Point-wise scaling of stochastic processes on $mathbb{R}$
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Consider a $n$-dimensional random vector $boldsymbol{X}$ with covariance matrix $mathbf{Sigma} = (sigma_{ij})$. We may apply a element-wise scaling to $boldsymbol{X}$ by multiplying it with a diagonal matrix $mathbf{A} = operatorname{diag}(a_1, dots, a_n)$. Then the covariance matrix of the scaled random vector $mathbf{A} boldsymbol{X}$ is
$$ operatorname{Cov}(mathbf{A} boldsymbol{X}) = mathbf{A} boldsymbol{X} mathbf{A}^top = (a_i a_j sigma_{ij}). $$
My question is: Does this result generalize to stochastic processes on $mathbb{R}$? In other words: Considering a stochastic process $X(t)$ with covariance function $c_X(t, t')$ and a (non-random) function $f(t)$, what is the covariance function of the stochastic process $Y(t) = f(t) X(t)$? Is it $c_Y(t, t') = f(t) f(t') c_X(t, t')$ or are things more complicated?
I am not very familiar with rigorous probability theory, but I appreciate any hints on the question or suggestions on where to read up on the topic. Thank you very much!
probability-theory stochastic-processes covariance
asked Dec 29 '18 at 16:57
bbrotbbrot
151110
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add a comment |
$begingroup$
Consider a $n$-dimensional random vector $boldsymbol{X}$ with covariance matrix $mathbf{Sigma} = (sigma_{ij})$. We may apply a element-wise scaling to $boldsymbol{X}$ by multiplying it with a diagonal matrix $mathbf{A} = operatorname{diag}(a_1, dots, a_n)$. Then the covariance matrix of the scaled random vector $mathbf{A} boldsymbol{X}$ is
$$ operatorname{Cov}(mathbf{A} boldsymbol{X}) = mathbf{A} boldsymbol{X} mathbf{A}^top = (a_i a_j sigma_{ij}). $$
My question is: Does this result generalize to stochastic processes on $mathbb{R}$? In other words: Considering a stochastic process $X(t)$ with covariance function $c_X(t, t')$ and a (non-random) function $f(t)$, what is the covariance function of the stochastic process $Y(t) = f(t) X(t)$? Is it $c_Y(t, t') = f(t) f(t') c_X(t, t')$ or are things more complicated?
I am not very familiar with rigorous probability theory, but I appreciate any hints on the question or suggestions on where to read up on the topic. Thank you very much!
probability-theory stochastic-processes covariance
asked Dec 29 '18 at 16:57
bbrotbbrot
151110
$endgroup$
1
$begingroup$
The generalization is correct. This is because the covariance of a process depends only on its two-dimensional distributions. Also more simply, $text{Cov}(f(t)X(t), f(t')X(t')) = f(t)f(t')text{Cov}(X(t),X(t'))$ by bilinearity of the covariance.
$endgroup$
– Michh
Dec 29 '18 at 17:44
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@Michh, thank you, that's really quite simple! Is it also true that a function of the form $f(t) f(t') c(t, t')$ is always a valid covariance function given that $c(t, t')$ is a valid covariance function?
$endgroup$
– bbrot
Dec 29 '18 at 18:22
$begingroup$
Absolutely, by setting $Y(t) = f(t) , X(t)$ where $X$ is a process with covariance $c$, $Y$ has the desired covariance function.
$endgroup$
– Michh
Dec 29 '18 at 18:28
add a comment |
$begingroup$
Consider a $n$-dimensional random vector $boldsymbol{X}$ with covariance matrix $mathbf{Sigma} = (sigma_{ij})$. We may apply a element-wise scaling to $boldsymbol{X}$ by multiplying it with a diagonal matrix $mathbf{A} = operatorname{diag}(a_1, dots, a_n)$. Then the covariance matrix of the scaled random vector $mathbf{A} boldsymbol{X}$ is
$$ operatorname{Cov}(mathbf{A} boldsymbol{X}) = mathbf{A} boldsymbol{X} mathbf{A}^top = (a_i a_j sigma_{ij}). $$
My question is: Does this result generalize to stochastic processes on $mathbb{R}$? In other words: Considering a stochastic process $X(t)$ with covariance function $c_X(t, t')$ and a (non-random) function $f(t)$, what is the covariance function of the stochastic process $Y(t) = f(t) X(t)$? Is it $c_Y(t, t') = f(t) f(t') c_X(t, t')$ or are things more complicated?
I am not very familiar with rigorous probability theory, but I appreciate any hints on the question or suggestions on where to read up on the topic. Thank you very much!
probability-theory stochastic-processes covariance
asked Dec 29 '18 at 16:57
bbrotbbrot
151110
$endgroup$
Consider a $n$-dimensional random vector $boldsymbol{X}$ with covariance matrix $mathbf{Sigma} = (sigma_{ij})$. We may apply a element-wise scaling to $boldsymbol{X}$ by multiplying it with a diagonal matrix $mathbf{A} = operatorname{diag}(a_1, dots, a_n)$. Then the covariance matrix of the scaled random vector $mathbf{A} boldsymbol{X}$ is
$$ operatorname{Cov}(mathbf{A} boldsymbol{X}) = mathbf{A} boldsymbol{X} mathbf{A}^top = (a_i a_j sigma_{ij}). $$
My question is: Does this result generalize to stochastic processes on $mathbb{R}$? In other words: Considering a stochastic process $X(t)$ with covariance function $c_X(t, t')$ and a (non-random) function $f(t)$, what is the covariance function of the stochastic process $Y(t) = f(t) X(t)$? Is it $c_Y(t, t') = f(t) f(t') c_X(t, t')$ or are things more complicated?
