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| HJ... |
 Posted: Fri Oct 16, 2009 11:16 am |
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Hello all,
I sense the following should be an elementary problem, but I've searched my reference books and notes and can't seem to find anything appropriate.
Suppose X is a continuous-time random process whose probability density is a standard normal Gaussian (mean=0, variance=1, i.e. white noise) and let T be a positive real number. Let Y be the integral of X from 0 to T. Is there a straightforward way to express the probability density function Y as a function of T (and the mean and variance of X)? Or is this problem ill-posed?
Thanks in advance for any pointers,
HJ |
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| robert... |
| Posted: Sat Oct 17, 2009 3:26 pm |
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| This problem is ill-posed, or you have left out relevant information, because the joint distributions of the X(t) process are not specified. Hope this helps |
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| HJ... |
| Posted: Sat Oct 17, 2009 4:46 pm |
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Hi Robert,
Thanks for the reply. I suspected my statement of the problem wasn't quite precise; let me try to clarify.
Suppose we have a Gaussian random process -- a useful physical example is the voltage appearing across a resistor at some temperature above absolute zero. If we neglect the inevitable bandlimiting effects that would be present in any real hardware, the time history of that voltage should look like a true zero-mean Gaussian whose variance is proportional to absolute temperature. Now integrate that voltage for a duration T (in my original example I specified the interval from t=0 to t=T, but the explicit value of the start point shouldn't matter -- just the duration).
My question is, how can I get a handle on the probability distribution function of the result of that integral?
My instinct is to break the interval into a finite number N of sub-intervals, each with a constant amplitude determined by a single realization of the random process in question. If the process is not bandlimited, these values should be completely independent regardless of how small the sub-intervals become. Then evaluating the sum of the areas under these intervals (with a delta_X term factored out) in the limit as N gets very large should yield the result I'm looking for. I have no idea if that's rigorous, though.
I hope this makes a little more sense (and I apologize in advance if it doesn't -- I'm not a theoretical mathematician by training...).
Thanks. |
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| karl... |
| Posted: Sun Oct 18, 2009 1:42 am |
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HJ schrieb:
[quote:1fe7f097ea]Hi Robert,
Thanks for the reply. I suspected my statement of the problem wasn't quite precise; let me try to clarify.
Suppose we have a Gaussian random process -- a useful physical example is the voltage appearing across a resistor at some temperature above absolute zero. If we neglect the inevitable bandlimiting effects that would be present in any real hardware, the time history of that voltage should look like a true zero-mean Gaussian whose variance is proportional to absolute temperature. Now integrate that voltage for a duration T (in my original example I specified the interval from t=0 to t=T, but the explicit value of the start point shouldn't matter -- just the duration).
My question is, how can I get a handle on the probability distribution function of the result of that integral?
My instinct is to break the interval into a finite number N of sub-intervals, each with a constant amplitude determined by a single realization of the random process in question. If the process is not bandlimited, these values should be completely independent regardless of how small the sub-intervals become. Then evaluating the sum of the areas under these intervals (with a delta_X term factored out) in the limit as N gets very large should yield the result I'm looking for. I have no idea if that's rigorous, though.
I hope this makes a little more sense (and I apologize in advance if it doesn't -- I'm not a theoretical mathematician by training...).
Thanks.
Have a look at A. Papoulis: Probability, Random Variables and[/quote:1fe7f097ea]
Stochastic Processes, McGraw-Hill, 1965, p.323-325. (The first edition
is the best, it appears also in the second and third somewhere).
Since the process is normal, its integral will be normal too. So if
you calculate mean and variance as described there, you are done. It
will not work, if you insist on white noise. Then all conditions one
makes for the existence of integrals are not fulfilled.
Ciao
Karl |
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| Henry... |
| Posted: Mon Oct 19, 2009 5:18 am |
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On 18 Oct, 03:46, HJ <heywoodj... at (no spam) yahoo.com> wrote:
[quote]Hi Robert,
Thanks for the reply. I suspected my statement of the problem wasn't quite precise; let me try to clarify.
Suppose we have a Gaussian random process -- a useful physical example is the voltage appearing across a resistor at some temperature above absolute zero. If we neglect the inevitable bandlimiting effects that would be present in any real hardware, the time history of that voltage should look like a true zero-mean Gaussian whose variance is proportional to absolute temperature. Now integrate that voltage for a duration T (in my original example I specified the interval from t=0 to t=T, but the explicit value of the start point shouldn't matter -- just the duration).
My question is, how can I get a handle on the probability distribution function of the result of that integral?
My instinct is to break the interval into a finite number N of sub-intervals, each with a constant amplitude determined by a single realization of the random process in question. If the process is not bandlimited, these values should be completely independent regardless of how small the sub-intervals become. Then evaluating the sum of the areas under these intervals (with a delta_X term factored out) in the limit as N gets very large should yield the result I'm looking for. I have no idea if that's rigorous, though.
