Mixing
Supremum of Brownian motion
$begingroup$
I am trying to understand the proof in "Roger and Williams" for the Lemma
Lemma: Let $B_t$ be a Brownian motion, then$$P(sup_t B_t=+infty,inf_t B_t=-infty)=1$$
Let $Z:=sup_t B_t$, they started with observing that, for $c>0$ and by scaling property of Brownian motion, $$cZ stackrel{d}{=} Z$$. Therefore, the law of $Z$ is concentrated on $0$ or $+infty$.
Question:
- Why is $cZstackrel{d}{=}Z$? I understand that, under scaling property, $cB_{t/c^2}$ is again a Brownian motion and the law for $B_t$ and $cB_{t/c^2}$ are the same. How can I see that its also hold for their supremum rigorously?
- Why is the law of $Z$ concentrated on ${0,infty}$? Why do they have to restrict $c>0$? What goes wrong with the argument if we use $cneq 0$ instead?
stochastic-processes stochastic-calculus brownian-motion stochastic-analysis
asked Jan 21 at 13:54
quallenjägerquallenjäger
454519
$endgroup$
add a comment |
$begingroup$
I am trying to understand the proof in "Roger and Williams" for the Lemma
Lemma: Let $B_t$ be a Brownian motion, then$$P(sup_t B_t=+infty,inf_t B_t=-infty)=1$$
Let $Z:=sup_t B_t$, they started with observing that, for $c>0$ and by scaling property of Brownian motion, $$cZ stackrel{d}{=} Z$$. Therefore, the law of $Z$ is concentrated on $0$ or $+infty$.
Question:
- Why is $cZstackrel{d}{=}Z$? I understand that, under scaling property, $cB_{t/c^2}$ is again a Brownian motion and the law for $B_t$ and $cB_{t/c^2}$ are the same. How can I see that its also hold for their supremum rigorously?
- Why is the law of $Z$ concentrated on ${0,infty}$? Why do they have to restrict $c>0$? What goes wrong with the argument if we use $cneq 0$ instead?
stochastic-processes stochastic-calculus brownian-motion stochastic-analysis
asked Jan 21 at 13:54
quallenjägerquallenjäger
454519
$endgroup$
add a comment |
$begingroup$
I am trying to understand the proof in "Roger and Williams" for the Lemma
Lemma: Let $B_t$ be a Brownian motion, then$$P(sup_t B_t=+infty,inf_t B_t=-infty)=1$$
Let $Z:=sup_t B_t$, they started with observing that, for $c>0$ and by scaling property of Brownian motion, $$cZ stackrel{d}{=} Z$$. Therefore, the law of $Z$ is concentrated on $0$ or $+infty$.
Question:
- Why is $cZstackrel{d}{=}Z$? I understand that, under scaling property, $cB_{t/c^2}$ is again a Brownian motion and the law for $B_t$ and $cB_{t/c^2}$ are the same. How can I see that its also hold for their supremum rigorously?
- Why is the law of $Z$ concentrated on ${0,infty}$? Why do they have to restrict $c>0$? What goes wrong with the argument if we use $cneq 0$ instead?
stochastic-processes stochastic-calculus brownian-motion stochastic-analysis
asked Jan 21 at 13:54
quallenjägerquallenjäger
454519
$endgroup$
I am trying to understand the proof in "Roger and Williams" for the Lemma
Lemma: Let $B_t$ be a Brownian motion, then$$P(sup_t B_t=+infty,inf_t B_t=-infty)=1$$
Let $Z:=sup_t B_t$, they started with observing that, for $c>0$ and by scaling property of Brownian motion, $$cZ stackrel{d}{=} Z$$. Therefore, the law of $Z$ is concentrated on $0$ or $+infty$.
Question:
- Why is $cZstackrel{d}{=}Z$? I understand that, under scaling property, $cB_{t/c^2}$ is again a Brownian motion and the law for $B_t$ and $cB_{t/c^2}$ are the same. How can I see that its also hold for their supremum rigorously?
