{mathbb{R}} left(- frac{1}{2} cdot 1_{{|u-y| leq x}} + 1_{[-x,0]}(u) right) exp left(- frac{u^2}{2t} right) , du.$$
On the other hand, the reflection principle also gives
$$mathbb{P}(T_x leq t) =2 mathbb{P}(B_t' leq -x) =sqrt{frac{2}{pi t}} int_{-infty}^{-x} exp left( - frac{u^2}{2t} right) , du tag{3}$$
Adding both expressions and using that
$$sqrt{frac{2}{pi t}} int_{-infty}^0 exp left(- frac{u^2}{2t} right) , du = frac{1}{sqrt{2pi t}} int_{mathbb{R}} exp left(- frac{u^2}{2t} right) , du = 1,$$
it follows from $(2)$ that
$$begin{align*} mathbb{P}(x+B_{t wedge T_x} leq y) = 1- frac{1}{sqrt{2pi t}} int_{|z-y| leq x} exp left( - frac{u^2}{2t} right) , du&= mathbb{P}(|B_t+y|>x) \ &= mathbb{P}(|B_t+y| geq x) end{align*}$$
which proves the assertion.
Some remarks on the reflection principle: The reflection principle states (...or implies, depending on the formulation...) that the running supremum $M_t := sup_{s leq t} B_s$ satisfies $$M_t stackrel{d}{=} |B_t|.$$ Applying this result to the Brownian motion $(-B_t)_{t geq 0}$ it follows immediately that $$- inf_{s leq t} B_s sim |B_t|.$$ In particular, $$mathbb{P}(T_x leq t) = mathbb{P} left( min_{s leq t} B_s leq x right) = mathbb{P}(|B_t| leq -x) = 2 mathbb{P}(B_t leq x)$$ for the stopping time $$T_x := inf{t geq 0; B_t = -x}$$ from the first part of this answer. Combining the reflection principle with the strong Markov property of Brownian motion, it is possible to compute the joint density of $(B_t,M_t)$:
Let $(B_t)_{t geq 0}$ be a Brownian motion. The joint distribution $(B_t,sup_{s leq t} B_s)$ equals $$mathbb{P} left[ B_t in dx, sup_{s leq t} B_s in dy right] = frac{2 (2y-x)}{sqrt{2pi t^3}} exp left(- frac{(2y-x)^2}{2t} right) 1_{[-infty,y]}(x) , dx , dy. $$
For a proof see e.g. René Schilling/Lothar Partzsch: Brownian Motion - An Introduction to Stochastic Processes, Exercise 6.8 (there are full solutions available on the web), or this question (and differentiate in order to obtain the density function).
As before, we can use the symmetry of Brownian motion and apply the result to $(-B_t)_{t geq 0}$ in order to get the joint distribution of $(B_t,inf_{s leq t} B_s)$.
answered Jan 24 at 11:12
sazsaz
81.5k861127
$endgroup$
$begingroup$
I have problems understanding the reflection principle. I do not see how $P(T_xle t)=2P(B_t'le-x)$ does hold. Also I do not understand how to calculate the joint distribution, since there is an infimum involved. I only can rewrite $P(inf B_s>-x)=P(sup B_s<x)=1-2P(B_t>x)$ with the reflection principle? But how do I conclude from there on? Thank you already for your very good answer!
$endgroup$
– user408858
Jan 24 at 19:59
$begingroup$
@user408858 Which version of the reflection principle do you know...?
$endgroup$
– saz
Jan 24 at 20:18
$begingroup$
I know, that $B_t'=B_{twedge T}-(B_t-B_{twedge T})$ is also a brownian motion. I was also looking up the article on wikipedia where it is stated, that $P(sup_{sin[0,t]} B_sge a)=2P(B_tge a)$, see en.wikipedia.org/wiki/Reflection_principle_(Wiener_process)
$endgroup$
– user408858
Jan 24 at 20:25
$begingroup$
@user408858 I see... well, I will add some comments on that tomorrow...
$endgroup$
– saz
Jan 24 at 20:27
$begingroup$
Well, I think I already can answer my first question: $P(T_xle t)=P(inf B_sle -x)=2P(xle B_t)$
$endgroup$
– user408858
Jan 24 at 20:50
|
show 3 more comments
$begingroup$
Perhaps we may go for the proof as follows.
First, note that
begin{align}
&mathbb{P}(W_{twedge T}le y|W_0=x)\
&=mathbb{P}(W_tle y|W_0=x,T>t),mathbb{P}(T>t)\
&+mathbb{P}(W_Tle y|W_0=x,Tle t),mathbb{P}(Tle t)\
&=mathbb{P}(W_tle y|W_0=x,T>t),mathbb{P}(T>t)+mathbb{P}(Tle t)tag{1},
end{align}
where $mathbb{P}(W_Tle y|W_0=x,Tle t)=1$ as $W_T=0<y$ always holds.
Second, note that
begin{align}
mathbb{P}(W_tle y|W_0=x)&=mathbb{P}(W_tle y|W_0=x,T>t),mathbb{P}(T>t)\
&+mathbb{P}(W_tle y|W_0=x,Tle t),mathbb{P}(Tle t).
end{align}
Substitute this result into $(1)$, and we obtain
begin{align}
&mathbb{P}(W_{twedge T}le y|W_0=x)\
&=mathbb{P}(W_tle y|W_0=x)+mathbb{P}(Tle t)\
&-mathbb{P}(W_tle y|W_0=x,Tle t),mathbb{P}(Tle t)\
&=mathbb{P}(W_tle y|W_0=x)\
&+mathbb{P}(W_t>y|W_0=x,Tle t),mathbb{P}(Tle t)tag{2}.
end{align}
Next, note that
$$
mathbb{P}(W_t>y|W_0=x,Tle t),mathbb{P}(Tle t)=mathbb{P}(W_t>y,Tle t|W_0=x)
$$
denotes the probability of a Brownian motion that starts from $x$, hits $0$ at time $T$, and then goes beyond $y$ at time $t$. By the reflection principle, it follows that
$$
mathbb{P}(W_t>y|W_0=x,Tle t),mathbb{P}(Tle t)=mathbb{P}(W_t<-y|W_0=x).