I hope this makes a little more sense (and I apologize in advance if it doesn't -- I'm not a theoretical mathematician by training...).
Thanks.
[/quote]
So you have N subintervals. The sum of N independent random variables
with mean 0 and variance v will be another random variable with mean 0
and variance Nv.
But each subinterval will last T/N seconds so the approximation to the
integral will have mean 0 and variance Nv*(T/N)^2 and that is v*T^2/
N.
With v and T fixed, the variance will decrease toward 0 as N increases
towards infinity. |
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| karl... |
| Posted: Mon Oct 19, 2009 9:51 am |
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Henry schrieb:
[quote]On 18 Oct, 03:46, HJ <heywoodj... at (no spam) yahoo.com> wrote:
Hi Robert,
Thanks for the reply. I suspected my statement of the problem wasn't quite precise; let me try to clarify.
Suppose we have a Gaussian random process -- a useful physical example is the voltage appearing across a resistor at some temperature above absolute zero. If we neglect the inevitable bandlimiting effects that would be present in any real hardware, the time history of that voltage should look like a true zero-mean Gaussian whose variance is proportional to absolute temperature. Now integrate that voltage for a duration T (in my original example I specified the interval from t=0 to t=T, but the explicit value of the start point shouldn't matter -- just the duration).
My question is, how can I get a handle on the probability distribution function of the result of that integral?
My instinct is to break the interval into a finite number N of sub-intervals, each with a constant amplitude determined by a single realization of the random process in question. If the process is not bandlimited, these values should be completely independent regardless of how small the sub-intervals become. Then evaluating the sum of the areas under these intervals (with a delta_X term factored out) in the limit as N gets very large should yield the result I'm looking for. I have no idea if that's rigorous, though.
I hope this makes a little more sense (and I apologize in advance if it doesn't -- I'm not a theoretical mathematician by training...).
Thanks.
So you have N subintervals. The sum of N independent random variables
with mean 0 and variance v will be another random variable with mean 0
and variance Nv.
[/quote]
How do you know that the integral of a white noise process over an
interval of length T/N has variance v?
[quote]But each subinterval will last T/N seconds so the approximation to the
integral will have mean 0 and variance Nv*(T/N)^2 and that is v*T^2/
N.
With v and T fixed, the variance will decrease toward 0 as N increases
towards infinity.
[/quote]
So following you the integral will be zero?
Ciao
Karl |
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| Henry... |
| Posted: Tue Oct 20, 2009 1:31 am |
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Guest
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karl wrote:
[quote]Henry schrieb:
On 18 Oct, 03:46, HJ <heywoodj... at (no spam) yahoo.com> wrote:
Hi Robert,
Thanks for the reply. I suspected my statement of the problem wasn't quite precise; let me try to clarify.
Suppose we have a Gaussian random process -- a useful physical example is the voltage appearing across a resistor at some temperature above absolute zero. If we neglect the inevitable bandlimiting effects that would be present in any real hardware, the time history of that voltage should look like a true zero-mean Gaussian whose variance is proportional to absolute temperature. Now integrate that voltage for a duration T (in my original example I specified the interval from t=0 to t=T, but the explicit value of the start point shouldn't matter -- just the duration).
My question is, how can I get a handle on the probability distribution function of the result of that integral?
My instinct is to break the interval into a finite number N of sub-intervals, each with a constant amplitude determined by a single realization of the random process in question. If the process is not bandlimited, these values should be completely independent regardless of how small the sub-intervals become. Then evaluating the sum of the areas under these intervals (with a delta_X term factored out) in the limit as N gets very large should yield the result I'm looking for. I have no idea if that's rigorous, though.
I hope this makes a little more sense (and I apologize in advance if it doesn't -- I'm not a theoretical mathematician by training...).
Thanks.
So you have N subintervals. The sum of N independent random variables
with mean 0 and variance v will be another random variable with mean 0
and variance Nv.
How do you know that the integral of a white noise process over an
interval of length T/N has variance v?
[/quote]
I don't say that. v is the variance of a single measurement of the
voltage (apparently a function of absolute temperature in the question).
[quote]
But each subinterval will last T/N seconds so the approximation to the
integral will have mean 0 and variance Nv*(T/N)^2 and that is v*T^2/
N.
With v and T fixed, the variance will decrease toward 0 as N increases
towards infinity.
So following you the integral will be zero?
[/quote]
The approximation to the integral will converge in distribution towards
a single point of zero as N increases.
