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I have a function $F(s)$ which is the Laplace transform of $f(t)$ (which is in itself a normally distributed random process), but I don't know what $f(t)$ is (this comes from solving a differential equation). I have also calculated the analytical mean and variance of $F(s)$, but now I need to find out the mean and variance of $f(t)$. So far, these are the things I have been using: \begin{equation} \langle F(s) \rangle = \int_{0}^{\infty} e^{-st} \langle f(t)\rangle dt \quad \Longrightarrow\quad \langle f(t) \rangle = \mathcal{L}^{-1}\left\{\langle F(s)\rangle \right\} \end{equation} which reads: "the mean of $f(t)$ is equal to the inverse Laplace transform of the mean of $F(s)$". Here $\langle \cdot \rangle$ is the expectation operator, which represents an ensemble average over all possible realizations of $f(t)$, rather than a time average. However, the variance is not as obvious: \begin{equation} \text{Var}[F(s)] = \langle F^2(s) \rangle - \langle F(s)\rangle^2, \end{equation} I have an expression for $\langle F^2(s)\rangle$, but if we write it formally we get \begin{equation} \langle F^2(s)\rangle =\left\langle \left(\int_{0}^{\infty} e^{-st} f(t) dt\right)^2\right\rangle = \int_{0}^{\infty} \int_{0}^{\infty} e^{-s(t+t')} \left\langle f(t)f(t')\right\rangle dt' dt \end{equation} since this turns my expression into a double integral (which in fact looks exactly like a double Laplace transform), does that mean that $\langle f(t) f(t')\rangle$ is the double inverse Laplace transform of $\langle F^2(s)\rangle$? Is there a straight foward connection between $\text{Var}[F(s)]$ and $\text{Var}[f(t)]$?

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  • $\begingroup$ Shouldn't $$\langle F(s) \rangle = \bigg \langle \int_{0}^{\infty} f(t) e^{-st} dt \bigg \rangle$$ ? I don't see why the average of the Laplace transform of $f$ should be equal to the Laplace transform of the average of $f$. $\endgroup$ – mattos May 22 '19 at 3:34
  • $\begingroup$ Yes but since $f(t)$ is the only random function in the integrand, by the linearity property of the expectation operator you get \begin{equation} \langle F(s) \rangle = \int_{0}^{\infty} \langle f(t) \rangle e^{-st} dt \end{equation} Here $\langle \cdot\rangle$ is an ensemble average of all possible paths for f(t), rather than a time average. $\endgroup$ – OscarNieves May 22 '19 at 3:57
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This is not an answer, but it's an observation that might point toward an answer.

We have $$ \langle F^2(s)\rangle = \int_{0}^{\infty} \int_{0}^{\infty} e^{-s(t+t')} \left\langle f(t)f(t')\right\rangle dt' dt $$ Make the change of variables $u = t + t', t = t$. Then \begin{align} \langle F^2(s)\rangle &= \int_{0}^{\infty} \int_{t}^{\infty} e^{-su} \left\langle f(t)f(u - t)\right\rangle du dt \\&= \int_{0}^{\infty} \int_{0}^{u} e^{-su} \left\langle f(t)f(u - t)\right\rangle dt du \\&= \int_{0}^{\infty} e^{-su} \left\langle \int_{0}^{u} f(t)f(u - t) dt \right\rangle du \end{align} The expression inside brackets kind of looks like a convolution of $f$ with itself. The part outside the brackets is just taking the Laplace transform of the bracketed expression. So it seems like the variance of the Laplace transform is connected to the Laplace transform of the mean of the original function convolved with itself.

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