$$
Substitute this result into $(2)$, and we obtain
begin{align}
mathbb{P}(W_{twedge T}le y|W_0=x)&=mathbb{P}(W_tle y|W_0=x)\
&+mathbb{P}(W_t<-y|W_0=x).tag{3}
end{align}
Thereafter, by $W_t'=W_t-x+y$, we have
begin{align}
&mathbb{P}(W_{twedge T}le y|W_0=x)\
&=mathbb{P}(W_t'le 2y-x|W_0'=y)+mathbb{P}(W_t'<-x|W_0'=y)\
&=mathbb{P}(W_t'ge x|W_0'=y)+mathbb{P}(W_t'<-x|W_0'=y),tag{4}
end{align}
where $mathbb{P}(W_t'le 2y-x|W_0'=y)=mathbb{P}(W_t'ge x|W_0'=y)$ due to the fact that $2y-x$ and $x$ are symmetric with respect to $y$.
Finally, note that $x>0$, and $(4)$ is obviously the desired equality
$$
mathbb{P}(W_{twedge T}le y|W_0=x)=mathbb{P}(left|W_t'right|ge x|W_0'=y).
$$
This completes the proof.
answered Jan 24 at 9:25
hypernovahypernova
4,834414
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
Let's write the assertion in term of a standard Brownian motion $(B_t)_{t geq 0}$.
If we set $T_{x} = inf{t geq 0; B_t = -x}$ then we need to show that
$$mathbb{P}(x+ B_{t wedge T_{x}} leq y) = mathbb{P}(|B_t+y| geq x) tag{1}$$
for any $x,y,t>0$.
Fix $x,y,t>0$. Since
$$B_{t wedge T_x} = begin{cases} -x & T_x leq t, \ B_t, & T_x>t end{cases}$$
we have
$$mathbb{P}(x+B_{t wedge T_x} leq y) = mathbb{P}(T_x leq t) + mathbb{P}(x+B_t leq y, T_x>t). tag{2}$$
Since the reflection principle allows us to calculate explicitly the joint distribution $(B_t,inf_{s leq t} B_s)$ (see below), we get
begin{align*}mathbb{P}(x+B_t leq y, T_x>t) &= mathbb{P} left( B_t leq y-x, inf_{s leq t} B_s>-x right) \ &= - frac{2}{sqrt{2pi t^3}} int_{-x}^0 int_{u leq z leq y-x} (2u-z) exp left(- frac{(2u-z)^2}{2t} right) , dz , du \ &= frac{2}{sqrt{2pi t}} int_{-x}^0 left[-exp left(- frac{(2u-y+x)^2}{2t} right) + exp left( -frac{u^2}{2t} right)right] , du.end{align*}
Using the linearity of the integral to split the integral on the right-hand side into two parts and performing a simple change of variables we get
$$mathbb{P}(x+B_t leq y, T_x>t) = sqrt{frac{2}{pi t}} int_{mathbb{R}} left(- frac{1}{2} cdot 1_{{|u-y| leq x}} + 1_{[-x,0]}(u) right) exp left(- frac{u^2}{2t} right) , du.$$
On the other hand, the reflection principle also gives
$$mathbb{P}(T_x leq t) =2 mathbb{P}(B_t' leq -x) =sqrt{frac{2}{pi t}} int_{-infty}^{-x} exp left( - frac{u^2}{2t} right) , du tag{3}$$
Adding both expressions and using that
$$sqrt{frac{2}{pi t}} int_{-infty}^0 exp left(- frac{u^2}{2t} right) , du = frac{1}{sqrt{2pi t}} int_{mathbb{R}} exp left(- frac{u^2}{2t} right) , du = 1,$$
it follows from $(2)$ that
$$begin{align*} mathbb{P}(x+B_{t wedge T_x} leq y) = 1- frac{1}{sqrt{2pi t}} int_{|z-y| leq x} exp left( - frac{u^2}{2t} right) , du&= mathbb{P}(|B_t+y|>x) \ &= mathbb{P}(|B_t+y| geq x) end{align*}$$
which proves the assertion.
Some remarks on the reflection principle: The reflection principle states (...or implies, depending on the formulation...) that the running supremum $M_t := sup_{s leq t} B_s$ satisfies $$M_t stackrel{d}{=} |B_t|.$$ Applying this result to the Brownian motion $(-B_t)_{t geq 0}$ it follows immediately that $$- inf_{s leq t} B_s sim |B_t|.$$ In particular, $$mathbb{P}(T_x leq t) = mathbb{P} left( min_{s leq t} B_s leq x right) = mathbb{P}(|B_t| leq -x) = 2 mathbb{P}(B_t leq x)$$ for the stopping time $$T_x := inf{t geq 0; B_t = -x}$$ from the first part of this answer. Combining the reflection principle with the strong Markov property of Brownian motion, it is possible to compute the joint density of $(B_t,M_t)$:
Let $(B_t)_{t geq 0}$ be a Brownian motion. The joint distribution $(B_t,sup_{s leq t} B_s)$ equals $$mathbb{P} left[ B_t in dx, sup_{s leq t} B_s in dy right] = frac{2 (2y-x)}{sqrt{2pi t^3}} exp left(- frac{(2y-x)^2}{2t} right) 1_{[-infty,y]}(x) , dx , dy. $$
For a proof see e.g. René Schilling/Lothar Partzsch: Brownian Motion - An Introduction to Stochastic Processes, Exercise 6.8 (there are full solutions available on the web), or this question (and differentiate in order to obtain the density function).
