Variance of the Wiener Increment
$begingroup$
I have a Wiener process $W(t)$, which is a normally distributed random variable with mean $langle W(t)rangle = mu = 0$ and variance $langle W(t)^2rangle = sigma^2 = t$. The angled brackets $langle rangle$ indicate an average over all realisations of the Wiener process.
The Wiener increment $Delta W$ is defined as:
$Delta W(t) = W(t + Delta t) - W(t) $
which corresponds to the time increment $Delta t$. The mean of $Delta W$ is zero, since the means of both $W(t + Delta t)$ and $ W(t) $ are zero.
I am trying to derive the variance of $Delta W$, which is $langle (Delta W)^2rangle$. So far I have:
$(Delta W)^2 = W(t + Delta t)^2 + W(t)^2 -2W(t + Delta t)W(t) $
$langle (Delta W)^2rangle = langle W(t + Delta t)^2rangle + langle W(t)^2rangle -2langle W(t + Delta t)W(t)rangle$
$langle (Delta W)^2rangle = t + Delta t + t -2langle W(t + Delta t)W(t)rangle$
I am not sure how to evaluate the last term (or if what I have done so far is correct) and would appreciate some help. I have been told that the correct answer is $Delta t$ and am trying to verify this. Many thanks!
statistics normal-distribution random-walk statistical-mechanics
edited Dec 7 '18 at 17:28
user10354138
7,4422925
asked Dec 7 '18 at 17:08
asphasph
32
$endgroup$
add a comment |
$begingroup$
I have a Wiener process $W(t)$, which is a normally distributed random variable with mean $langle W(t)rangle = mu = 0$ and variance $langle W(t)^2rangle = sigma^2 = t$. The angled brackets $langle rangle$ indicate an average over all realisations of the Wiener process.
The Wiener increment $Delta W$ is defined as:
$Delta W(t) = W(t + Delta t) - W(t) $
which corresponds to the time increment $Delta t$. The mean of $Delta W$ is zero, since the means of both $W(t + Delta t)$ and $ W(t) $ are zero.
I am trying to derive the variance of $Delta W$, which is $langle (Delta W)^2rangle$. So far I have:
$(Delta W)^2 = W(t + Delta t)^2 + W(t)^2 -2W(t + Delta t)W(t) $
$langle (Delta W)^2rangle = langle W(t + Delta t)^2rangle + langle W(t)^2rangle -2langle W(t + Delta t)W(t)rangle$
$langle (Delta W)^2rangle = t + Delta t + t -2langle W(t + Delta t)W(t)rangle$
I am not sure how to evaluate the last term (or if what I have done so far is correct) and would appreciate some help. I have been told that the correct answer is $Delta t$ and am trying to verify this. Many thanks!
statistics normal-distribution random-walk statistical-mechanics
edited Dec 7 '18 at 17:28
user10354138
7,4422925
asked Dec 7 '18 at 17:08
asphasph
32
$endgroup$
add a comment |
$begingroup$
I have a Wiener process $W(t)$, which is a normally distributed random variable with mean $langle W(t)rangle = mu = 0$ and variance $langle W(t)^2rangle = sigma^2 = t$. The angled brackets $langle rangle$ indicate an average over all realisations of the Wiener process.
The Wiener increment $Delta W$ is defined as:
$Delta W(t) = W(t + Delta t) - W(t) $
which corresponds to the time increment $Delta t$. The mean of $Delta W$ is zero, since the means of both $W(t + Delta t)$ and $ W(t) $ are zero.