[quote]
Ciao
Karl[/quote] |
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| karl... |
| Posted: Tue Oct 20, 2009 2:00 am |
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Guest
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Henry schrieb:
[quote]karl wrote:
Henry schrieb:
On 18 Oct, 03:46, HJ <heywoodj... at (no spam) yahoo.com> wrote:
Hi Robert,
Thanks for the reply. I suspected my statement of the problem wasn't
quite precise; let me try to clarify.
Suppose we have a Gaussian random process -- a useful physical
example is the voltage appearing across a resistor at some
temperature above absolute zero. If we neglect the inevitable
bandlimiting effects that would be present in any real hardware, the
time history of that voltage should look like a true zero-mean
Gaussian whose variance is proportional to absolute temperature. Now
integrate that voltage for a duration T (in my original example I
specified the interval from t=0 to t=T, but the explicit value of
the start point shouldn't matter -- just the duration).
My question is, how can I get a handle on the probability
distribution function of the result of that integral?
My instinct is to break the interval into a finite number N of
sub-intervals, each with a constant amplitude determined by a single
realization of the random process in question. If the process is not
bandlimited, these values should be completely independent
regardless of how small the sub-intervals become. Then evaluating
the sum of the areas under these intervals (with a delta_X term
factored out) in the limit as N gets very large should yield the
result I'm looking for. I have no idea if that's rigorous, though.
I hope this makes a little more sense (and I apologize in advance if
it doesn't -- I'm not a theoretical mathematician by training...).
Thanks.
So you have N subintervals. The sum of N independent random variables
with mean 0 and variance v will be another random variable with mean 0
and variance Nv.
How do you know that the integral of a white noise process over an
interval of length T/N has variance v?
I don't say that. v is the variance of a single measurement of the
voltage (apparently a function of absolute temperature in the question).
But each subinterval will last T/N seconds so the approximation to the
integral will have mean 0 and variance Nv*(T/N)^2 and that is v*T^2/
N.
With v and T fixed, the variance will decrease toward 0 as N increases
towards infinity.
So following you the integral will be zero?
The approximation to the integral will converge in distribution towards
a single point of zero as N increases.
Nonsense.[/quote]
Ciao
Karl |
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| Henry... |
| Posted: Thu Oct 22, 2009 3:31 pm |
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Guest
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karl wrote:
[quote]Henry schrieb:
karl wrote:
Henry schrieb:
So you have N subintervals. The sum of N independent random variables
with mean 0 and variance v will be another random variable with mean 0
and variance Nv.
How do you know that the integral of a white noise process over an
interval of length T/N has variance v?
I don't say that. v is the variance of a single measurement of the
voltage (apparently a function of absolute temperature in the question).
But each subinterval will last T/N seconds so the approximation to the
integral will have mean 0 and variance Nv*(T/N)^2 and that is v*T^2/
N.
With v and T fixed, the variance will decrease toward 0 as N increases
towards infinity.
So following you the integral will be zero?
The approximation to the integral will converge in distribution towards
a single point of zero as N increases.
Nonsense.
[/quote]
Here is something using R: T is an arbitrary 5, the standard deviation
of the individual samples is an arbitrary 2 (so v=4), N varies, and the
experiment is repeated 1000 times to learn something about the
distribution of each approximation to the integral:
T <- 5
X <- rep(0,times=1000)
N <- 1
for(i in 1:1000) X[i] <- sum(T/N * rnorm(N, mean = 0, sd = 2))
mean(X); sd(X)
#[1] -0.422043
#[1] 10.10662
N <- 100
for(i in 1:1000) X[i] <- sum(T/N * rnorm(N, mean = 0, sd = 2))
mean(X); sd(X)
#[1] 0.0394404
#[1] 0.9726805
N <- 10000
for(i in 1:1000) X[i] <- sum(T/N * rnorm(N, mean = 0, sd = 2))
mean(X); sd(X)
#[1] -0.0005572816
#[1] 0.1063677
N <- 1000000
for(i in 1:1000) X[i] <- sum(T/N * rnorm(N, mean = 0, sd = 2))
mean(X); sd(X)
#[1] -8.628373e-05
#[1] 0.01025021
That is pretty much what I expected, converging towards a single point
of zero as N increased, with a standard deviation close to sqrt(100/N),
although with N=1000000 it took more time than I had hoped. |
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| karl... |
| Posted: Thu Oct 22, 2009 6:54 pm |
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Guest
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Henry schrieb:
[quote]karl wrote:
Henry schrieb:
karl wrote:
Henry schrieb:
So you have N subintervals. The sum of N independent random variables
with mean 0 and variance v will be another random variable with mean 0
and variance Nv.
How do you know that the integral of a white noise process over an
interval of length T/N has variance v?
I don't say that. v is the variance of a single measurement of the
voltage (apparently a function of absolute temperature in the question).
But each subinterval will last T/N seconds so the approximation to the
integral will have mean 0 and variance Nv*(T/N)^2 and that is v*T^2/
N.