As before, we can use the symmetry of Brownian motion and apply the result to $(-B_t)_{t geq 0}$ in order to get the joint distribution of $(B_t,inf_{s leq t} B_s)$.
answered Jan 24 at 11:12
sazsaz
81.5k861127
$endgroup$
$begingroup$
I have problems understanding the reflection principle. I do not see how $P(T_xle t)=2P(B_t'le-x)$ does hold. Also I do not understand how to calculate the joint distribution, since there is an infimum involved. I only can rewrite $P(inf B_s>-x)=P(sup B_s<x)=1-2P(B_t>x)$ with the reflection principle? But how do I conclude from there on? Thank you already for your very good answer!
$endgroup$
– user408858
Jan 24 at 19:59
$begingroup$
@user408858 Which version of the reflection principle do you know...?
$endgroup$
– saz
Jan 24 at 20:18
$begingroup$
I know, that $B_t'=B_{twedge T}-(B_t-B_{twedge T})$ is also a brownian motion. I was also looking up the article on wikipedia where it is stated, that $P(sup_{sin[0,t]} B_sge a)=2P(B_tge a)$, see en.wikipedia.org/wiki/Reflection_principle_(Wiener_process)
$endgroup$
– user408858
Jan 24 at 20:25
$begingroup$
@user408858 I see... well, I will add some comments on that tomorrow...
$endgroup$
– saz
Jan 24 at 20:27
$begingroup$
Well, I think I already can answer my first question: $P(T_xle t)=P(inf B_sle -x)=2P(xle B_t)$
$endgroup$
– user408858
Jan 24 at 20:50
|
show 3 more comments
$begingroup$
Let's write the assertion in term of a standard Brownian motion $(B_t)_{t geq 0}$.
If we set $T_{x} = inf{t geq 0; B_t = -x}$ then we need to show that
$$mathbb{P}(x+ B_{t wedge T_{x}} leq y) = mathbb{P}(|B_t+y| geq x) tag{1}$$
for any $x,y,t>0$.
Fix $x,y,t>0$. Since
$$B_{t wedge T_x} = begin{cases} -x & T_x leq t, \ B_t, & T_x>t end{cases}$$
we have
$$mathbb{P}(x+B_{t wedge T_x} leq y) = mathbb{P}(T_x leq t) + mathbb{P}(x+B_t leq y, T_x>t). tag{2}$$
Since the reflection principle allows us to calculate explicitly the joint distribution $(B_t,inf_{s leq t} B_s)$ (see below), we get
begin{align*}mathbb{P}(x+B_t leq y, T_x>t) &= mathbb{P} left( B_t leq y-x, inf_{s leq t} B_s>-x right) \ &= - frac{2}{sqrt{2pi t^3}} int_{-x}^0 int_{u leq z leq y-x} (2u-z) exp left(- frac{(2u-z)^2}{2t} right) , dz , du \ &= frac{2}{sqrt{2pi t}} int_{-x}^0 left[-exp left(- frac{(2u-y+x)^2}{2t} right) + exp left( -frac{u^2}{2t} right)right] , du.end{align*}
Using the linearity of the integral to split the integral on the right-hand side into two parts and performing a simple change of variables we get
$$mathbb{P}(x+B_t leq y, T_x>t) = sqrt{frac{2}{pi t}} int_{mathbb{R}} left(- frac{1}{2} cdot 1_{{|u-y| leq x}} + 1_{[-x,0]}(u) right) exp left(- frac{u^2}{2t} right) , du.$$
On the other hand, the reflection principle also gives
$$mathbb{P}(T_x leq t) =2 mathbb{P}(B_t' leq -x) =sqrt{frac{2}{pi t}} int_{-infty}^{-x} exp left( - frac{u^2}{2t} right) , du tag{3}$$
Adding both expressions and using that
$$sqrt{frac{2}{pi t}} int_{-infty}^0 exp left(- frac{u^2}{2t} right) , du = frac{1}{sqrt{2pi t}} int_{mathbb{R}} exp left(- frac{u^2}{2t} right) , du = 1,$$
it follows from $(2)$ that
$$begin{align*} mathbb{P}(x+B_{t wedge T_x} leq y) = 1- frac{1}{sqrt{2pi t}} int_{|z-y| leq x} exp left( - frac{u^2}{2t} right) , du&= mathbb{P}(|B_t+y|>x) \ &= mathbb{P}(|B_t+y| geq x) end{align*}$$
which proves the assertion.
Some remarks on the reflection principle: The reflection principle states (...or implies, depending on the formulation...) that the running supremum $M_t := sup_{s leq t} B_s$ satisfies $$M_t stackrel{d}{=} |B_t|.$$ Applying this result to the Brownian motion $(-B_t)_{t geq 0}$ it follows immediately that $$- inf_{s leq t} B_s sim |B_t|.$$ In particular, $$mathbb{P}(T_x leq t) = mathbb{P} left( min_{s leq t} B_s leq x right) = mathbb{P}(|B_t| leq -x) = 2 mathbb{P}(B_t leq x)$$ for the stopping time $$T_x := inf{t geq 0; B_t = -x}$$ from the first part of this answer. Combining the reflection principle with the strong Markov property of Brownian motion, it is possible to compute the joint density of $(B_t,M_t)$:
Let $(B_t)_{t geq 0}$ be a Brownian motion. The joint distribution $(B_t,sup_{s leq t} B_s)$ equals $$mathbb{P} left[ B_t in dx, sup_{s leq t} B_s in dy right] = frac{2 (2y-x)}{sqrt{2pi t^3}} exp left(- frac{(2y-x)^2}{2t} right) 1_{[-infty,y]}(x) , dx , dy. $$
For a proof see e.g. René Schilling/Lothar Partzsch: Brownian Motion - An Introduction to Stochastic Processes, Exercise 6.8 (there are full solutions available on the web), or this question (and differentiate in order to obtain the density function).