- Why is the law of $Z$ concentrated on ${0,infty}$? Why do they have to restrict $c>0$? What goes wrong with the argument if we use $cneq 0$ instead?
stochastic-processes stochastic-calculus brownian-motion stochastic-analysis
stochastic-processes stochastic-calculus brownian-motion stochastic-analysis
asked Jan 21 at 13:54
quallenjägerquallenjäger
454519
asked Jan 21 at 13:54
quallenjägerquallenjäger
454519
asked Jan 21 at 13:54
quallenjägerquallenjäger
454519
asked Jan 21 at 13:54
quallenjägerquallenjäger
454519
asked Jan 21 at 13:54
quallenjägerquallenjäger
454519
454519
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
For 1., consider the function $f$ that takes a continuous function on $[0, infty)$ and gives its supremum. If you already convinced yourself that $B_t$ and $tilde{B}_t = c B_{t/c^2}$ have the same law as processes then you have that $Z = f(B_cdot)$ and $tilde{Z} = f(tilde{B}_cdot)$ have the same law too. Then, for any fixed $c > 0$ we have
$$
P[Z in A] = P[tilde{Z} in A] = P[f(c B_{cdot/c^2}) in A] = P[c f(B_{cdot/c^2}) in A] = P[c Z in A],
$$
where the last equality follow from noticing that the supremum of a function with domain $[0, infty)$ is not altered by a temporal rescaling.
For 2., just notice that this property holds for all $c > 0$. Thus, for any fixed $M > 0$ if you make $c to infty$
$$
P[Z geq M] = P[Z geq M/c] = P[Z > 0],
$$
implying that $P[Z = infty] = P[Z geq M, M in mathbb{N}] = lim_M P[Z geq M] = P[Z > 0]$. Since $Z$ is supported on $[0, infty]$ we have $P[Z = 0] + P[Z = infty] = 1$. Finally, you need to argue why one of these probabilities is zero. In general you can have a random variable $Z$ on $[0, infty]$ that assumes ie, $0$ or $infty$ with probability $1/2$ and it would satisfy $Z = cZ$ in distribution for every $c > 0$. In your specific case you should use some property of the Brownian motion to justify this last step.
Finally, I think it should be clearer now why we consider $c > 0$. If we use $c < 0$ then we would get an infimum when we make the first step. This is valid, and amounts to the fact the infimum of a Brownian motion has the same distribution as the supremum.
answered Jan 21 at 14:46
DanielDaniel
1,541210
$endgroup$
$begingroup$
Why can you say $P(Z=infty)=P(Z geq 1/M,M in Bbb N)$?
$endgroup$
– quallenjäger
Jan 21 at 15:48
$begingroup$
Sorry, my mistake. I'll edit the answer to fix it. The correct is obviously $P[Z = infty] = P[Z geq M, M in mathbb{N}]$ and then you use that these events are nested.
$endgroup$
– Daniel
Jan 21 at 17:13
add a comment |
$begingroup$
Since the Brownian motion has continuous sample paths, it holds that $$Z = sup_{t geq 0} B_t > a iff exists t in (0,infty) cap mathbb{Q}: B_t>a.$$
If we denote by $(t_k)_{k in mathbb{N}}$ an enumeration of $(0,infty) cap mathbb{Q}$, this gives $$mathbb{P}(Z>a) = lim_{n to infty}mathbb{P} left( bigcup_{k=1}^n {B_{t_k}>a} right)$$ which shows that the distribution of $Z$ depends only on the finite dimensional distributions of $(B_{t_1},ldots,B_{t_n})$, $n in mathbb{N}$. Consequently, if $(W_t)_{t geq 0}$ is another process with continuous sample paths which has the same finite-dimensional distributions as $(B_t)_{t geq 0}$, then $$sup_{t geq 0} B_t stackrel{d}{=} sup_{t geq 0} W_t.$$
Applying this for $W_t := c W_{t/c^2}$ with $c>0$ we get
$$c sup_{s geq 0} B_s = sup_{t geq 0} W_t stackrel{d}{=} sup_{t geq 0} B_t$$
which proves $c Z stackrel{d}{=} Z$. Hence,
begin{align*} mathbb{E}left(1_{{Z<infty}} (1- e^{-Z}) right) = mathbb{E} left( (1-e^{-cZ}) 1_{{cZ<infty}} right) &= mathbb{E} left( (1-e^{-cZ}) 1_{{Z<infty}} right) \ &xrightarrow{c downarrow 0} 0. end{align*}
As $1-e^{-Z} geq 0$ on ${Z<infty}$ this implies $1-e^{-Z}=0$ almost surely on ${Z<infty}$, i.e. $Z=0$ on ${Z<infty}$. Consequently, we have shown that the law of $Z$ is concentrated on $0$ and $+infty$.
