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:
$$\operatorname{Var}\left( \sum_{i=1}^N X_i\right) = N\sigma^2+2\sigma^2\sum_{1\le i<j\le N}\mathbf{\rho}(X_i,X_j)$$
Source: Wikipedia – Variance – Sum of correlated variables
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 $I$?
Best Answer
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 $\operatorname{Var}I =E(I^2)-(EI)^2$.