# Skewness of the integral of a stochastic process

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)?

• Is $X(t)$ a Gaussian process? If so $I$ is also Gaussian with zero skew. Aug 8, 2016 at 16:52
• I am interested in a lognormal process $X(t)$. My understanding is that it is not possible to obtain the probability density function of the sum of lognormal random variables. So I am trying to use the moments (mean, variance etc) instead. It is my suspicion that as the autocorrelation of $X(t)$ decreases the skewness of the distribution of $I$ decreases. Heuristically the distribution of $I$ looks more normal and less lognormal as the autocorrelation of $X(t)$ decreases. If the $\rho(s_1,s_2)$ is the Dirac delta function then $X(t)$ is a white-noise process and $I$ will be a constant.
– egg
Aug 9, 2016 at 8:50

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.
• $$E[X(t)X(u)X(v)] = M(1,1,1) = \exp\left(\mu(1+1+1) + \frac{1}{2}(1,1,1)^T\boldsymbol\Sigma(1,1,1)\right)$$ $$= \exp\left(3\mu + \frac{\sigma^2}{2}(1,1,1)^T\boldsymbol\rho(1,1,1)\right)$$ $$\boldsymbol\rho = \begin{bmatrix} 1 & \rho(t-u) & \rho(t-v)\\ \rho(t-u) & 1 & \rho(u-v) \\ \rho(t-v) & \rho(u-v) & 1 \end{bmatrix}$$ $$= \exp\left(3\mu + \frac{\sigma^2}{2}\left(3+2(\rho(t-u) + \rho(t-v)+\rho(v-u)\right)\right)$$ I think I need to triple integrate this now.
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.
• There are too many expectation operators and misplaced square brackets. Also, you cannot separate $E[X(t)X(u)X(v)]$ into $E_X\left[E[X(t)X(u)]X(v)\right]$ Aug 30, 2016 at 15:05