Skewness of the integral of a stochastic process

Question

Consider the integral of a stationary random process $X(t)$ with mean $\mu$, variance $\sigma^2$ and stationary correlation function $\rho(t_1 - t_2)$:

$$I = \int_0^L X(t)\,\mathrm{dt}$$

In my previous post the moments of $I$ was found to be:

$$E[I] = \mu_I = L\mu$$

$$\text{Var}[I] = \sigma_I^2 =\sigma^2 \int_0^L \int_0^L \rho(t_1-t_2)\,\mathrm{dt_1\,dt_2} = 2\sigma^2\int_{0}^{L} \rho(\tau)(L-|\tau|)\,\mathrm d\tau$$

Where $\tau = t_1 - t_2$. I would now like the calculate the skewness of $I$:

$$\text{Skew}[I] = E\left[\left(\frac{I-\mu_I}{\sigma_I}\right)^3\right] = \frac{E[I^3] - 3\mu_I\sigma_I^2 - \mu_I^3}{\sigma_I^3}$$

I know $\mu_I$ and $\sigma_I^2$ so the problem is to evaluate $E[I^3]$:

$$E[I^3] = E\left[\int_0^L X(t)\,\mathrm{dt}\int_0^L X(u)\,\mathrm{du} \int_0^L X(v)\,\mathrm{dv}\right]$$

$$ = E\left[\int_0^L \int_0^L \int_0^L X(t) X(u)X(v)\,\mathrm{dt}\,\mathrm{du}\,\mathrm{dv}\right]$$

$$ = \int_0^L \int_0^L \int_0^L E\left[X(t) X(u)X(v)\right]\,\mathrm{dt}\,\mathrm{du}\,\mathrm{dv}$$

In the previous post the expected value was then written in terms of the covariance function however I don't know how to proceed here.

Can you show me how to evaluate the 3rd moment (if possible)?

Answer 1

The above expectation inside your triple integral can be found via the following trick.

When $X(t)$, as you say, is a log-normal process, then $(Y(r),Y(s), Y(t)) = (\ln X(r), \ln X(s), \ln X(t))$ is multinormal with moment generating function $$ M(u,v,w) = E(e^{u Y(r) + v Y(s) + w Y(t)})=e^{\mu(u+v+w) + \frac12(u,v,w)^T\Sigma (u,v,w)} $$ where $\Sigma$ is the covariance matrix of $Y(r),Y(s),Y(t)$. Presumably you know the elements of $\Sigma$ if you know the autocovariance function of $Y(t)=\ln X(t)$? From this you can work out the above expectation since this equals $$ E(X(r)X(s)X(t)) = E(e^{Y(r)+Y(s)+Y(t)}) = M(1,1,1), $$ by definition of the moment generating function.

Answer 2

You could appeal to Fubini's theorem to rewrite the third moment as: $$E[X^3] = E_X\left[\int_0^L\int_0^L\int_0^L E[X(t)X(u)]X(v) dt dudv \right],$$ and apply what you have already calculated for $E[X(t)X(u)]$. You may end up with a not-so-nice expression.