Moment Function of Gaussian Process
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Let $ {Y_t,0 leq t leq T} $ be a Gaussian process and $ f(.) $ is a measurable function. How can I represent the $n$th moment of $ f(Y_t) $ which is $ E[f(Y_t)^n] $?
Is it
$$ E[f(Y_t)^n] = int_{mathbb{R}_n} frac{f(y_1) ... f(y_n)}{(2pi)^{z/2}; |Sigma|^{1/2}} expleft(frac{1}{2} (mathbf{y}-mu)^intercal Sigma^{-1}; (mathbf{y}-mu)right) mathbf{dy} $$
where $ mathbf{y} = (y_1,...,y_n) $, $ mathbf{dy} = dy_1 times ... times dy_n $, and $ mu = (mu_1,...,mu_n) $ or
$$ E[f(Y_t)^n] = int_{-infty}^infty frac{f(y)^n}{sigma sqrt{2pi}} expleft(-frac{(y-mu)^2}{2sigma^2}right) dy $$
where $ mu = E[Y_t] $ and $ sigma^2 = operatorname{Var}(Y_t) $.
Please help me. If the answer is one of the equations then why is the reason behind the answer, if the answer is both of them, then why and when to use each equations. This confuses me a lot as Gaussian process is a multivariate normal but when we only take $ Y_t $ then it would be just normal.
stochastic-processes normal-distribution expected-value
asked Nov 29 at 5:41
Ben
526
add a comment |
up vote
1
down vote
favorite
Let $ {Y_t,0 leq t leq T} $ be a Gaussian process and $ f(.) $ is a measurable function. How can I represent the $n$th moment of $ f(Y_t) $ which is $ E[f(Y_t)^n] $?
Is it
$$ E[f(Y_t)^n] = int_{mathbb{R}_n} frac{f(y_1) ... f(y_n)}{(2pi)^{z/2}; |Sigma|^{1/2}} expleft(frac{1}{2} (mathbf{y}-mu)^intercal Sigma^{-1}; (mathbf{y}-mu)right) mathbf{dy} $$
where $ mathbf{y} = (y_1,...,y_n) $, $ mathbf{dy} = dy_1 times ... times dy_n $, and $ mu = (mu_1,...,mu_n) $ or
$$ E[f(Y_t)^n] = int_{-infty}^infty frac{f(y)^n}{sigma sqrt{2pi}} expleft(-frac{(y-mu)^2}{2sigma^2}right) dy $$
where $ mu = E[Y_t] $ and $ sigma^2 = operatorname{Var}(Y_t) $.
Please help me. If the answer is one of the equations then why is the reason behind the answer, if the answer is both of them, then why and when to use each equations. This confuses me a lot as Gaussian process is a multivariate normal but when we only take $ Y_t $ then it would be just normal.
stochastic-processes normal-distribution expected-value
asked Nov 29 at 5:41
Ben
526
2
$Y_t$ is simply a [univariate] Gaussian random variable, so it should be the second (if you fix the typo by changing $x$ to $y$).
– angryavian
Nov 29 at 5:55
I thought so too, but I found this arxiv.org/pdf/1510.01848.pdf. The moment of $ bar{sigma}^2_0 $, $ m_z $, is using the second equation in my question
– Ben
Nov 29 at 13:36
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $ {Y_t,0 leq t leq T} $ be a Gaussian process and $ f(.) $ is a measurable function. How can I represent the $n$th moment of $ f(Y_t) $ which is $ E[f(Y_t)^n] $?
Is it
$$ E[f(Y_t)^n] = int_{mathbb{R}_n} frac{f(y_1) ... f(y_n)}{(2pi)^{z/2}; |Sigma|^{1/2}} expleft(frac{1}{2} (mathbf{y}-mu)^intercal Sigma^{-1}; (mathbf{y}-mu)right) mathbf{dy} $$
where $ mathbf{y} = (y_1,...,y_n) $, $ mathbf{dy} = dy_1 times ... times dy_n $, and $ mu = (mu_1,...,mu_n) $ or
$$ E[f(Y_t)^n] = int_{-infty}^infty frac{f(y)^n}{sigma sqrt{2pi}} expleft(-frac{(y-mu)^2}{2sigma^2}right) dy $$
where $ mu = E[Y_t] $ and $ sigma^2 = operatorname{Var}(Y_t) $.