I am trying to derive the variance of $Delta W$, which is $langle (Delta W)^2rangle$. So far I have:
$(Delta W)^2 = W(t + Delta t)^2 + W(t)^2 -2W(t + Delta t)W(t) $
$langle (Delta W)^2rangle = langle W(t + Delta t)^2rangle + langle W(t)^2rangle -2langle W(t + Delta t)W(t)rangle$
$langle (Delta W)^2rangle = t + Delta t + t -2langle W(t + Delta t)W(t)rangle$
I am not sure how to evaluate the last term (or if what I have done so far is correct) and would appreciate some help. I have been told that the correct answer is $Delta t$ and am trying to verify this. Many thanks!
statistics normal-distribution random-walk statistical-mechanics
edited Dec 7 '18 at 17:28
user10354138
7,4422925
asked Dec 7 '18 at 17:08
asphasph
32
$endgroup$
I have a Wiener process $W(t)$, which is a normally distributed random variable with mean $langle W(t)rangle = mu = 0$ and variance $langle W(t)^2rangle = sigma^2 = t$. The angled brackets $langle rangle$ indicate an average over all realisations of the Wiener process.
The Wiener increment $Delta W$ is defined as:
$Delta W(t) = W(t + Delta t) - W(t) $
which corresponds to the time increment $Delta t$. The mean of $Delta W$ is zero, since the means of both $W(t + Delta t)$ and $ W(t) $ are zero.
I am trying to derive the variance of $Delta W$, which is $langle (Delta W)^2rangle$. So far I have:
$(Delta W)^2 = W(t + Delta t)^2 + W(t)^2 -2W(t + Delta t)W(t) $
$langle (Delta W)^2rangle = langle W(t + Delta t)^2rangle + langle W(t)^2rangle -2langle W(t + Delta t)W(t)rangle$
$langle (Delta W)^2rangle = t + Delta t + t -2langle W(t + Delta t)W(t)rangle$
I am not sure how to evaluate the last term (or if what I have done so far is correct) and would appreciate some help. I have been told that the correct answer is $Delta t$ and am trying to verify this. Many thanks!
statistics normal-distribution random-walk statistical-mechanics
statistics normal-distribution random-walk statistical-mechanics
edited Dec 7 '18 at 17:28
user10354138
7,4422925
asked Dec 7 '18 at 17:08
asphasph
32
edited Dec 7 '18 at 17:28
user10354138
7,4422925
asked Dec 7 '18 at 17:08
asphasph
32
edited Dec 7 '18 at 17:28
user10354138
7,4422925
edited Dec 7 '18 at 17:28
user10354138
7,4422925
edited Dec 7 '18 at 17:28
user10354138
7,4422925
7,4422925
asked Dec 7 '18 at 17:08
asphasph
32
asked Dec 7 '18 at 17:08
asphasph
32
asked Dec 7 '18 at 17:08
asphasph
32
32
add a comment |
add a comment |
1 Answer
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$begingroup$
So you can take a shortcut if you know that the Wiener process has independent increments. This is stronger than merely being Markov, because it follows that the distribution of the increments starting from a time $t$ do not depend even on $W_t$. Once you know that (which in some definitions of the Wiener process is built into the definition), it follows that the distribution and thus moments of $Delta W$ do not depend on $W$ at the initial time. Thus in particular moments of $W_t - W_s$ for $t geq s$ are just the same moments of $W_{t-s}$.
But the alternative is to show that the covariance function of the Wiener process is $E[W_t W_s] = min { t,s }$. Exactly how to do this will depend on what you already have available to you to use.
answered Dec 7 '18 at 17:18
IanIan
68.1k25388
$endgroup$
$begingroup$
Thanks, that was helpful. I don't really understand the second part, but the first part of your answer is a much clearer way of thinking about the problem and has cleared up my problem.
$endgroup$
– asph
Dec 9 '18 at 15:42
add a comment |
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1 Answer
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$begingroup$
So you can take a shortcut if you know that the Wiener process has independent increments. This is stronger than merely being Markov, because it follows that the distribution of the increments starting from a time $t$ do not depend even on $W_t$. Once you know that (which in some definitions of the Wiener process is built into the definition), it follows that the distribution and thus moments of $Delta W$ do not depend on $W$ at the initial time. Thus in particular moments of $W_t - W_s$ for $t geq s$ are just the same moments of $W_{t-s}$.
