What is meant by a continuous-time white noise process?

In a discussion following a question a few months ago, I stated that as an engineer, I am used to thinking of a continuous-time wide-sense-stationary white noise process $\{X(t) \colon -\infty < t < \infty\}$ as a zero-mean process having autocorrelation function $R_X(\tau) = E[X(t)X(t+\tau)] = \sigma^2\delta(\tau)$ where $\delta(\tau)$ is the Dirac delta or impulse, and power spectral density $S_X(f) = \sigma^2, -\infty < f < \infty$. At that time, several people with very high reputation on Math.SE assured me that this was an unduly restrictive notion, and that no difficulties arise if one takes the autocorrelation function to be $$E[X(t)X(t+\tau)] = \begin{cases}\sigma^2, & \tau = 0,\\ 0, & \tau \neq 0. \end{cases}$$

What engineers like to call a white noise process is a hypothetical beast that is never observed directly in any physical system, but which can be used to account for the fact that the output of a linear time-invariant system whose input is thermal noise is well-modeled by a wide-sense-stationary Gaussian process whose power spectral density is proportional to $|H(f)|^2$ where $H(f)$ is the transfer function of the linear system. Standard second-order random process theory says that the input and output power spectral densities $S_X(f)$ nd $S_Y(f)$ are related as $$S_Y(f) = S_X(f)|H(f)|^2.$$ Thus, pretending that thermal noise is a white Gaussian noise process in the engineering sense and pretending that the second-order theory extends to white noise processes (even though their variance is not finite) allows us to get to the result that the output power spectral density is proportional to $|H(f)|^2$.

My query about the definition of a white noise process is occasioned by a more recent question regarding the variance of a random variable $Y$ defined as $$Y = \int_0^T h(t)X(t)\ \mathrm dt$$ where $\{X(t)\}$ is a white Gaussian noise process. The answer given by Nate Eldredge leads to $$\operatorname{var}(Y) = \sigma^2 \int_0^T |h(t)|^2\ \mathrm dt$$ (as I pointed out in a comment on the answer) if the autocorrelation function is taken to be $R_X(\tau) = \sigma^2\delta(\tau)$ (the engineering definition). However, the OP on that question specified $R_X(0) = \sigma^2$, not $\sigma^2\delta(\tau)$, that is, the definition accepted by mathematicians. For this autocorrelation function, the variance is $$\int_0^T \int_0^T E[X(t)X(s)]h(t)h(s)\mathrm dt\mathrm ds = 0$$ since the integrand is nonzero only on a set of measure $0$.

So, what is the variance of the random variable $Y$? and what do readers of Math.SE understand by the phrase white noise process?

Perhaps this question should be converted to a Community wiki?

  • $\begingroup$ CW-hammered per OP request. $\endgroup$ Commented Apr 24, 2012 at 12:38
  • 9
    $\begingroup$ Short answer: (1.) $Y$ is undefined; (2.) the RHS of $Y$ is a shorthand for a mathematically elaborate object called stochastic integral; (3.) applying to this object operations valid for classical (deterministic) integrals can lead to chaos. $\endgroup$
    – Did
    Commented Jan 15, 2013 at 7:16
  • $\begingroup$ I’m not the best at this, but I feel that there may be a confusion about $\delta$. My understanding is that $\delta(x)$ is infinity if $x=0$ and $0$ otherwise, such that $\int_a^b\delta(x)dx=1$ if $a<0<b$. This function should be consistent with the mathematical definition, but is usually not used by mathematicians. $\endgroup$
    – Teepeemm
    Commented Jun 8, 2014 at 20:27
  • $\begingroup$ I think the notation $\sigma^2$ in the definition of the autocorrelation function is somewhat misleading because $\sigma$ is usually used to denote the standard deviation, and the $\sigma$ of a continuous-time white noise is necessarily infinite. The coefficient can be replaced by $A$ or something else to denote the power spectral density of the white noise, which is constant. $\endgroup$
    – Hermis14
    Commented Apr 4, 2022 at 16:08
  • $\begingroup$ The (mathematically not rigorous) formulas from this question were just mentioned in another one as a reference. Perhaps would be good to have some disclaimer about them being used in engineering, but not in mathematics, in case I understand the issue correctly. $\endgroup$
    – SBF
    Commented May 25, 2023 at 12:16

2 Answers 2


This is a bit late, but I see that the main points in this question have not been completely addressed. I'll set \begin{equation} \sigma = 1 \end{equation} for this answer.

The definition of white noise may be context-dependent: How you define it depends on what you want to do with it. There's nothing inherently wrong with saying that white noise (indexed by a set $T$) is just the process of iid standard normal random variables indexed by $T$, i.e. $E[X(t)X(s)] = \begin{cases} 1 & t = s \\ 0 & t \neq s \end{cases}.$ However, as cardinal noted here, Example 1.2.5 of Kallianpur's text shows that this process is not measurable (as a function of $(t, \omega)$). This is why, as Did commented above, $Y$ is undefined (with this definition of $X$). Thus, this definition of white noise is not appropriate for defining objects like $Y$.