With v and T fixed, the variance will decrease toward 0 as N increases
towards infinity.
So following you the integral will be zero?
The approximation to the integral will converge in distribution towards
a single point of zero as N increases.
Nonsense.
Here is something using R: T is an arbitrary 5, the standard deviation
of the individual samples is an arbitrary 2 (so v=4), N varies, and the
experiment is repeated 1000 times to learn something about the
distribution of each approximation to the integral:
T <- 5
X <- rep(0,times=1000)
N <- 1
for(i in 1:1000) X[i] <- sum(T/N * rnorm(N, mean = 0, sd = 2))
mean(X); sd(X)
#[1] -0.422043
#[1] 10.10662
N <- 100
for(i in 1:1000) X[i] <- sum(T/N * rnorm(N, mean = 0, sd = 2))
mean(X); sd(X)
#[1] 0.0394404
#[1] 0.9726805
N <- 10000
for(i in 1:1000) X[i] <- sum(T/N * rnorm(N, mean = 0, sd = 2))
mean(X); sd(X)
#[1] -0.0005572816
#[1] 0.1063677
N <- 1000000
for(i in 1:1000) X[i] <- sum(T/N * rnorm(N, mean = 0, sd = 2))
mean(X); sd(X)
#[1] -8.628373e-05
#[1] 0.01025021
That is pretty much what I expected, converging towards a single point
of zero as N increased, with a standard deviation close to sqrt(100/N),
although with N=1000000 it took more time than I had hoped.
[/quote]
The only thing you have to show is now that these approximations
converge to white noise.
Saying that this is obvious is no mathematical proof.
Maybe you should read some text about stochastic processes before
making bold claims.
Ciao
Karl |
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| Henry... |
| Posted: Fri Oct 23, 2009 6:06 am |
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Guest
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On 23 Oct, 01:54, karl <oud... at (no spam) nononet.com> wrote:
[quote]The only thing you have to show is now that these approximations
converge to white noise.
Saying that this is obvious is no mathematical proof.
Maybe you should read some text about stochastic processes before
making bold claims.
[/quote]
Earlier I proved that HJ's approximation to the integral would be a
random variable with mean 0 and variance v*T^2/N. Then I illustrated
this with a simulation.
When you asked:
[quote]So following you the integral will be zero?
I replied:
The approximation to the integral will converge in distribution towards a single point of zero as N increases.
[/quote]
I have made no claim about whether white noise itself is integrable,
so I fail to see why you are so unhappy. |
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| Henry... |
| Posted: Fri Oct 23, 2009 11:53 am |
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Guest
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On 23 Oct, 19:16, karl <oud... at (no spam) nononet.com> wrote:
[quote]HJ said definitely white noise.
[/quote]
HJ's second post in the thread suggested an approximation, and that is
what I replied to |
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| karl... |
| Posted: Fri Oct 23, 2009 12:16 pm |
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Guest
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Henry schrieb:
[quote]On 23 Oct, 01:54, karl <oud... at (no spam) nononet.com> wrote:
The only thing you have to show is now that these approximations
converge to white noise.
Saying that this is obvious is no mathematical proof.
Maybe you should read some text about stochastic processes before
making bold claims.
Earlier I proved that HJ's approximation to the integral would be a
random variable with mean 0 and variance v*T^2/N. Then I illustrated
this with a simulation.
When you asked:
So following you the integral will be zero?
I replied:
The approximation to the integral will converge in distribution towards a single point of zero as N increases.
I have made no claim about whether white noise itself is integrable,
so I fail to see why you are so unhappy.
[/quote]
HJ said definitely white noise.
Ciao
Karl |
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| karl... |
| Posted: Fri Oct 23, 2009 10:18 pm |
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Guest
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Henry schrieb:
[quote]On 23 Oct, 19:16, karl <oud... at (no spam) nononet.com> wrote:
HJ said definitely white noise.
HJ's second post in the thread suggested an approximation, and that is
what I replied to
[/quote]
So you have shown that a wrong approximation gives a wrong result.
Ciao
Kalr |
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| Henry... |
| Posted: Sat Oct 24, 2009 9:45 am |
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Guest
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On 24 Oct, 05:18, karl <oud... at (no spam) nononet.com> wrote:
[quote]Henry schrieb:
On 23 Oct, 19:16, karl <oud... at (no spam) nononet.com> wrote:
HJ said definitely white noise.
HJ's second post in the thread suggested an approximation, and that is
what I replied to
So you have shown that a wrong approximation gives a wrong result.
[/quote]
That is your assertion, not mine. And you need to back it up with
something more substantial than your current comments.
If you are going to say that white noise with a fixed finite variance
integrates to Brownian noise, I would interested to see what you think
the variance of the integral is. |
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