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

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

The variance of $I$ is given by (see StackExchange):

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

Where $\tau = t_1 - t_2$. The variance of $I$ is maximised when $X(t)$ is perfectly correlated i.e. $\rho(t_1-t_2)=1$:

$$\text{Var}[I_{corr}] = \sigma^2 \int_0^L \int_0^L 1\,\mathrm{dt_1\,dt_2} = \sigma^2L^2$$

The variance of $I$ is minimised when $X(t)$ is uncorrelated (white noise process) i.e. $\rho(t_1-t_2) = \delta(t_1-t_2)$ where $\delta$ is the Dirac delta function. Can you evaluate the following integral?

$$\text{Var}[I_{uncorr}] = \sigma^2 \int_0^L \int_0^L \delta(t_1-t_2)\,\mathrm{dt_1\,dt_2} = $$


1 Answer 1


Not sure this is actually in-topic for Cross-validated (sounds more appropriate for the Mathematics forum), anyway the answer is $L$. Explanation: $$\int_0^L \delta(t_1-a)dt_1 = \bf 1_{[0,L]}(a)$$ Thus $$ \int_0^L \int_0^L \delta(t_1-t_2)dt_1 dt_2= \int_0^L\bf 1_{[0,L]}(t_2)dt_2=L $$

  • $\begingroup$ Interesting. $\text{Var}[I_{uncorr}] = \sigma^2L$? $\endgroup$
    – egg
    Commented Aug 15, 2016 at 11:21
  • $\begingroup$ Yes, that's it. $\endgroup$
    – DeltaIV
    Commented Aug 15, 2016 at 11:44
  • $\begingroup$ This seems strange to me. I find it hard to believe that if $L = 1$ the variance of the integral is independent of the correlation structure of the process. $\endgroup$
    – egg
    Commented Aug 15, 2016 at 12:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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