Please help me. If the answer is one of the equations then why is the reason behind the answer, if the answer is both of them, then why and when to use each equations. This confuses me a lot as Gaussian process is a multivariate normal but when we only take $ Y_t $ then it would be just normal.
stochastic-processes normal-distribution expected-value
asked Nov 29 at 5:41
Ben
526
Let $ {Y_t,0 leq t leq T} $ be a Gaussian process and $ f(.) $ is a measurable function. How can I represent the $n$th moment of $ f(Y_t) $ which is $ E[f(Y_t)^n] $?
Is it
$$ E[f(Y_t)^n] = int_{mathbb{R}_n} frac{f(y_1) ... f(y_n)}{(2pi)^{z/2}; |Sigma|^{1/2}} expleft(frac{1}{2} (mathbf{y}-mu)^intercal Sigma^{-1}; (mathbf{y}-mu)right) mathbf{dy} $$
where $ mathbf{y} = (y_1,...,y_n) $, $ mathbf{dy} = dy_1 times ... times dy_n $, and $ mu = (mu_1,...,mu_n) $ or
$$ E[f(Y_t)^n] = int_{-infty}^infty frac{f(y)^n}{sigma sqrt{2pi}} expleft(-frac{(y-mu)^2}{2sigma^2}right) dy $$
where $ mu = E[Y_t] $ and $ sigma^2 = operatorname{Var}(Y_t) $.
Please help me. If the answer is one of the equations then why is the reason behind the answer, if the answer is both of them, then why and when to use each equations. This confuses me a lot as Gaussian process is a multivariate normal but when we only take $ Y_t $ then it would be just normal.
stochastic-processes normal-distribution expected-value
stochastic-processes normal-distribution expected-value
asked Nov 29 at 5:41
Ben
526
asked Nov 29 at 5:41
Ben
526
edited Nov 30 at 16:50
asked Nov 29 at 5:41
Ben
526
asked Nov 29 at 5:41
Ben
526
asked Nov 29 at 5:41
Ben
526
526
2
$Y_t$ is simply a [univariate] Gaussian random variable, so it should be the second (if you fix the typo by changing $x$ to $y$).
– angryavian
Nov 29 at 5:55
I thought so too, but I found this arxiv.org/pdf/1510.01848.pdf. The moment of $ bar{sigma}^2_0 $, $ m_z $, is using the second equation in my question
– Ben
Nov 29 at 13:36
add a comment |
2
$Y_t$ is simply a [univariate] Gaussian random variable, so it should be the second (if you fix the typo by changing $x$ to $y$).
– angryavian
Nov 29 at 5:55
I thought so too, but I found this arxiv.org/pdf/1510.01848.pdf. The moment of $ bar{sigma}^2_0 $, $ m_z $, is using the second equation in my question
– Ben
Nov 29 at 13:36
2
2
$Y_t$ is simply a [univariate] Gaussian random variable, so it should be the second (if you fix the typo by changing $x$ to $y$).
– angryavian
Nov 29 at 5:55
$Y_t$ is simply a [univariate] Gaussian random variable, so it should be the second (if you fix the typo by changing $x$ to $y$).
– angryavian
Nov 29 at 5:55
I thought so too, but I found this arxiv.org/pdf/1510.01848.pdf. The moment of $ bar{sigma}^2_0 $, $ m_z $, is using the second equation in my question
– Ben
Nov 29 at 13:36
I thought so too, but I found this arxiv.org/pdf/1510.01848.pdf. The moment of $ bar{sigma}^2_0 $, $ m_z $, is using the second equation in my question
– Ben
Nov 29 at 13:36
add a comment |
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2
$Y_t$ is simply a [univariate] Gaussian random variable, so it should be the second (if you fix the typo by changing $x$ to $y$).
– angryavian
Nov 29 at 5:55
I thought so too, but I found this arxiv.org/pdf/1510.01848.pdf. The moment of $ bar{sigma}^2_0 $, $ m_z $, is using the second equation in my question
– Ben
Nov 29 at 13:36