Mixing
Wiener process equality
$begingroup$
I came across such a problem that I can't solve
Prove that with probability $=1$ the equality
$$W_{t}^{4} = W_{t}^{3} - 3W_{t}^{2} +1$$ holds for infinite number of $t ge 0$ but it doesn't hold for any $t in mathbb{Q}_{+}$. $W_{t}$ is a standard Wiener process.
What I did:
I computed expected value of both sides and end up with $$3t^2 = -3t+1$$ which has two solutions $notin mathbb{Q}$, one of which is positive. It makes me think that there may exist one $t$ satisfying conditions but no more than one.
Is there a mistake in my attempt?
If so, what is the solution of the problem?
stochastic-processes brownian-motion
asked Jan 6 at 21:37
GuestttGuesttt
538
$endgroup$
add a comment |
$begingroup$
I came across such a problem that I can't solve
Prove that with probability $=1$ the equality
$$W_{t}^{4} = W_{t}^{3} - 3W_{t}^{2} +1$$ holds for infinite number of $t ge 0$ but it doesn't hold for any $t in mathbb{Q}_{+}$. $W_{t}$ is a standard Wiener process.
What I did:
I computed expected value of both sides and end up with $$3t^2 = -3t+1$$ which has two solutions $notin mathbb{Q}$, one of which is positive. It makes me think that there may exist one $t$ satisfying conditions but no more than one.
Is there a mistake in my attempt?
If so, what is the solution of the problem?
stochastic-processes brownian-motion
asked Jan 6 at 21:37
GuestttGuesttt
538
$endgroup$
add a comment |
$begingroup$
I came across such a problem that I can't solve
Prove that with probability $=1$ the equality
$$W_{t}^{4} = W_{t}^{3} - 3W_{t}^{2} +1$$ holds for infinite number of $t ge 0$ but it doesn't hold for any $t in mathbb{Q}_{+}$. $W_{t}$ is a standard Wiener process.
What I did:
I computed expected value of both sides and end up with $$3t^2 = -3t+1$$ which has two solutions $notin mathbb{Q}$, one of which is positive. It makes me think that there may exist one $t$ satisfying conditions but no more than one.
Is there a mistake in my attempt?
If so, what is the solution of the problem?
stochastic-processes brownian-motion
asked Jan 6 at 21:37
GuestttGuesttt
538
$endgroup$
I came across such a problem that I can't solve
Prove that with probability $=1$ the equality
$$W_{t}^{4} = W_{t}^{3} - 3W_{t}^{2} +1$$ holds for infinite number of $t ge 0$ but it doesn't hold for any $t in mathbb{Q}_{+}$. $W_{t}$ is a standard Wiener process.
What I did:
I computed expected value of both sides and end up with $$3t^2 = -3t+1$$ which has two solutions $notin mathbb{Q}$, one of which is positive. It makes me think that there may exist one $t$ satisfying conditions but no more than one.
Is there a mistake in my attempt?
If so, what is the solution of the problem?
stochastic-processes brownian-motion
stochastic-processes brownian-motion
asked Jan 6 at 21:37
GuestttGuesttt
538
asked Jan 6 at 21:37
GuestttGuesttt
538
edited Jan 6 at 22:11
Guesttt
asked Jan 6 at 21:37
GuestttGuesttt
538
asked Jan 6 at 21:37
GuestttGuesttt
538
asked Jan 6 at 21:37
GuestttGuesttt
538
538
add a comment |
add a comment |
1 Answer
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$begingroup$
Your paragraph "What I did" is way off base.
First, consider any real number $x$ solving $x^4=x^3-3x^2+1$, then the process $(W_t)$ is recurrent hence the (random) set ${tgeqslant0mid W_t=x}$ is almost surely unbounded. This solves the first part of your question. However...
Consider $X$ the (non random) set of real numbers $x$ such that $x^4=x^3-3x^2+1$, then $X$ is finite hence, for every fixed $t$, $P(W_tin X)=0$. By countable union, $P(exists tinmathbb Q_+,,W_tin X)=0$, qed.
answered Jan 6 at 22:40
DidDid
247k23223459
$endgroup$
$begingroup$
Sorry I understood the question that $t$ is fixed, and we measure probability that $P(W_t in X)$, but it's obviously $0$ like you said. Is $W_t$ self recurrent for the same reason why a discrete Markov process on the integers line comes back to origin in finite steps with probability $1$?
$endgroup$
– Jakobian
Jan 7 at 15:49
$begingroup$
@Jakobian Delete comment on main, then? Anyway, not every discrete Markov process on the integers line is recurrent but the simple random walk is, and indeed this can be used to show the linear Brownian motion is.
$endgroup$
– Did
Jan 7 at 16:05
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your paragraph "What I did" is way off base.
First, consider any real number $x$ solving $x^4=x^3-3x^2+1$, then the process $(W_t)$ is recurrent hence the (random) set ${tgeqslant0mid W_t=x}$ is almost surely unbounded. This solves the first part of your question. However...
Consider $X$ the (non random) set of real numbers $x$ such that $x^4=x^3-3x^2+1$, then $X$ is finite hence, for every fixed $t$, $P(W_tin X)=0$. By countable union, $P(exists tinmathbb Q_+,,W_tin X)=0$, qed.
answered Jan 6 at 22:40
DidDid
247k23223459
$endgroup$
$begingroup$
Sorry I understood the question that $t$ is fixed, and we measure probability that $P(W_t in X)$, but it's obviously $0$ like you said. Is $W_t$ self recurrent for the same reason why a discrete Markov process on the integers line comes back to origin in finite steps with probability $1$?