Regarding the 2nd part of your second question: For $c=0$ the process $(W_t)_{t geq 0}$ is clearly not well-defined; for $c<0$ we still have $cZstackrel{d}{=} Z$ (which is not surprising because of the symmetry of Brownian motion).
answered Jan 21 at 14:47
sazsaz
81.5k861127
$endgroup$
$begingroup$
Thank you for your answer! Why can I infer from $cZ stackrel{d}{=} Z$ that the law of $Z$ is concentrated on ${0,+infty}$
$endgroup$
– quallenjäger
Jan 21 at 14:53
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@quallenjäger $cZ stackrel{d}{=} Z$ implies $1_{{cZ<infty}} cZ stackrel{d}{=} Z 1_{{Z<infty}} =: tilde{Z}$. If you consider the characteristic function of $tilde{Z}$ , you will thus find that $phi(c) = phi(1)$ for all $c$, i.e. $tilde{Z}=0$.
$endgroup$
– saz
Jan 21 at 14:58
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sorry, I accepted the other answer. Your answer is great as well as always. Thank you
$endgroup$
– quallenjäger
Jan 21 at 15:34
$begingroup$
@quallenjäger YOu are welcome. I've just added some more details about infering that the law of $Z$ is concentrated on $0$ and $+infty$.
$endgroup$
– saz
Jan 21 at 17:14
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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$begingroup$
For 1., consider the function $f$ that takes a continuous function on $[0, infty)$ and gives its supremum. If you already convinced yourself that $B_t$ and $tilde{B}_t = c B_{t/c^2}$ have the same law as processes then you have that $Z = f(B_cdot)$ and $tilde{Z} = f(tilde{B}_cdot)$ have the same law too. Then, for any fixed $c > 0$ we have
$$
P[Z in A] = P[tilde{Z} in A] = P[f(c B_{cdot/c^2}) in A] = P[c f(B_{cdot/c^2}) in A] = P[c Z in A],
$$
where the last equality follow from noticing that the supremum of a function with domain $[0, infty)$ is not altered by a temporal rescaling.
For 2., just notice that this property holds for all $c > 0$. Thus, for any fixed $M > 0$ if you make $c to infty$
$$
P[Z geq M] = P[Z geq M/c] = P[Z > 0],
$$
implying that $P[Z = infty] = P[Z geq M, M in mathbb{N}] = lim_M P[Z geq M] = P[Z > 0]$. Since $Z$ is supported on $[0, infty]$ we have $P[Z = 0] + P[Z = infty] = 1$. Finally, you need to argue why one of these probabilities is zero. In general you can have a random variable $Z$ on $[0, infty]$ that assumes ie, $0$ or $infty$ with probability $1/2$ and it would satisfy $Z = cZ$ in distribution for every $c > 0$. In your specific case you should use some property of the Brownian motion to justify this last step.
Finally, I think it should be clearer now why we consider $c > 0$. If we use $c < 0$ then we would get an infimum when we make the first step. This is valid, and amounts to the fact the infimum of a Brownian motion has the same distribution as the supremum.
answered Jan 21 at 14:46
DanielDaniel
1,541210
$endgroup$
$begingroup$
Why can you say $P(Z=infty)=P(Z geq 1/M,M in Bbb N)$?
$endgroup$
– quallenjäger
Jan 21 at 15:48
$begingroup$
Sorry, my mistake. I'll edit the answer to fix it. The correct is obviously $P[Z = infty] = P[Z geq M, M in mathbb{N}]$ and then you use that these events are nested.
$endgroup$
– Daniel
Jan 21 at 17:13
add a comment |
$begingroup$
For 1., consider the function $f$ that takes a continuous function on $[0, infty)$ and gives its supremum. If you already convinced yourself that $B_t$ and $tilde{B}_t = c B_{t/c^2}$ have the same law as processes then you have that $Z = f(B_cdot)$ and $tilde{Z} = f(tilde{B}_cdot)$ have the same law too. Then, for any fixed $c > 0$ we have
$$
P[Z in A] = P[tilde{Z} in A] = P[f(c B_{cdot/c^2}) in A] = P[c f(B_{cdot/c^2}) in A] = P[c Z in A],
$$
where the last equality follow from noticing that the supremum of a function with domain $[0, infty)$ is not altered by a temporal rescaling.
