# Approximating the variance of the integral of a white noise Gaussian process

Let $X(t)$ be a stationary Gaussian process with mean $\mu$, variance $\sigma^2$ and stationary correlation function $\rho(t_1-t_2)$. If $X(t)$ is a white noise process the correlation function is given by the Dirac delta function $\rho(t_1-t_2) = \delta(t_1-t_2)$.

The integral of this process is given by:

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

According to this CrossValidated post the variance of $I$ is given by:

$$\text{Var}[I] = L\sigma^2$$

However this does not agree with the results I obtained through simulation. The approach is to discretise the white noise Gaussian process into $N$ independent normal variables. The integral can then be approximated through:

$$I = \int_0^L X(t) \approx \frac{L}{N}\sum_{i=1}^NX_i$$

Where $X_i$ are indepedent random variables $X_i \sim \mathcal{N}(\mu,\sigma^2)$. In simulation I find that as $N$ grows large then $\text{Var}[I] \rightarrow 0$. Why does it not approach $L\sigma^2$? What is the problem with my approximation?

The disparity arises from the fact that your discretization of the continuous process does not assign the appropriate variance to the $X_i$. Here's the key (for heuristics, see here, Section 3.2):

If $\{X(t)\}_{t\in\mathbb{R}}$ is a continuous Gaussian process such that \begin{align}&E[X(t)]=0\\ &E[X(t)X(t')]=\sigma^2 \delta(t-t')\end{align} then a discrete sampling of the process, with a uniform sampling interval $\Delta$, viz., $$X(t),X(t+1\Delta),X(t+2\Delta),...$$ is simulated as an i.i.d sequence of Gaussian$(\text{mean}=0,\text{variance}=\frac{\sigma^2}{\Delta})$ random variables, the quality of the simulation increasing as $\Delta\to 0$.

(The power spectral density of the i.i.d. sequence approaches that of the continuous process that it simulates -- flat and equal to the same constant value $\sigma^2$ -- except that for the i.i.d. sequence it is "band limited", i.e. vanishing outside of a finite-width interval.)

Thus, in the present case, to simulate $$I = \int_0^L X(t) \, dt \approx \sum_{i=1}^N X(i\Delta)\,\Delta$$ where $\Delta=\frac{L}{N}$, one would use $$\hat{I} = \Delta\sum_{i=1}^N X_i$$ with $X_1,X_2,\ldots$ i.i.d. $\text{Gaussian}(\text{mean}=\mu,\text{variance}=\frac{\sigma^2}{\Delta})$. Then \begin{align} \text{Var}[\hat{I}] = \Delta^2\,N\,\frac{\sigma^2}{\Delta} = (N\Delta)\sigma^2 = L\sigma^2. \end{align}

I think somewhere between your first StackExchange post reference in your CrossValidated post there's been some confusion about the process in question.

It seems to me the answer to your CrossValidated post regards a Wiener process/Brownian motion with independent increments in the process. Or perhaps the person providing that answer mistook the integral of the process you describe for such a process.

To me your numerical result seems correct.

• This Wikipage states that the Wiener process $W(t)$ is the integral of a white noise Gaussian process (with unit variance?). So I am interested in the Wiener process at point L i.e. $W(L)$. The variance of the Wiener process is $t$ and so $\text{Var}[W[L]} = L$ and so if the white noise process has variance $\sigma^2$ then $\text{Var} = L\sigma^2$ which agrees with the cross-validated post. I think there is a problem with my approximation. @KeithWM
– egg
Aug 15 '16 at 13:50