As before, we can use the symmetry of Brownian motion and apply the result to $(-B_t)_{t geq 0}$ in order to get the joint distribution of $(B_t,inf_{s leq t} B_s)$.
answered Jan 24 at 11:12
sazsaz
81.5k861127
$endgroup$
$begingroup$
I have problems understanding the reflection principle. I do not see how $P(T_xle t)=2P(B_t'le-x)$ does hold. Also I do not understand how to calculate the joint distribution, since there is an infimum involved. I only can rewrite $P(inf B_s>-x)=P(sup B_s<x)=1-2P(B_t>x)$ with the reflection principle? But how do I conclude from there on? Thank you already for your very good answer!
$endgroup$
– user408858
Jan 24 at 19:59
$begingroup$
@user408858 Which version of the reflection principle do you know...?
$endgroup$
– saz
Jan 24 at 20:18
$begingroup$
I know, that $B_t'=B_{twedge T}-(B_t-B_{twedge T})$ is also a brownian motion. I was also looking up the article on wikipedia where it is stated, that $P(sup_{sin[0,t]} B_sge a)=2P(B_tge a)$, see en.wikipedia.org/wiki/Reflection_principle_(Wiener_process)
$endgroup$
– user408858
Jan 24 at 20:25
$begingroup$
@user408858 I see... well, I will add some comments on that tomorrow...
$endgroup$
– saz
Jan 24 at 20:27
$begingroup$
Well, I think I already can answer my first question: $P(T_xle t)=P(inf B_sle -x)=2P(xle B_t)$
$endgroup$
– user408858
Jan 24 at 20:50
|
show 3 more comments
$begingroup$
Let's write the assertion in term of a standard Brownian motion $(B_t)_{t geq 0}$.
If we set $T_{x} = inf{t geq 0; B_t = -x}$ then we need to show that
$$mathbb{P}(x+ B_{t wedge T_{x}} leq y) = mathbb{P}(|B_t+y| geq x) tag{1}$$
for any $x,y,t>0$.
Fix $x,y,t>0$. Since
$$B_{t wedge T_x} = begin{cases} -x & T_x leq t, \ B_t, & T_x>t end{cases}$$
we have
$$mathbb{P}(x+B_{t wedge T_x} leq y) = mathbb{P}(T_x leq t) + mathbb{P}(x+B_t leq y, T_x>t). tag{2}$$
Since the reflection principle allows us to calculate explicitly the joint distribution $(B_t,inf_{s leq t} B_s)$ (see below), we get
begin{align*}mathbb{P}(x+B_t leq y, T_x>t) &= mathbb{P} left( B_t leq y-x, inf_{s leq t} B_s>-x right) \ &= - frac{2}{sqrt{2pi t^3}} int_{-x}^0 int_{u leq z leq y-x} (2u-z) exp left(- frac{(2u-z)^2}{2t} right) , dz , du \ &= frac{2}{sqrt{2pi t}} int_{-x}^0 left[-exp left(- frac{(2u-y+x)^2}{2t} right) + exp left( -frac{u^2}{2t} right)right] , du.end{align*}
Using the linearity of the integral to split the integral on the right-hand side into two parts and performing a simple change of variables we get
$$mathbb{P}(x+B_t leq y, T_x>t) = sqrt{frac{2}{pi t}} int_{mathbb{R}} left(- frac{1}{2} cdot 1_{{|u-y| leq x}} + 1_{[-x,0]}(u) right) exp left(- frac{u^2}{2t} right) , du.$$
On the other hand, the reflection principle also gives
$$mathbb{P}(T_x leq t) =2 mathbb{P}(B_t' leq -x) =sqrt{frac{2}{pi t}} int_{-infty}^{-x} exp left( - frac{u^2}{2t} right) , du tag{3}$$
Adding both expressions and using that
$$sqrt{frac{2}{pi t}} int_{-infty}^0 exp left(- frac{u^2}{2t} right) , du = frac{1}{sqrt{2pi t}} int_{mathbb{R}} exp left(- frac{u^2}{2t} right) , du = 1,$$
it follows from $(2)$ that
$$begin{align*} mathbb{P}(x+B_{t wedge T_x} leq y) = 1- frac{1}{sqrt{2pi t}} int_{|z-y| leq x} exp left( - frac{u^2}{2t} right) , du&= mathbb{P}(|B_t+y|>x) \ &= mathbb{P}(|B_t+y| geq x) end{align*}$$
which proves the assertion.
Some remarks on the reflection principle: The reflection principle states (...or implies, depending on the formulation...) that the running supremum $M_t := sup_{s leq t} B_s$ satisfies $$M_t stackrel{d}{=} |B_t|.$$ Applying this result to the Brownian motion $(-B_t)_{t geq 0}$ it follows immediately that $$- inf_{s leq t} B_s sim |B_t|.$$ In particular, $$mathbb{P}(T_x leq t) = mathbb{P} left( min_{s leq t} B_s leq x right) = mathbb{P}(|B_t| leq -x) = 2 mathbb{P}(B_t leq x)$$ for the stopping time $$T_x := inf{t geq 0; B_t = -x}$$ from the first part of this answer. Combining the reflection principle with the strong Markov property of Brownian motion, it is possible to compute the joint density of $(B_t,M_t)$:
Let $(B_t)_{t geq 0}$ be a Brownian motion. The joint distribution $(B_t,sup_{s leq t} B_s)$ equals $$mathbb{P} left[ B_t in dx, sup_{s leq t} B_s in dy right] = frac{2 (2y-x)}{sqrt{2pi t^3}} exp left(- frac{(2y-x)^2}{2t} right) 1_{[-infty,y]}(x) , dx , dy. $$
For a proof see e.g. René Schilling/Lothar Partzsch: Brownian Motion - An Introduction to Stochastic Processes, Exercise 6.8 (there are full solutions available on the web), or this question (and differentiate in order to obtain the density function).
As before, we can use the symmetry of Brownian motion and apply the result to $(-B_t)_{t geq 0}$ in order to get the joint distribution of $(B_t,inf_{s leq t} B_s)$.
answered Jan 24 at 11:12
sazsaz
81.5k861127
$endgroup$
Let's write the assertion in term of a standard Brownian motion $(B_t)_{t geq 0}$.