For 2., just notice that this property holds for all $c > 0$. Thus, for any fixed $M > 0$ if you make $c to infty$
$$
P[Z geq M] = P[Z geq M/c] = P[Z > 0],
$$
implying that $P[Z = infty] = P[Z geq M, M in mathbb{N}] = lim_M P[Z geq M] = P[Z > 0]$. Since $Z$ is supported on $[0, infty]$ we have $P[Z = 0] + P[Z = infty] = 1$. Finally, you need to argue why one of these probabilities is zero. In general you can have a random variable $Z$ on $[0, infty]$ that assumes ie, $0$ or $infty$ with probability $1/2$ and it would satisfy $Z = cZ$ in distribution for every $c > 0$. In your specific case you should use some property of the Brownian motion to justify this last step.
Finally, I think it should be clearer now why we consider $c > 0$. If we use $c < 0$ then we would get an infimum when we make the first step. This is valid, and amounts to the fact the infimum of a Brownian motion has the same distribution as the supremum.
answered Jan 21 at 14:46
DanielDaniel
1,541210
$endgroup$
$begingroup$
Why can you say $P(Z=infty)=P(Z geq 1/M,M in Bbb N)$?
$endgroup$
– quallenjäger
Jan 21 at 15:48
$begingroup$
Sorry, my mistake. I'll edit the answer to fix it. The correct is obviously $P[Z = infty] = P[Z geq M, M in mathbb{N}]$ and then you use that these events are nested.
$endgroup$
– Daniel
Jan 21 at 17:13
add a comment |
$begingroup$
For 1., consider the function $f$ that takes a continuous function on $[0, infty)$ and gives its supremum. If you already convinced yourself that $B_t$ and $tilde{B}_t = c B_{t/c^2}$ have the same law as processes then you have that $Z = f(B_cdot)$ and $tilde{Z} = f(tilde{B}_cdot)$ have the same law too. Then, for any fixed $c > 0$ we have
$$
P[Z in A] = P[tilde{Z} in A] = P[f(c B_{cdot/c^2}) in A] = P[c f(B_{cdot/c^2}) in A] = P[c Z in A],
$$
where the last equality follow from noticing that the supremum of a function with domain $[0, infty)$ is not altered by a temporal rescaling.
For 2., just notice that this property holds for all $c > 0$. Thus, for any fixed $M > 0$ if you make $c to infty$
$$
P[Z geq M] = P[Z geq M/c] = P[Z > 0],
$$
implying that $P[Z = infty] = P[Z geq M, M in mathbb{N}] = lim_M P[Z geq M] = P[Z > 0]$. Since $Z$ is supported on $[0, infty]$ we have $P[Z = 0] + P[Z = infty] = 1$. Finally, you need to argue why one of these probabilities is zero. In general you can have a random variable $Z$ on $[0, infty]$ that assumes ie, $0$ or $infty$ with probability $1/2$ and it would satisfy $Z = cZ$ in distribution for every $c > 0$. In your specific case you should use some property of the Brownian motion to justify this last step.
Finally, I think it should be clearer now why we consider $c > 0$. If we use $c < 0$ then we would get an infimum when we make the first step. This is valid, and amounts to the fact the infimum of a Brownian motion has the same distribution as the supremum.
answered Jan 21 at 14:46
DanielDaniel
1,541210
$endgroup$
For 1., consider the function $f$ that takes a continuous function on $[0, infty)$ and gives its supremum. If you already convinced yourself that $B_t$ and $tilde{B}_t = c B_{t/c^2}$ have the same law as processes then you have that $Z = f(B_cdot)$ and $tilde{Z} = f(tilde{B}_cdot)$ have the same law too. Then, for any fixed $c > 0$ we have
$$
P[Z in A] = P[tilde{Z} in A] = P[f(c B_{cdot/c^2}) in A] = P[c f(B_{cdot/c^2}) in A] = P[c Z in A],
$$
where the last equality follow from noticing that the supremum of a function with domain $[0, infty)$ is not altered by a temporal rescaling.
For 2., just notice that this property holds for all $c > 0$. Thus, for any fixed $M > 0$ if you make $c to infty$
$$
P[Z geq M] = P[Z geq M/c] = P[Z > 0],
$$
implying that $P[Z = infty] = P[Z geq M, M in mathbb{N}] = lim_M P[Z geq M] = P[Z > 0]$. Since $Z$ is supported on $[0, infty]$ we have $P[Z = 0] + P[Z = infty] = 1$. Finally, you need to argue why one of these probabilities is zero. In general you can have a random variable $Z$ on $[0, infty]$ that assumes ie, $0$ or $infty$ with probability $1/2$ and it would satisfy $Z = cZ$ in distribution for every $c > 0$. In your specific case you should use some property of the Brownian motion to justify this last step.
