Assume a very simple LTI system. Assume $x$ is white Gaussian i.i.d. with variance $\sigma^2$.

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The output variance is straightforward to obtain. For example, for a continuous-time system:

$$\mbox{Var}(y) = \mathbb{E}(y^2) = \sigma^2 \int_0^{\infty}|H(s)|^2 \,\mathrm d s$$

However, I am interested in higher order moments, for example, $\mathbb{E}(y^4)$. How can this be obtained (is there a closed form solution in the first place)?

PS: Note that $\mathbb{E}(y^4)\neq\sigma^4 \int_0^{\infty}|H(s)|^4 \, \mathrm d s$.


1 Answer 1


For the case of input process $\{X(t)\}$ being white Gaussian noise with two-sided power spectral density $\frac{N_0}{2}$, the output process $\{Y(t)\}$ is a strictly stationary zero-mean Gaussian process in which all the random variables have the same variance $\frac{N_0}{2}\int_{-\infty}^\infty |H(f)|^2 \,\mathrm df$ almost as you say. But the key point here is that the $Y$'s are zero-mean Gaussian random variables with known variance, and it is known that if $Y \sim N(0,\sigma_Y^2)$ then $$E[Y^n] = \begin{cases}\sigma_Y^n \cdot (n-1) \times (n-3) \times (n-5)\times \cdots \times 3 \times 1, & n ~ \text{even},\\ 0, & n ~ \text{odd}.\end{cases}$$

Now, if the input process is just white noise (not necessarily Gaussian white noise), then the output process is not necessarily Gaussian, and while the mean and variance are as specified above, we cannot say anything about the distribution of the $Y(t)$'s and cannot infer anything about the higher moments of the $Y(t)$'s the way we could for Gaussian output processes.

If the input process is a zero-mean WSS Gaussian process (which is also a strictly stationary process) but not necessarily a _white noise process, than $\{Y(t)\}$ is also a zero-mean WSS Gaussian process (and so strictly stationary too) with variance $$\sigma_Y^2 = \int_{-\infty}^\infty S_X(f)|H)f)|^2 \,\mathrm df$$ where $S_X(f)$ is the power spectral density of the input process and so once again we are in business and can use $$E[Y^n] = \begin{cases}\sigma_Y^n \cdot (n-1) \times (n-3) \times (n-5)\times \cdots \times 3 \times 1, & n ~ \text{even},\\ 0, & n ~ \text{odd}.\end{cases}$$

  • $\begingroup$ Why not mention that $X$ is also zero-mean? I am referring to the 1st sentence. $\endgroup$ Mar 15, 2019 at 21:52
  • $\begingroup$ @RodrigodeAzevedo. I very specifically avoided making the statement that $\{X(t)\}$ is a zero-mean process but I will make the statement about $\{Y(t)\}$ more explicit. $\endgroup$ Mar 16, 2019 at 20:36
  • $\begingroup$ If I may ask, why is $Y$ zero-mean if $X$ is not necessarily so? $\endgroup$ Mar 16, 2019 at 20:40
  • $\begingroup$ @RodrigodeAzevedo White Gaussian noise is a mythical beast defined by its action: the noise at output of a LTI system with transfer function is a zero-mean Gaussian process with PSD proportional to $|H(f)|^2$.We pretend that we can fit this observation into theory of WSS process by imagining that the input process is a zero-mean Gaussian process with autocorrelation function $K\delta(t)$ but that is different from asserting that the input process has zero mean or that the random variables comprising the input process are Gaussian etc. $\endgroup$ Mar 17, 2019 at 1:04

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