Rather, you want $X$ to have covariance given by the Dirac delta. But the $\delta$ function is not a function but rather a measure and the best context for understanding it is the theory of distributions (or generalized functions---these are not to be confused with "probability distributions"). Likewise, the appropriate context for white noise is the theory of random distributions.

Let's warm up with a heuristic explanation: We'll think of white noise as the "derivative" of Brownian motion: "$dB_t/dt = X_t$". So ignoring rigor for a moment, we could write \begin{equation} \int_0^T h(t) X(t) dt = \int_0^T h(t) \frac{dB_t}{dt} dt = \int_0^T h(t) dB_t. \end{equation}

The reason this isn't rigorous is that Brownian motion is nowhere differentiable. However, the theory of distributions allows us to "differentiate" non-differentiable functions. First of all, a distribution is a linear functional (linear map taking values in the real numbers) on a space of "test functions" (usually smooth functions of compact support). A continuous function $F$ can be viewed as a distribution via the pairing \begin{equation} (F, f) = \int_0^\infty F(t) f(t) dt. \end{equation} The distributional derivative of $F$ is the distribution $F'$ whose pairing with a test function $f$ is defined by \begin{equation} (F', f) = -(F, f'). \end{equation}

Thinking of Brownian motion as a random function, we can define white noise $X$ as its distributional derivative. Thus, $X$ is a random distribution whose pairing with a test function $f$ is the random variable \begin{equation} (X, f) = -(B, f') = -\int_0^\infty B(t) f'(t) dt. \end{equation} By stochastic integration by parts, \begin{equation} (X, f) = \int_0^\infty f(t) dB_t; \end{equation} this is the Itô integral of $f$ with respect to $B$.

Now a well-known fact in stochastic calculus is that $M_T = \int_0^T f(t) dB_t$ is a martingale starting at $M_0 = 0$, so $E (X, f) = 0$. Moreover, by the Itô isometry, \begin{equation} \mathrm{Var}((X, f)) = E (X, f)^2 = \int_0^\infty f(t)^2 dt. \end{equation} It can also be verified that $(X, f)$ is Gaussian.

My main point is that a more appropriate definition of $Y$ might be \begin{equation} Y = \int_0^T h(t) dB_t. \end{equation}

As a last note, because of the way $X$ was defined above, $X_t$ is not defined but $(X, f)$ is. That is, $X$ is a stochastic process but whose index set is given by $T = \{ \text{test functions} \}$ rather than $T = [0, \infty)$. Moreover, again by the Itô isometry, \begin{equation} E (X, f) (X, g) = \int_0^\infty f(t) g(t) dt. \end{equation} Abandoning rigor again, this becomes \begin{equation} E (X, f) (X, g) = \int_0^\infty \int_0^\infty f(s) \delta(s - t) g(t) ds dt \end{equation} and it is in this sense that the covariance of $X$ is the Dirac delta.

Edit: Note that we could leave the definition of $(X, f)$ in terms of the ordinary integral and do all the above calculations using Fubini's theorem and (ordinary) integration by parts (it's just a bit messier).

  • $\begingroup$ Is the cross variation term that pops up in the stochastic integration by parts equal $0$? And if so why? $\endgroup$
    – alpastor
    Commented Sep 4, 2019 at 23:43
  • $\begingroup$ Yes, because any cross-variation involving a process that is continuous and of finite total variation is 0. Typically, a test function $f$ would be smooth with bounded support, so would satisfy this requirement. $\endgroup$
    – Ben
    Commented Sep 5, 2019 at 13:02
  • $\begingroup$ @BenCW So, in your def $(X,f) = \int_{0}^{T} f(t) dB_{t}$, how should I choose the function $f(t)$? What if I choose it as $f(t) = t$ or $f(t) = 1$ or $f(t) = \delta(t)$? These give all different answers, but honestly, I can't see the physical intuition. I want to choose $f(t)$ such that I get the derivative of Brownian motion back out of integral -- ie I want to get white noise out. So what should I do? $\endgroup$ Commented Nov 27, 2019 at 12:27
  • $\begingroup$ @BenCW So for example if I want to define a white noise process with variance $\sigma^{2}$, how would I do that using this distributional definition? Would I define $f(t) = \sigma^{2}$? $\endgroup$ Commented Nov 27, 2019 at 12:49
  • $\begingroup$ @the_src_dude I'm not sure I understand what you're asking about the pairing. When $X$ is fixed (as in this case, where $X$ is a white noise process), you can think of $(X, f)$ as defining a mapping on possible test functions $f$. So $f$ here is just a dummy variable and you don't have to choose it to be anything. In other words, $(X, f)$ should depend on $f$ and typically would have different values for different choices of $f$. The thing that stays constant is $X$. $\endgroup$
    – Ben
    Commented Dec 1, 2019 at 15:18

Actually X is the digital signal and Y is the average analog signal power generated in the time-domain. The variance shows how deviation goes if signal Y goes for a long time.

Moreover, you should learn more about the delta function (your function becomes one if x = 0 ) . The engineering and mathematical aspects to explain the autocorrelation function are also correct. There is no contracdiction between them.


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