Finally, I think it should be clearer now why we consider $c > 0$. If we use $c < 0$ then we would get an infimum when we make the first step. This is valid, and amounts to the fact the infimum of a Brownian motion has the same distribution as the supremum.
answered Jan 21 at 14:46
DanielDaniel
1,541210
edited Jan 21 at 17:16
answered Jan 21 at 14:46
DanielDaniel
1,541210
answered Jan 21 at 14:46
DanielDaniel
1,541210
answered Jan 21 at 14:46
DanielDaniel
1,541210
1,541210
$begingroup$
Why can you say $P(Z=infty)=P(Z geq 1/M,M in Bbb N)$?
$endgroup$
– quallenjäger
Jan 21 at 15:48
$begingroup$
Sorry, my mistake. I'll edit the answer to fix it. The correct is obviously $P[Z = infty] = P[Z geq M, M in mathbb{N}]$ and then you use that these events are nested.
$endgroup$
– Daniel
Jan 21 at 17:13
add a comment |
$begingroup$
Why can you say $P(Z=infty)=P(Z geq 1/M,M in Bbb N)$?
$endgroup$
– quallenjäger
Jan 21 at 15:48
$begingroup$
Sorry, my mistake. I'll edit the answer to fix it. The correct is obviously $P[Z = infty] = P[Z geq M, M in mathbb{N}]$ and then you use that these events are nested.
$endgroup$
– Daniel
Jan 21 at 17:13
$begingroup$
Why can you say $P(Z=infty)=P(Z geq 1/M,M in Bbb N)$?
$endgroup$
– quallenjäger
Jan 21 at 15:48
$begingroup$
Why can you say $P(Z=infty)=P(Z geq 1/M,M in Bbb N)$?
$endgroup$
– quallenjäger
Jan 21 at 15:48
$begingroup$
Sorry, my mistake. I'll edit the answer to fix it. The correct is obviously $P[Z = infty] = P[Z geq M, M in mathbb{N}]$ and then you use that these events are nested.
$endgroup$
– Daniel
Jan 21 at 17:13
$begingroup$
Sorry, my mistake. I'll edit the answer to fix it. The correct is obviously $P[Z = infty] = P[Z geq M, M in mathbb{N}]$ and then you use that these events are nested.
$endgroup$
– Daniel
Jan 21 at 17:13
add a comment |
$begingroup$
Since the Brownian motion has continuous sample paths, it holds that $$Z = sup_{t geq 0} B_t > a iff exists t in (0,infty) cap mathbb{Q}: B_t>a.$$
If we denote by $(t_k)_{k in mathbb{N}}$ an enumeration of $(0,infty) cap mathbb{Q}$, this gives $$mathbb{P}(Z>a) = lim_{n to infty}mathbb{P} left( bigcup_{k=1}^n {B_{t_k}>a} right)$$ which shows that the distribution of $Z$ depends only on the finite dimensional distributions of $(B_{t_1},ldots,B_{t_n})$, $n in mathbb{N}$. Consequently, if $(W_t)_{t geq 0}$ is another process with continuous sample paths which has the same finite-dimensional distributions as $(B_t)_{t geq 0}$, then $$sup_{t geq 0} B_t stackrel{d}{=} sup_{t geq 0} W_t.$$
Applying this for $W_t := c W_{t/c^2}$ with $c>0$ we get
$$c sup_{s geq 0} B_s = sup_{t geq 0} W_t stackrel{d}{=} sup_{t geq 0} B_t$$
which proves $c Z stackrel{d}{=} Z$. Hence,
begin{align*} mathbb{E}left(1_{{Z<infty}} (1- e^{-Z}) right) = mathbb{E} left( (1-e^{-cZ}) 1_{{cZ<infty}} right) &= mathbb{E} left( (1-e^{-cZ}) 1_{{Z<infty}} right) \ &xrightarrow{c downarrow 0} 0. end{align*}
As $1-e^{-Z} geq 0$ on ${Z<infty}$ this implies $1-e^{-Z}=0$ almost surely on ${Z<infty}$, i.e. $Z=0$ on ${Z<infty}$. Consequently, we have shown that the law of $Z$ is concentrated on $0$ and $+infty$.
