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I want to compute the autocovariance of a Gausian zero-mean white noise process convolved with a function $f(t) \in L^2$. Could anyone show me how I'd do this or provide a reference?

$L^2$ is the space of square integrable functions on the real line $\mathbb{R}$ and the autocovariance of a zero mean process, $X(t)$, is $E[X(t_1)X(t_2)]$. where E is expectation.

Felix Benning
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ncRubert
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2 Answers2

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If the autocorrelation function of a zero-mean (stationary) Gaussian white noise process is $R_X(\tau) = E[X(t)X(t+\tau)] = K\delta(\tau)$, then the result $Y(t)$ of passing $X(t)$ through a filter with impulse response $h(t)$ is also a zero-mean stationary Gaussian process with autocorrelation function (and autocovariance function) given by $$R_Y(\tau) = E[Y(\lambda)Y(\lambda+\tau)] = K\int_{-\infty}^{\infty} h(t)h(t+\tau)\, \mathrm dt.$$

The above applies to Gaussian white noise as engineers understand the term. But, as I noted in this unanswered question, mathematicians also use the term (stationary) white noise to mean a process whose autocovariance function is given by $$\text{cov}(X(t), X(t+\tau)) = \begin{cases} \sigma^2, &\tau = 0,\\ 0, &\tau \neq 0, \end{cases}$$ in which case the description above does not apply. I would suppose that nonstationary white noise would mean $\text{var}(X(t))= \text{cov}(X(t), X(t))$ varies with $t$ instead of having fixed value $\sigma^2$ but I am not too sure about this.

Community
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Dilip Sarwate
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  • +1 Perhaps nitpicking: this assumes the white noise is stationary – leonbloy Aug 28 '12 at 20:39
  • @Dilip Sarwate I'm a bit confused by your response. I did mean stationary Gaussian white noise in the sense referred to by the question you link. However I expected the resulting process to be non-stationary, so I don't see how you can write the resulting autocorrelation in terms of a single variable t. Also, it's a bit confusing to use $\tau$ as a dummy variable of integration when writing about the autocorrelation as you usually reserve this to mean a time lag $t_1$ - $t_2$ – ncRubert Aug 28 '12 at 21:40
  • @ncRubert If the input to a linear time-invariant system is a wide-sense-stationary process, the output is also a wide-sense-stationary process. Since you specified convolution with a function in $L^2(\mathcal R)$, the natural assumption is that a time-invariant system is under consideration. Wide-sense-stationary Gaussian processes are also strictly stationary. Finally, I fixed the notation to suit your tastes. – Dilip Sarwate Aug 28 '12 at 23:24
  • @leonbloy Yes, I assumed stationarity. I am not sure how nonstationary white noise is defined. – Dilip Sarwate Aug 28 '12 at 23:48
  • @DilipSarwate What is $K$ in your equation ? Is it $\sigma^{2}$ – raK1 Sep 02 '16 at 02:12
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While Sarwate's answer does provide the correct result, the autocovariance function using a delta function definition is a bit problematic (as already mentioned in their answer). But the more formal definition offered afterwards is not correct either as it implies white noise is an uncountable collection of iid random variables (the induced function is not measurable with probability one so integration against it does not make sense).

My favourite formal definition is given by Adler and Taylor (2007, sec. 1.4.3) see also this anwer.

Using this definition and stochastic integration allows us to get the same result $$ \mathbb{E}[X(t_1)X(t_2)] = \int f(t_1 -s) f(t_2-s) ds $$ for the convolution $$ X(t) := (f * W) (t) = \int f(t-s) W(ds) $$ of white noise $W$, cf. Scalar product of a deterministic function with Gaussian white noise. Note that we do not require the domain of $f$ to be the real numbers here.

Felix Benning
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