I am not very familiar with rigorous probability theory, but I appreciate any hints on the question or suggestions on where to read up on the topic. Thank you very much!
probability-theory stochastic-processes covariance
probability-theory stochastic-processes covariance
asked Dec 29 '18 at 16:57
bbrotbbrot
151110
asked Dec 29 '18 at 16:57
bbrotbbrot
151110
asked Dec 29 '18 at 16:57
bbrotbbrot
151110
asked Dec 29 '18 at 16:57
bbrotbbrot
151110
asked Dec 29 '18 at 16:57
bbrotbbrot
151110
151110
1
$begingroup$
The generalization is correct. This is because the covariance of a process depends only on its two-dimensional distributions. Also more simply, $text{Cov}(f(t)X(t), f(t')X(t')) = f(t)f(t')text{Cov}(X(t),X(t'))$ by bilinearity of the covariance.
$endgroup$
– Michh
Dec 29 '18 at 17:44
$begingroup$
@Michh, thank you, that's really quite simple! Is it also true that a function of the form $f(t) f(t') c(t, t')$ is always a valid covariance function given that $c(t, t')$ is a valid covariance function?
$endgroup$
– bbrot
Dec 29 '18 at 18:22
$begingroup$
Absolutely, by setting $Y(t) = f(t) , X(t)$ where $X$ is a process with covariance $c$, $Y$ has the desired covariance function.
$endgroup$
– Michh
Dec 29 '18 at 18:28
add a comment |
1
$begingroup$
The generalization is correct. This is because the covariance of a process depends only on its two-dimensional distributions. Also more simply, $text{Cov}(f(t)X(t), f(t')X(t')) = f(t)f(t')text{Cov}(X(t),X(t'))$ by bilinearity of the covariance.
$endgroup$
– Michh
Dec 29 '18 at 17:44
$begingroup$
@Michh, thank you, that's really quite simple! Is it also true that a function of the form $f(t) f(t') c(t, t')$ is always a valid covariance function given that $c(t, t')$ is a valid covariance function?
$endgroup$
– bbrot
Dec 29 '18 at 18:22
$begingroup$
Absolutely, by setting $Y(t) = f(t) , X(t)$ where $X$ is a process with covariance $c$, $Y$ has the desired covariance function.
$endgroup$
– Michh
Dec 29 '18 at 18:28
1
1
$begingroup$
The generalization is correct. This is because the covariance of a process depends only on its two-dimensional distributions. Also more simply, $text{Cov}(f(t)X(t), f(t')X(t')) = f(t)f(t')text{Cov}(X(t),X(t'))$ by bilinearity of the covariance.
$endgroup$
– Michh
Dec 29 '18 at 17:44
$begingroup$
The generalization is correct. This is because the covariance of a process depends only on its two-dimensional distributions. Also more simply, $text{Cov}(f(t)X(t), f(t')X(t')) = f(t)f(t')text{Cov}(X(t),X(t'))$ by bilinearity of the covariance.
$endgroup$
– Michh
Dec 29 '18 at 17:44
$begingroup$
@Michh, thank you, that's really quite simple! Is it also true that a function of the form $f(t) f(t') c(t, t')$ is always a valid covariance function given that $c(t, t')$ is a valid covariance function?
$endgroup$
– bbrot
Dec 29 '18 at 18:22
$begingroup$
@Michh, thank you, that's really quite simple! Is it also true that a function of the form $f(t) f(t') c(t, t')$ is always a valid covariance function given that $c(t, t')$ is a valid covariance function?
$endgroup$
– bbrot
Dec 29 '18 at 18:22
$begingroup$
Absolutely, by setting $Y(t) = f(t) , X(t)$ where $X$ is a process with covariance $c$, $Y$ has the desired covariance function.
$endgroup$
– Michh
Dec 29 '18 at 18:28
$begingroup$
Absolutely, by setting $Y(t) = f(t) , X(t)$ where $X$ is a process with covariance $c$, $Y$ has the desired covariance function.
$endgroup$
– Michh
Dec 29 '18 at 18:28
add a comment |
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$begingroup$
The generalization is correct. This is because the covariance of a process depends only on its two-dimensional distributions. Also more simply, $text{Cov}(f(t)X(t), f(t')X(t')) = f(t)f(t')text{Cov}(X(t),X(t'))$ by bilinearity of the covariance.
$endgroup$
– Michh
Dec 29 '18 at 17:44
$begingroup$
@Michh, thank you, that's really quite simple! Is it also true that a function of the form $f(t) f(t') c(t, t')$ is always a valid covariance function given that $c(t, t')$ is a valid covariance function?
$endgroup$
– bbrot
Dec 29 '18 at 18:22
$begingroup$
Absolutely, by setting $Y(t) = f(t) , X(t)$ where $X$ is a process with covariance $c$, $Y$ has the desired covariance function.
$endgroup$
– Michh
Dec 29 '18 at 18:28