Regarding the 2nd part of your second question: For $c=0$ the process $(W_t)_{t geq 0}$ is clearly not well-defined; for $c<0$ we still have $cZstackrel{d}{=} Z$ (which is not surprising because of the symmetry of Brownian motion).
answered Jan 21 at 14:47
sazsaz
81.5k861127
$endgroup$
$begingroup$
Thank you for your answer! Why can I infer from $cZ stackrel{d}{=} Z$ that the law of $Z$ is concentrated on ${0,+infty}$
$endgroup$
– quallenjäger
Jan 21 at 14:53
$begingroup$
@quallenjäger $cZ stackrel{d}{=} Z$ implies $1_{{cZ<infty}} cZ stackrel{d}{=} Z 1_{{Z<infty}} =: tilde{Z}$. If you consider the characteristic function of $tilde{Z}$ , you will thus find that $phi(c) = phi(1)$ for all $c$, i.e. $tilde{Z}=0$.
$endgroup$
– saz
Jan 21 at 14:58
$begingroup$
sorry, I accepted the other answer. Your answer is great as well as always. Thank you
$endgroup$
– quallenjäger
Jan 21 at 15:34
$begingroup$
@quallenjäger YOu are welcome. I've just added some more details about infering that the law of $Z$ is concentrated on $0$ and $+infty$.
$endgroup$
– saz
Jan 21 at 17:14
add a comment |
$begingroup$
Since the Brownian motion has continuous sample paths, it holds that $$Z = sup_{t geq 0} B_t > a iff exists t in (0,infty) cap mathbb{Q}: B_t>a.$$
If we denote by $(t_k)_{k in mathbb{N}}$ an enumeration of $(0,infty) cap mathbb{Q}$, this gives $$mathbb{P}(Z>a) = lim_{n to infty}mathbb{P} left( bigcup_{k=1}^n {B_{t_k}>a} right)$$ which shows that the distribution of $Z$ depends only on the finite dimensional distributions of $(B_{t_1},ldots,B_{t_n})$, $n in mathbb{N}$. Consequently, if $(W_t)_{t geq 0}$ is another process with continuous sample paths which has the same finite-dimensional distributions as $(B_t)_{t geq 0}$, then $$sup_{t geq 0} B_t stackrel{d}{=} sup_{t geq 0} W_t.$$
Applying this for $W_t := c W_{t/c^2}$ with $c>0$ we get
$$c sup_{s geq 0} B_s = sup_{t geq 0} W_t stackrel{d}{=} sup_{t geq 0} B_t$$
which proves $c Z stackrel{d}{=} Z$. Hence,
begin{align*} mathbb{E}left(1_{{Z<infty}} (1- e^{-Z}) right) = mathbb{E} left( (1-e^{-cZ}) 1_{{cZ<infty}} right) &= mathbb{E} left( (1-e^{-cZ}) 1_{{Z<infty}} right) \ &xrightarrow{c downarrow 0} 0. end{align*}
As $1-e^{-Z} geq 0$ on ${Z<infty}$ this implies $1-e^{-Z}=0$ almost surely on ${Z<infty}$, i.e. $Z=0$ on ${Z<infty}$. Consequently, we have shown that the law of $Z$ is concentrated on $0$ and $+infty$.
Regarding the 2nd part of your second question: For $c=0$ the process $(W_t)_{t geq 0}$ is clearly not well-defined; for $c<0$ we still have $cZstackrel{d}{=} Z$ (which is not surprising because of the symmetry of Brownian motion).
answered Jan 21 at 14:47
sazsaz
81.5k861127
$endgroup$
$begingroup$
Thank you for your answer! Why can I infer from $cZ stackrel{d}{=} Z$ that the law of $Z$ is concentrated on ${0,+infty}$
$endgroup$
– quallenjäger
Jan 21 at 14:53
$begingroup$
@quallenjäger $cZ stackrel{d}{=} Z$ implies $1_{{cZ<infty}} cZ stackrel{d}{=} Z 1_{{Z<infty}} =: tilde{Z}$. If you consider the characteristic function of $tilde{Z}$ , you will thus find that $phi(c) = phi(1)$ for all $c$, i.e. $tilde{Z}=0$.
$endgroup$
– saz
Jan 21 at 14:58
$begingroup$
sorry, I accepted the other answer. Your answer is great as well as always. Thank you
$endgroup$
– quallenjäger
Jan 21 at 15:34
$begingroup$
@quallenjäger YOu are welcome. I've just added some more details about infering that the law of $Z$ is concentrated on $0$ and $+infty$.