$endgroup$
– Jakobian
Jan 7 at 15:49
$begingroup$
@Jakobian Delete comment on main, then? Anyway, not every discrete Markov process on the integers line is recurrent but the simple random walk is, and indeed this can be used to show the linear Brownian motion is.
$endgroup$
– Did
Jan 7 at 16:05
add a comment |
$begingroup$
Your paragraph "What I did" is way off base.
First, consider any real number $x$ solving $x^4=x^3-3x^2+1$, then the process $(W_t)$ is recurrent hence the (random) set ${tgeqslant0mid W_t=x}$ is almost surely unbounded. This solves the first part of your question. However...
Consider $X$ the (non random) set of real numbers $x$ such that $x^4=x^3-3x^2+1$, then $X$ is finite hence, for every fixed $t$, $P(W_tin X)=0$. By countable union, $P(exists tinmathbb Q_+,,W_tin X)=0$, qed.
answered Jan 6 at 22:40
DidDid
247k23223459
$endgroup$
$begingroup$
Sorry I understood the question that $t$ is fixed, and we measure probability that $P(W_t in X)$, but it's obviously $0$ like you said. Is $W_t$ self recurrent for the same reason why a discrete Markov process on the integers line comes back to origin in finite steps with probability $1$?
$endgroup$
– Jakobian
Jan 7 at 15:49
$begingroup$
@Jakobian Delete comment on main, then? Anyway, not every discrete Markov process on the integers line is recurrent but the simple random walk is, and indeed this can be used to show the linear Brownian motion is.
$endgroup$
– Did
Jan 7 at 16:05
add a comment |
$begingroup$
Your paragraph "What I did" is way off base.
First, consider any real number $x$ solving $x^4=x^3-3x^2+1$, then the process $(W_t)$ is recurrent hence the (random) set ${tgeqslant0mid W_t=x}$ is almost surely unbounded. This solves the first part of your question. However...
Consider $X$ the (non random) set of real numbers $x$ such that $x^4=x^3-3x^2+1$, then $X$ is finite hence, for every fixed $t$, $P(W_tin X)=0$. By countable union, $P(exists tinmathbb Q_+,,W_tin X)=0$, qed.
answered Jan 6 at 22:40
DidDid
247k23223459
$endgroup$
Your paragraph "What I did" is way off base.
First, consider any real number $x$ solving $x^4=x^3-3x^2+1$, then the process $(W_t)$ is recurrent hence the (random) set ${tgeqslant0mid W_t=x}$ is almost surely unbounded. This solves the first part of your question. However...
Consider $X$ the (non random) set of real numbers $x$ such that $x^4=x^3-3x^2+1$, then $X$ is finite hence, for every fixed $t$, $P(W_tin X)=0$. By countable union, $P(exists tinmathbb Q_+,,W_tin X)=0$, qed.
answered Jan 6 at 22:40
DidDid
247k23223459
edited Jan 8 at 9:54
answered Jan 6 at 22:40
DidDid
247k23223459
answered Jan 6 at 22:40
DidDid
247k23223459
answered Jan 6 at 22:40
DidDid
247k23223459
247k23223459
$begingroup$
Sorry I understood the question that $t$ is fixed, and we measure probability that $P(W_t in X)$, but it's obviously $0$ like you said. Is $W_t$ self recurrent for the same reason why a discrete Markov process on the integers line comes back to origin in finite steps with probability $1$?
$endgroup$
– Jakobian
Jan 7 at 15:49
$begingroup$
@Jakobian Delete comment on main, then? Anyway, not every discrete Markov process on the integers line is recurrent but the simple random walk is, and indeed this can be used to show the linear Brownian motion is.
$endgroup$
– Did
Jan 7 at 16:05
add a comment |
$begingroup$
Sorry I understood the question that $t$ is fixed, and we measure probability that $P(W_t in X)$, but it's obviously $0$ like you said. Is $W_t$ self recurrent for the same reason why a discrete Markov process on the integers line comes back to origin in finite steps with probability $1$?
$endgroup$
– Jakobian
Jan 7 at 15:49
$begingroup$
@Jakobian Delete comment on main, then? Anyway, not every discrete Markov process on the integers line is recurrent but the simple random walk is, and indeed this can be used to show the linear Brownian motion is.
$endgroup$
– Did
Jan 7 at 16:05
$begingroup$
Sorry I understood the question that $t$ is fixed, and we measure probability that $P(W_t in X)$, but it's obviously $0$ like you said. Is $W_t$ self recurrent for the same reason why a discrete Markov process on the integers line comes back to origin in finite steps with probability $1$?
$endgroup$
– Jakobian
Jan 7 at 15:49
$begingroup$
Sorry I understood the question that $t$ is fixed, and we measure probability that $P(W_t in X)$, but it's obviously $0$ like you said. Is $W_t$ self recurrent for the same reason why a discrete Markov process on the integers line comes back to origin in finite steps with probability $1$?
$endgroup$
– Jakobian
Jan 7 at 15:49
$begingroup$
@Jakobian Delete comment on main, then? Anyway, not every discrete Markov process on the integers line is recurrent but the simple random walk is, and indeed this can be used to show the linear Brownian motion is.
$endgroup$
– Did
Jan 7 at 16:05
$begingroup$
@Jakobian Delete comment on main, then? Anyway, not every discrete Markov process on the integers line is recurrent but the simple random walk is, and indeed this can be used to show the linear Brownian motion is.
$endgroup$
– Did
Jan 7 at 16:05
add a comment |
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