# Variance of the integral of a stochastic process

I understand the discete case i.e. the sum of $N$ identically distributed random variables $X_i$ with variance $\sigma^2$. The correlation between these random variables is given by the correlation matrix $\mathbf{\rho}(X_i,X_j)$.

The variance of the linear combination of random variables $X_i$ is given by:

$$\operatorname{Var}\left( \sum_{i=1}^N X_i\right) = N\sigma^2+2\sigma^2\sum_{1\le i<j\le N}\mathbf{\rho}(X_i,X_j)$$

I would like to consider the continuous case of a stochastic process which will be denoted as $X(t)$. The process is stationary with constant variance $\sigma^2$ and correlation function $\rho(X(t),X(h)$. Similar to above I would like to calculate the variance of the linear combination of the random variables $X(t)$. I think that the linear combination over some domain $t \in [0,L]$ can be expressed as

$$I = \int_0^L X(t) dt$$

I would like to know the variance of $I$:

$$\operatorname{Var}(I)=?$$

I speculate that if the process $Z(X_i)$ is completely correlated i.e. $\rho(X(t),X(h)) = 1$ then the variance of $I$ is minimised maximised and is given by: $$\operatorname{Var}(I)=L^2\sigma^2$$

If the variables are uncorrelated i.e. $\rho(X(t),X(h))=0$ then I suspect that the variance of $I$ is maximised minimised. It may be infinite zero?

I find the continuous case (i.e. infinite random variables over some domain [0,L]) difficult to understand. Could anyone provide me an expression for the variance of $I$?

Interchanging the order of integration and expectation you get $$E(I)=E\int_0^L X(t) dt = \int_0^L EX(t) dt = \int_0^L \mu dt = L\mu$$ and similarly, the second moment of $$I$$ becomes \begin{align} E(I^2)&=E\left(\int_0^LX(t)dt\int_0^LX(u)du\right) \\ &= E \int_0^L \int_0^L X(t)X(u)du dt \\ &= \int_0^L \int_0^L E[X(t)X(u)]du dt \\ &= \int_0^L \int_0^L [\operatorname{Cov}(X(t),X(u))+EX(t)EX(u)] du dt \\ &= \sigma^2 \int_0^L \int_0^L \rho(t-u)dudt + L^2 \mu^2. \end{align} If the correlation function $$\rho(h)$$ is for example exponential the double intergral can be easily solved. The variance of $$I$$ can be found from the general formula $$\operatorname{Var}I =E(I^2)-(EI)^2$$.

• Excellent. This makes sense to me. I will try to verify this with simulation. – egg Aug 1 '16 at 13:43
• This assumes that the process is wide-sense stationary or weakly stationary. Also, for this special case, the double integral simplifies to a single integral $$\int_{-L}^{L} \rho(s)(L-|s|)\,\mathrm ds$$ which is easier to work with. – Dilip Sarwate Aug 2 '16 at 17:52
• Good comment. Integration using Wolfram Mathematica gives the same result using both expressions for $$\rho(s) = \exp\left(-\pi\left(\frac{s}{\theta}\right)^2\right)$$ Is it correct that your expression works for all isotropic correlation functions i.e. $s = |t - u|$? Could you explain how you got to your expression? My understanding of double integrals is poor. – egg Aug 8 '16 at 14:33
• Even if unrelated to the question itself, I find a little confusing the statement $\text{Var}[I]=\text{E}[I^2]-\text{E}[I]^2$ as a consequence of the exponential correlation. Isn't it true for any random variable? – Riccardo Buscicchio Jan 10 at 11:49
• @RiccardoBuscicchio Yes – Jarle Tufto Jan 10 at 16:06