$endgroup$
– saz
Jan 21 at 17:14
add a comment |
$begingroup$
Since the Brownian motion has continuous sample paths, it holds that $$Z = sup_{t geq 0} B_t > a iff exists t in (0,infty) cap mathbb{Q}: B_t>a.$$
If we denote by $(t_k)_{k in mathbb{N}}$ an enumeration of $(0,infty) cap mathbb{Q}$, this gives $$mathbb{P}(Z>a) = lim_{n to infty}mathbb{P} left( bigcup_{k=1}^n {B_{t_k}>a} right)$$ which shows that the distribution of $Z$ depends only on the finite dimensional distributions of $(B_{t_1},ldots,B_{t_n})$, $n in mathbb{N}$. Consequently, if $(W_t)_{t geq 0}$ is another process with continuous sample paths which has the same finite-dimensional distributions as $(B_t)_{t geq 0}$, then $$sup_{t geq 0} B_t stackrel{d}{=} sup_{t geq 0} W_t.$$
Applying this for $W_t := c W_{t/c^2}$ with $c>0$ we get
$$c sup_{s geq 0} B_s = sup_{t geq 0} W_t stackrel{d}{=} sup_{t geq 0} B_t$$
which proves $c Z stackrel{d}{=} Z$. Hence,
begin{align*} mathbb{E}left(1_{{Z<infty}} (1- e^{-Z}) right) = mathbb{E} left( (1-e^{-cZ}) 1_{{cZ<infty}} right) &= mathbb{E} left( (1-e^{-cZ}) 1_{{Z<infty}} right) \ &xrightarrow{c downarrow 0} 0. end{align*}
As $1-e^{-Z} geq 0$ on ${Z<infty}$ this implies $1-e^{-Z}=0$ almost surely on ${Z<infty}$, i.e. $Z=0$ on ${Z<infty}$. Consequently, we have shown that the law of $Z$ is concentrated on $0$ and $+infty$.
Regarding the 2nd part of your second question: For $c=0$ the process $(W_t)_{t geq 0}$ is clearly not well-defined; for $c<0$ we still have $cZstackrel{d}{=} Z$ (which is not surprising because of the symmetry of Brownian motion).
answered Jan 21 at 14:47
sazsaz
81.5k861127
$endgroup$
Since the Brownian motion has continuous sample paths, it holds that $$Z = sup_{t geq 0} B_t > a iff exists t in (0,infty) cap mathbb{Q}: B_t>a.$$
If we denote by $(t_k)_{k in mathbb{N}}$ an enumeration of $(0,infty) cap mathbb{Q}$, this gives $$mathbb{P}(Z>a) = lim_{n to infty}mathbb{P} left( bigcup_{k=1}^n {B_{t_k}>a} right)$$ which shows that the distribution of $Z$ depends only on the finite dimensional distributions of $(B_{t_1},ldots,B_{t_n})$, $n in mathbb{N}$. Consequently, if $(W_t)_{t geq 0}$ is another process with continuous sample paths which has the same finite-dimensional distributions as $(B_t)_{t geq 0}$, then $$sup_{t geq 0} B_t stackrel{d}{=} sup_{t geq 0} W_t.$$
Applying this for $W_t := c W_{t/c^2}$ with $c>0$ we get
$$c sup_{s geq 0} B_s = sup_{t geq 0} W_t stackrel{d}{=} sup_{t geq 0} B_t$$
which proves $c Z stackrel{d}{=} Z$. Hence,
begin{align*} mathbb{E}left(1_{{Z<infty}} (1- e^{-Z}) right) = mathbb{E} left( (1-e^{-cZ}) 1_{{cZ<infty}} right) &= mathbb{E} left( (1-e^{-cZ}) 1_{{Z<infty}} right) \ &xrightarrow{c downarrow 0} 0. end{align*}
As $1-e^{-Z} geq 0$ on ${Z<infty}$ this implies $1-e^{-Z}=0$ almost surely on ${Z<infty}$, i.e. $Z=0$ on ${Z<infty}$. Consequently, we have shown that the law of $Z$ is concentrated on $0$ and $+infty$.
