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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 $\mathcal{R}$ and the autocovariance of a zero mean process, $X(t)$, is $E[X(t_1)X(t_2)]$. where E is expectation.

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up vote 3 down vote accepted

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.

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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

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