If we set $T_{x} = inf{t geq 0; B_t = -x}$ then we need to show that
$$mathbb{P}(x+ B_{t wedge T_{x}} leq y) = mathbb{P}(|B_t+y| geq x) tag{1}$$
for any $x,y,t>0$.
Fix $x,y,t>0$. Since
$$B_{t wedge T_x} = begin{cases} -x & T_x leq t, \ B_t, & T_x>t end{cases}$$
we have
$$mathbb{P}(x+B_{t wedge T_x} leq y) = mathbb{P}(T_x leq t) + mathbb{P}(x+B_t leq y, T_x>t). tag{2}$$
Since the reflection principle allows us to calculate explicitly the joint distribution $(B_t,inf_{s leq t} B_s)$ (see below), we get
begin{align*}mathbb{P}(x+B_t leq y, T_x>t) &= mathbb{P} left( B_t leq y-x, inf_{s leq t} B_s>-x right) \ &= - frac{2}{sqrt{2pi t^3}} int_{-x}^0 int_{u leq z leq y-x} (2u-z) exp left(- frac{(2u-z)^2}{2t} right) , dz , du \ &= frac{2}{sqrt{2pi t}} int_{-x}^0 left[-exp left(- frac{(2u-y+x)^2}{2t} right) + exp left( -frac{u^2}{2t} right)right] , du.end{align*}
Using the linearity of the integral to split the integral on the right-hand side into two parts and performing a simple change of variables we get
$$mathbb{P}(x+B_t leq y, T_x>t) = sqrt{frac{2}{pi t}} int_{mathbb{R}} left(- frac{1}{2} cdot 1_{{|u-y| leq x}} + 1_{[-x,0]}(u) right) exp left(- frac{u^2}{2t} right) , du.$$
On the other hand, the reflection principle also gives
$$mathbb{P}(T_x leq t) =2 mathbb{P}(B_t' leq -x) =sqrt{frac{2}{pi t}} int_{-infty}^{-x} exp left( - frac{u^2}{2t} right) , du tag{3}$$
Adding both expressions and using that
$$sqrt{frac{2}{pi t}} int_{-infty}^0 exp left(- frac{u^2}{2t} right) , du = frac{1}{sqrt{2pi t}} int_{mathbb{R}} exp left(- frac{u^2}{2t} right) , du = 1,$$
it follows from $(2)$ that
$$begin{align*} mathbb{P}(x+B_{t wedge T_x} leq y) = 1- frac{1}{sqrt{2pi t}} int_{|z-y| leq x} exp left( - frac{u^2}{2t} right) , du&= mathbb{P}(|B_t+y|>x) \ &= mathbb{P}(|B_t+y| geq x) end{align*}$$
which proves the assertion.
Some remarks on the reflection principle: The reflection principle states (...or implies, depending on the formulation...) that the running supremum $M_t := sup_{s leq t} B_s$ satisfies $$M_t stackrel{d}{=} |B_t|.$$ Applying this result to the Brownian motion $(-B_t)_{t geq 0}$ it follows immediately that $$- inf_{s leq t} B_s sim |B_t|.$$ In particular, $$mathbb{P}(T_x leq t) = mathbb{P} left( min_{s leq t} B_s leq x right) = mathbb{P}(|B_t| leq -x) = 2 mathbb{P}(B_t leq x)$$ for the stopping time $$T_x := inf{t geq 0; B_t = -x}$$ from the first part of this answer. Combining the reflection principle with the strong Markov property of Brownian motion, it is possible to compute the joint density of $(B_t,M_t)$:
Let $(B_t)_{t geq 0}$ be a Brownian motion. The joint distribution $(B_t,sup_{s leq t} B_s)$ equals $$mathbb{P} left[ B_t in dx, sup_{s leq t} B_s in dy right] = frac{2 (2y-x)}{sqrt{2pi t^3}} exp left(- frac{(2y-x)^2}{2t} right) 1_{[-infty,y]}(x) , dx , dy. $$
For a proof see e.g. René Schilling/Lothar Partzsch: Brownian Motion - An Introduction to Stochastic Processes, Exercise 6.8 (there are full solutions available on the web), or this question (and differentiate in order to obtain the density function).
As before, we can use the symmetry of Brownian motion and apply the result to $(-B_t)_{t geq 0}$ in order to get the joint distribution of $(B_t,inf_{s leq t} B_s)$.
answered Jan 24 at 11:12
sazsaz
81.5k861127
edited Jan 25 at 7:50
answered Jan 24 at 11:12
sazsaz
81.5k861127
answered Jan 24 at 11:12
sazsaz
81.5k861127
answered Jan 24 at 11:12
sazsaz
81.5k861127
81.5k861127
$begingroup$
I have problems understanding the reflection principle. I do not see how $P(T_xle t)=2P(B_t'le-x)$ does hold. Also I do not understand how to calculate the joint distribution, since there is an infimum involved. I only can rewrite $P(inf B_s>-x)=P(sup B_s<x)=1-2P(B_t>x)$ with the reflection principle? But how do I conclude from there on? Thank you already for your very good answer!
$endgroup$
– user408858
Jan 24 at 19:59
$begingroup$
@user408858 Which version of the reflection principle do you know...?
$endgroup$
– saz
Jan 24 at 20:18
$begingroup$
I know, that $B_t'=B_{twedge T}-(B_t-B_{twedge T})$ is also a brownian motion. I was also looking up the article on wikipedia where it is stated, that $P(sup_{sin[0,t]} B_sge a)=2P(B_tge a)$, see en.wikipedia.org/wiki/Reflection_principle_(Wiener_process)
$endgroup$
– user408858
Jan 24 at 20:25
$begingroup$
@user408858 I see... well, I will add some comments on that tomorrow...