Regarding the 2nd part of your second question: For $c=0$ the process $(W_t)_{t geq 0}$ is clearly not well-defined; for $c<0$ we still have $cZstackrel{d}{=} Z$ (which is not surprising because of the symmetry of Brownian motion).
answered Jan 21 at 14:47
sazsaz
81.5k861127
edited Jan 21 at 17:13
answered Jan 21 at 14:47
sazsaz
81.5k861127
answered Jan 21 at 14:47
sazsaz
81.5k861127
answered Jan 21 at 14:47
sazsaz
81.5k861127
81.5k861127
$begingroup$
Thank you for your answer! Why can I infer from $cZ stackrel{d}{=} Z$ that the law of $Z$ is concentrated on ${0,+infty}$
$endgroup$
– quallenjäger
Jan 21 at 14:53
$begingroup$
@quallenjäger $cZ stackrel{d}{=} Z$ implies $1_{{cZ<infty}} cZ stackrel{d}{=} Z 1_{{Z<infty}} =: tilde{Z}$. If you consider the characteristic function of $tilde{Z}$ , you will thus find that $phi(c) = phi(1)$ for all $c$, i.e. $tilde{Z}=0$.
$endgroup$
– saz
Jan 21 at 14:58
$begingroup$
sorry, I accepted the other answer. Your answer is great as well as always. Thank you
$endgroup$
– quallenjäger
Jan 21 at 15:34
$begingroup$
@quallenjäger YOu are welcome. I've just added some more details about infering that the law of $Z$ is concentrated on $0$ and $+infty$.
$endgroup$
– saz
Jan 21 at 17:14
add a comment |
$begingroup$
Thank you for your answer! Why can I infer from $cZ stackrel{d}{=} Z$ that the law of $Z$ is concentrated on ${0,+infty}$
$endgroup$
– quallenjäger
Jan 21 at 14:53
$begingroup$
@quallenjäger $cZ stackrel{d}{=} Z$ implies $1_{{cZ<infty}} cZ stackrel{d}{=} Z 1_{{Z<infty}} =: tilde{Z}$. If you consider the characteristic function of $tilde{Z}$ , you will thus find that $phi(c) = phi(1)$ for all $c$, i.e. $tilde{Z}=0$.
$endgroup$
– saz
Jan 21 at 14:58
$begingroup$
sorry, I accepted the other answer. Your answer is great as well as always. Thank you
$endgroup$
– quallenjäger
Jan 21 at 15:34
$begingroup$
@quallenjäger YOu are welcome. I've just added some more details about infering that the law of $Z$ is concentrated on $0$ and $+infty$.
$endgroup$
– saz
Jan 21 at 17:14
$begingroup$
Thank you for your answer! Why can I infer from $cZ stackrel{d}{=} Z$ that the law of $Z$ is concentrated on ${0,+infty}$
$endgroup$
– quallenjäger
Jan 21 at 14:53
$begingroup$
Thank you for your answer! Why can I infer from $cZ stackrel{d}{=} Z$ that the law of $Z$ is concentrated on ${0,+infty}$
$endgroup$
– quallenjäger
Jan 21 at 14:53
$begingroup$
@quallenjäger $cZ stackrel{d}{=} Z$ implies $1_{{cZ<infty}} cZ stackrel{d}{=} Z 1_{{Z<infty}} =: tilde{Z}$. If you consider the characteristic function of $tilde{Z}$ , you will thus find that $phi(c) = phi(1)$ for all $c$, i.e. $tilde{Z}=0$.
$endgroup$
– saz
Jan 21 at 14:58
$begingroup$
@quallenjäger $cZ stackrel{d}{=} Z$ implies $1_{{cZ<infty}} cZ stackrel{d}{=} Z 1_{{Z<infty}} =: tilde{Z}$. If you consider the characteristic function of $tilde{Z}$ , you will thus find that $phi(c) = phi(1)$ for all $c$, i.e. $tilde{Z}=0$.
$endgroup$
– saz
Jan 21 at 14:58
$begingroup$
sorry, I accepted the other answer. Your answer is great as well as always. Thank you
$endgroup$
– quallenjäger
Jan 21 at 15:34
$begingroup$
sorry, I accepted the other answer. Your answer is great as well as always. Thank you
$endgroup$
– quallenjäger
Jan 21 at 15:34
$begingroup$
@quallenjäger YOu are welcome. I've just added some more details about infering that the law of $Z$ is concentrated on $0$ and $+infty$.
$endgroup$
– saz
Jan 21 at 17:14
$begingroup$
@quallenjäger YOu are welcome. I've just added some more details about infering that the law of $Z$ is concentrated on $0$ and $+infty$.
$endgroup$
– saz
Jan 21 at 17:14
add a comment |
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