Max and min variance of the integral of a stationary stochastic process

by Jack Pierce-Brown   Last Updated August 15, 2016 08:08 AM

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?

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



Answers 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 $$ \int_0^L \int_0^L \delta(t_1-t_2)dt_1 dt_2= \int_0^L\bf 1_{[0,L]}(t_2)dt_2=L $$

DeltaIV
DeltaIV
August 15, 2016 11:17 AM

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