But the alternative is to show that the covariance function of the Wiener process is $E[W_t W_s] = min { t,s }$. Exactly how to do this will depend on what you already have available to you to use.
answered Dec 7 '18 at 17:18
IanIan
68.1k25388
$endgroup$
$begingroup$
Thanks, that was helpful. I don't really understand the second part, but the first part of your answer is a much clearer way of thinking about the problem and has cleared up my problem.
$endgroup$
– asph
Dec 9 '18 at 15:42
add a comment |
$begingroup$
So you can take a shortcut if you know that the Wiener process has independent increments. This is stronger than merely being Markov, because it follows that the distribution of the increments starting from a time $t$ do not depend even on $W_t$. Once you know that (which in some definitions of the Wiener process is built into the definition), it follows that the distribution and thus moments of $Delta W$ do not depend on $W$ at the initial time. Thus in particular moments of $W_t - W_s$ for $t geq s$ are just the same moments of $W_{t-s}$.
But the alternative is to show that the covariance function of the Wiener process is $E[W_t W_s] = min { t,s }$. Exactly how to do this will depend on what you already have available to you to use.
answered Dec 7 '18 at 17:18
IanIan
68.1k25388
$endgroup$
$begingroup$
Thanks, that was helpful. I don't really understand the second part, but the first part of your answer is a much clearer way of thinking about the problem and has cleared up my problem.
$endgroup$
– asph
Dec 9 '18 at 15:42
add a comment |
$begingroup$
So you can take a shortcut if you know that the Wiener process has independent increments. This is stronger than merely being Markov, because it follows that the distribution of the increments starting from a time $t$ do not depend even on $W_t$. Once you know that (which in some definitions of the Wiener process is built into the definition), it follows that the distribution and thus moments of $Delta W$ do not depend on $W$ at the initial time. Thus in particular moments of $W_t - W_s$ for $t geq s$ are just the same moments of $W_{t-s}$.
But the alternative is to show that the covariance function of the Wiener process is $E[W_t W_s] = min { t,s }$. Exactly how to do this will depend on what you already have available to you to use.
answered Dec 7 '18 at 17:18
IanIan
68.1k25388
$endgroup$
So you can take a shortcut if you know that the Wiener process has independent increments. This is stronger than merely being Markov, because it follows that the distribution of the increments starting from a time $t$ do not depend even on $W_t$. Once you know that (which in some definitions of the Wiener process is built into the definition), it follows that the distribution and thus moments of $Delta W$ do not depend on $W$ at the initial time. Thus in particular moments of $W_t - W_s$ for $t geq s$ are just the same moments of $W_{t-s}$.
But the alternative is to show that the covariance function of the Wiener process is $E[W_t W_s] = min { t,s }$. Exactly how to do this will depend on what you already have available to you to use.
answered Dec 7 '18 at 17:18
IanIan
68.1k25388
answered Dec 7 '18 at 17:18
IanIan
68.1k25388
answered Dec 7 '18 at 17:18
IanIan
68.1k25388
answered Dec 7 '18 at 17:18
IanIan
68.1k25388
68.1k25388
$begingroup$
Thanks, that was helpful. I don't really understand the second part, but the first part of your answer is a much clearer way of thinking about the problem and has cleared up my problem.
$endgroup$
– asph
Dec 9 '18 at 15:42
add a comment |
$begingroup$
Thanks, that was helpful. I don't really understand the second part, but the first part of your answer is a much clearer way of thinking about the problem and has cleared up my problem.
$endgroup$
– asph
Dec 9 '18 at 15:42
$begingroup$
Thanks, that was helpful. I don't really understand the second part, but the first part of your answer is a much clearer way of thinking about the problem and has cleared up my problem.
$endgroup$
– asph
Dec 9 '18 at 15:42
$begingroup$
Thanks, that was helpful. I don't really understand the second part, but the first part of your answer is a much clearer way of thinking about the problem and has cleared up my problem.
$endgroup$
– asph
Dec 9 '18 at 15:42
add a comment |
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