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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!

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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

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