$endgroup$
– saz
Jan 24 at 20:27
$begingroup$
Well, I think I already can answer my first question: $P(T_xle t)=P(inf B_sle -x)=2P(xle B_t)$
$endgroup$
– user408858
Jan 24 at 20:50
|
show 3 more comments
$begingroup$
I have problems understanding the reflection principle. I do not see how $P(T_xle t)=2P(B_t'le-x)$ does hold. Also I do not understand how to calculate the joint distribution, since there is an infimum involved. I only can rewrite $P(inf B_s>-x)=P(sup B_s<x)=1-2P(B_t>x)$ with the reflection principle? But how do I conclude from there on? Thank you already for your very good answer!
$endgroup$
– user408858
Jan 24 at 19:59
$begingroup$
@user408858 Which version of the reflection principle do you know...?
$endgroup$
– saz
Jan 24 at 20:18
$begingroup$
I know, that $B_t'=B_{twedge T}-(B_t-B_{twedge T})$ is also a brownian motion. I was also looking up the article on wikipedia where it is stated, that $P(sup_{sin[0,t]} B_sge a)=2P(B_tge a)$, see en.wikipedia.org/wiki/Reflection_principle_(Wiener_process)
$endgroup$
– user408858
Jan 24 at 20:25
$begingroup$
@user408858 I see... well, I will add some comments on that tomorrow...
$endgroup$
– saz
Jan 24 at 20:27
$begingroup$
Well, I think I already can answer my first question: $P(T_xle t)=P(inf B_sle -x)=2P(xle B_t)$
$endgroup$
– user408858
Jan 24 at 20:50
$begingroup$
I have problems understanding the reflection principle. I do not see how $P(T_xle t)=2P(B_t'le-x)$ does hold. Also I do not understand how to calculate the joint distribution, since there is an infimum involved. I only can rewrite $P(inf B_s>-x)=P(sup B_s<x)=1-2P(B_t>x)$ with the reflection principle? But how do I conclude from there on? Thank you already for your very good answer!
$endgroup$
– user408858
Jan 24 at 19:59
$begingroup$
I have problems understanding the reflection principle. I do not see how $P(T_xle t)=2P(B_t'le-x)$ does hold. Also I do not understand how to calculate the joint distribution, since there is an infimum involved. I only can rewrite $P(inf B_s>-x)=P(sup B_s<x)=1-2P(B_t>x)$ with the reflection principle? But how do I conclude from there on? Thank you already for your very good answer!
$endgroup$
– user408858
Jan 24 at 19:59
$begingroup$
@user408858 Which version of the reflection principle do you know...?
$endgroup$
– saz
Jan 24 at 20:18
$begingroup$
@user408858 Which version of the reflection principle do you know...?
$endgroup$
– saz
Jan 24 at 20:18
$begingroup$
I know, that $B_t'=B_{twedge T}-(B_t-B_{twedge T})$ is also a brownian motion. I was also looking up the article on wikipedia where it is stated, that $P(sup_{sin[0,t]} B_sge a)=2P(B_tge a)$, see en.wikipedia.org/wiki/Reflection_principle_(Wiener_process)
$endgroup$
– user408858
Jan 24 at 20:25
$begingroup$
I know, that $B_t'=B_{twedge T}-(B_t-B_{twedge T})$ is also a brownian motion. I was also looking up the article on wikipedia where it is stated, that $P(sup_{sin[0,t]} B_sge a)=2P(B_tge a)$, see en.wikipedia.org/wiki/Reflection_principle_(Wiener_process)
$endgroup$
– user408858
Jan 24 at 20:25
$begingroup$
@user408858 I see... well, I will add some comments on that tomorrow...
$endgroup$
– saz
Jan 24 at 20:27
$begingroup$
@user408858 I see... well, I will add some comments on that tomorrow...
$endgroup$
– saz
Jan 24 at 20:27
$begingroup$
Well, I think I already can answer my first question: $P(T_xle t)=P(inf B_sle -x)=2P(xle B_t)$
$endgroup$
– user408858
Jan 24 at 20:50
$begingroup$
Well, I think I already can answer my first question: $P(T_xle t)=P(inf B_sle -x)=2P(xle B_t)$
$endgroup$
– user408858
Jan 24 at 20:50
|
show 3 more comments
$begingroup$
Perhaps we may go for the proof as follows.
First, note that
begin{align}
&mathbb{P}(W_{twedge T}le y|W_0=x)\
&=mathbb{P}(W_tle y|W_0=x,T>t),mathbb{P}(T>t)\
&+mathbb{P}(W_Tle y|W_0=x,Tle t),mathbb{P}(Tle t)\
&=mathbb{P}(W_tle y|W_0=x,T>t),mathbb{P}(T>t)+mathbb{P}(Tle t)tag{1},
end{align}
where $mathbb{P}(W_Tle y|W_0=x,Tle t)=1$ as $W_T=0<y$ always holds.
Second, note that
begin{align}
mathbb{P}(W_tle y|W_0=x)&=mathbb{P}(W_tle y|W_0=x,T>t),mathbb{P}(T>t)\
&+mathbb{P}(W_tle y|W_0=x,Tle t),mathbb{P}(Tle t).
end{align}
Substitute this result into $(1)$, and we obtain
begin{align}
&mathbb{P}(W_{twedge T}le y|W_0=x)\
&=mathbb{P}(W_tle y|W_0=x)+mathbb{P}(Tle t)\
&-mathbb{P}(W_tle y|W_0=x,Tle t),mathbb{P}(Tle t)\
&=mathbb{P}(W_tle y|W_0=x)\
&+mathbb{P}(W_t>y|W_0=x,Tle t),mathbb{P}(Tle t)tag{2}.
end{align}
Next, note that
$$
mathbb{P}(W_t>y|W_0=x,Tle t),mathbb{P}(Tle t)=mathbb{P}(W_t>y,Tle t|W_0=x)
$$
denotes the probability of a Brownian motion that starts from $x$, hits $0$ at time $T$, and then goes beyond $y$ at time $t$. By the reflection principle, it follows that
$$
mathbb{P}(W_t>y|W_0=x,Tle t),mathbb{P}(Tle t)=mathbb{P}(W_t<-y|W_0=x).
$$
Substitute this result into $(2)$, and we obtain
begin{align}
mathbb{P}(W_{twedge T}le y|W_0=x)&=mathbb{P}(W_tle y|W_0=x)\
&+mathbb{P}(W_t<-y|W_0=x).tag{3}
end{align}
Thereafter, by $W_t'=W_t-x+y$, we have
begin{align}
&mathbb{P}(W_{twedge T}le y|W_0=x)\
&=mathbb{P}(W_t'le 2y-x|W_0'=y)+mathbb{P}(W_t'<-x|W_0'=y)\
&=mathbb{P}(W_t'ge x|W_0'=y)+mathbb{P}(W_t'<-x|W_0'=y),tag{4}
end{align}
where $mathbb{P}(W_t'le 2y-x|W_0'=y)=mathbb{P}(W_t'ge x|W_0'=y)$ due to the fact that $2y-x$ and $x$ are symmetric with respect to $y$.
