5
$\begingroup$

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 $I$?

$\endgroup$
5
$\begingroup$

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 and the variance of $I$ is then $\operatorname{Var}I =E(I^2)-(EI)^2$.

$\endgroup$
  • $\begingroup$ Excellent. This makes sense to me. I will try to verify this with simulation. $\endgroup$ – 7Jack Aug 1 '16 at 13:43
  • 2
    $\begingroup$ 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. $\endgroup$ – Dilip Sarwate Aug 2 '16 at 17:52
  • 1
    $\begingroup$ 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. $\endgroup$ – 7Jack Aug 8 '16 at 14:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.