Finally, note that $x>0$, and $(4)$ is obviously the desired equality
$$
mathbb{P}(W_{twedge T}le y|W_0=x)=mathbb{P}(left|W_t'right|ge x|W_0'=y).
$$
This completes the proof.
answered Jan 24 at 9:25
hypernovahypernova
4,834414
$endgroup$
add a comment |
$begingroup$
Perhaps we may go for the proof as follows.
First, note that
begin{align}
&mathbb{P}(W_{twedge T}le y|W_0=x)\
&=mathbb{P}(W_tle y|W_0=x,T>t),mathbb{P}(T>t)\
&+mathbb{P}(W_Tle y|W_0=x,Tle t),mathbb{P}(Tle t)\
&=mathbb{P}(W_tle y|W_0=x,T>t),mathbb{P}(T>t)+mathbb{P}(Tle t)tag{1},
end{align}
where $mathbb{P}(W_Tle y|W_0=x,Tle t)=1$ as $W_T=0<y$ always holds.
Second, note that
begin{align}
mathbb{P}(W_tle y|W_0=x)&=mathbb{P}(W_tle y|W_0=x,T>t),mathbb{P}(T>t)\
&+mathbb{P}(W_tle y|W_0=x,Tle t),mathbb{P}(Tle t).
end{align}
Substitute this result into $(1)$, and we obtain
begin{align}
&mathbb{P}(W_{twedge T}le y|W_0=x)\
&=mathbb{P}(W_tle y|W_0=x)+mathbb{P}(Tle t)\
&-mathbb{P}(W_tle y|W_0=x,Tle t),mathbb{P}(Tle t)\
&=mathbb{P}(W_tle y|W_0=x)\
&+mathbb{P}(W_t>y|W_0=x,Tle t),mathbb{P}(Tle t)tag{2}.
end{align}
Next, note that
$$
mathbb{P}(W_t>y|W_0=x,Tle t),mathbb{P}(Tle t)=mathbb{P}(W_t>y,Tle t|W_0=x)
$$
denotes the probability of a Brownian motion that starts from $x$, hits $0$ at time $T$, and then goes beyond $y$ at time $t$. By the reflection principle, it follows that
$$
mathbb{P}(W_t>y|W_0=x,Tle t),mathbb{P}(Tle t)=mathbb{P}(W_t<-y|W_0=x).
$$
Substitute this result into $(2)$, and we obtain
begin{align}
mathbb{P}(W_{twedge T}le y|W_0=x)&=mathbb{P}(W_tle y|W_0=x)\
&+mathbb{P}(W_t<-y|W_0=x).tag{3}
end{align}
Thereafter, by $W_t'=W_t-x+y$, we have
begin{align}
&mathbb{P}(W_{twedge T}le y|W_0=x)\
&=mathbb{P}(W_t'le 2y-x|W_0'=y)+mathbb{P}(W_t'<-x|W_0'=y)\
&=mathbb{P}(W_t'ge x|W_0'=y)+mathbb{P}(W_t'<-x|W_0'=y),tag{4}
end{align}
where $mathbb{P}(W_t'le 2y-x|W_0'=y)=mathbb{P}(W_t'ge x|W_0'=y)$ due to the fact that $2y-x$ and $x$ are symmetric with respect to $y$.
Finally, note that $x>0$, and $(4)$ is obviously the desired equality
$$
mathbb{P}(W_{twedge T}le y|W_0=x)=mathbb{P}(left|W_t'right|ge x|W_0'=y).
$$
This completes the proof.
answered Jan 24 at 9:25
hypernovahypernova
4,834414
$endgroup$
add a comment |
$begingroup$
Perhaps we may go for the proof as follows.
First, note that
begin{align}
&mathbb{P}(W_{twedge T}le y|W_0=x)\
&=mathbb{P}(W_tle y|W_0=x,T>t),mathbb{P}(T>t)\
&+mathbb{P}(W_Tle y|W_0=x,Tle t),mathbb{P}(Tle t)\
&=mathbb{P}(W_tle y|W_0=x,T>t),mathbb{P}(T>t)+mathbb{P}(Tle t)tag{1},
end{align}
where $mathbb{P}(W_Tle y|W_0=x,Tle t)=1$ as $W_T=0<y$ always holds.
Second, note that
begin{align}
mathbb{P}(W_tle y|W_0=x)&=mathbb{P}(W_tle y|W_0=x,T>t),mathbb{P}(T>t)\
&+mathbb{P}(W_tle y|W_0=x,Tle t),mathbb{P}(Tle t).
end{align}
Substitute this result into $(1)$, and we obtain
begin{align}
&mathbb{P}(W_{twedge T}le y|W_0=x)\
&=mathbb{P}(W_tle y|W_0=x)+mathbb{P}(Tle t)\
&-mathbb{P}(W_tle y|W_0=x,Tle t),mathbb{P}(Tle t)\
&=mathbb{P}(W_tle y|W_0=x)\
&+mathbb{P}(W_t>y|W_0=x,Tle t),mathbb{P}(Tle t)tag{2}.
end{align}
Next, note that
$$
mathbb{P}(W_t>y|W_0=x,Tle t),mathbb{P}(Tle t)=mathbb{P}(W_t>y,Tle t|W_0=x)
$$
denotes the probability of a Brownian motion that starts from $x$, hits $0$ at time $T$, and then goes beyond $y$ at time $t$. By the reflection principle, it follows that
$$
mathbb{P}(W_t>y|W_0=x,Tle t),mathbb{P}(Tle t)=mathbb{P}(W_t<-y|W_0=x).
$$
Substitute this result into $(2)$, and we obtain
begin{align}
mathbb{P}(W_{twedge T}le y|W_0=x)&=mathbb{P}(W_tle y|W_0=x)\
&+mathbb{P}(W_t<-y|W_0=x).tag{3}
end{align}
Thereafter, by $W_t'=W_t-x+y$, we have
begin{align}
&mathbb{P}(W_{twedge T}le y|W_0=x)\
&=mathbb{P}(W_t'le 2y-x|W_0'=y)+mathbb{P}(W_t'<-x|W_0'=y)\
&=mathbb{P}(W_t'ge x|W_0'=y)+mathbb{P}(W_t'<-x|W_0'=y),tag{4}
end{align}
where $mathbb{P}(W_t'le 2y-x|W_0'=y)=mathbb{P}(W_t'ge x|W_0'=y)$ due to the fact that $2y-x$ and $x$ are symmetric with respect to $y$.
Finally, note that $x>0$, and $(4)$ is obviously the desired equality
$$
mathbb{P}(W_{twedge T}le y|W_0=x)=mathbb{P}(left|W_t'right|ge x|W_0'=y).
$$
This completes the proof.
answered Jan 24 at 9:25
hypernovahypernova
4,834414
$endgroup$
Perhaps we may go for the proof as follows.
First, note that
begin{align}
&mathbb{P}(W_{twedge T}le y|W_0=x)\
&=mathbb{P}(W_tle y|W_0=x,T>t),mathbb{P}(T>t)\
&+mathbb{P}(W_Tle y|W_0=x,Tle t),mathbb{P}(Tle t)\
&=mathbb{P}(W_tle y|W_0=x,T>t),mathbb{P}(T>t)+mathbb{P}(Tle t)tag{1},
end{align}
where $mathbb{P}(W_Tle y|W_0=x,Tle t)=1$ as $W_T=0<y$ always holds.
Second, note that
begin{align}
mathbb{P}(W_tle y|W_0=x)&=mathbb{P}(W_tle y|W_0=x,T>t),mathbb{P}(T>t)\
&+mathbb{P}(W_tle y|W_0=x,Tle t),mathbb{P}(Tle t).
end{align}
Substitute this result into $(1)$, and we obtain
begin{align}
&mathbb{P}(W_{twedge T}le y|W_0=x)\
&=mathbb{P}(W_tle y|W_0=x)+mathbb{P}(Tle t)\
&-mathbb{P}(W_tle y|W_0=x,Tle t),mathbb{P}(Tle t)\
&=mathbb{P}(W_tle y|W_0=x)\
&+mathbb{P}(W_t>y|W_0=x,Tle t),mathbb{P}(Tle t)tag{2}.
end{align}
Next, note that
$$
mathbb{P}(W_t>y|W_0=x,Tle t),mathbb{P}(Tle t)=mathbb{P}(W_t>y,Tle t|W_0=x)
$$
denotes the probability of a Brownian motion that starts from $x$, hits $0$ at time $T$, and then goes beyond $y$ at time $t$. By the reflection principle, it follows that
$$
mathbb{P}(W_t>y|W_0=x,Tle t),mathbb{P}(Tle t)=mathbb{P}(W_t<-y|W_0=x).
$$
Substitute this result into $(2)$, and we obtain
begin{align}
mathbb{P}(W_{twedge T}le y|W_0=x)&=mathbb{P}(W_tle y|W_0=x)\
&+mathbb{P}(W_t<-y|W_0=x).tag{3}
end{align}
Thereafter, by $W_t'=W_t-x+y$, we have
begin{align}
&mathbb{P}(W_{twedge T}le y|W_0=x)\
&=mathbb{P}(W_t'le 2y-x|W_0'=y)+mathbb{P}(W_t'<-x|W_0'=y)\
&=mathbb{P}(W_t'ge x|W_0'=y)+mathbb{P}(W_t'<-x|W_0'=y),tag{4}
end{align}
where $mathbb{P}(W_t'le 2y-x|W_0'=y)=mathbb{P}(W_t'ge x|W_0'=y)$ due to the fact that $2y-x$ and $x$ are symmetric with respect to $y$.
Finally, note that $x>0$, and $(4)$ is obviously the desired equality
$$
mathbb{P}(W_{twedge T}le y|W_0=x)=mathbb{P}(left|W_t'right|ge x|W_0'=y).
$$
This completes the proof.
answered Jan 24 at 9:25
hypernovahypernova
4,834414
answered Jan 24 at 9:25
hypernovahypernova
4,834414
answered Jan 24 at 9:25
hypernovahypernova
4,834414
answered Jan 24 at 9:25
hypernovahypernova
4,834414
4,834414
add a comment |
add a comment |
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$begingroup$
What does $t wedge T$ mean? For some $t<T$ ? I guess you assume that $x>y$...
$endgroup$
– Diger
Jan 24 at 1:48
$begingroup$
$twedge T=min(t,T)$
$endgroup$
– user408858
Jan 24 at 1:49
$begingroup$
Never seen it, thanks.
$endgroup$
– Diger
Jan 24 at 1:50
$begingroup$
Where did you encounter this problem?
$endgroup$
– saz
Jan 24 at 7:54
$begingroup$
Probably you are telling me that you need rigorous proof, but from a heuristic perspective isn't it fairly clear that this is so, since the Normal-distribution is symmetric i.e. it doesn't matter whether you go left (from x to y) or right (from y to x); they appear with equal probability?
$endgroup$
– Diger
Jan 24 at 12:55