Personal tools
You are here: Home Electrical and Computer Engineering Stochastic Processes Analytic Properties of Random Processes

Analytic Properties of Random Processes

Document Actions
  • Content View
  • Bookmarks
  • CourseFeed

Properties  ::  Continuity  ::  Differentiation  ::  Integration


Recall that $f: T \rightarrow \Rbb$ is differentiable at $t_0
\in T$ if

\begin{displaymath}\lim_{t \rightarrow t_0} \frac{f(t) - f(t_0)}{t-t_0} = f'(t_0)


Similarly, a r.p. $\{X_t, t\in T\}$ is mean-square differentiable at $t_0$ if

\begin{displaymath}X_{t_0}' = \lim_{t \rightarrow t_0} \frac{X_t - X_{t_0}}{t-t_0}

exists in the mean-square sense, that is,

\begin{displaymath}E[\left\vert X_{t_0}' - \frac{X_t - X_{t_0}}{t-t_0}\right\vert^2]
\rightarrow 0 \text{ as } t \rightarrow t_0.

If $X_t$ is mean-square differentiable at every $t_0
\in T$ , then $X_t'$ defines another random process on the underlying probability space $(\omega,\Fc,P)$ .

Suppose $Y_t$ is a second-order random process, and $\lim_{
t\rightarrow t_0} Y_t = Z$ (m.s.). Then:

  1. $E[Z^2] < \infty$ ;
  2. and if $E[X^2] < \infty$ then

    \begin{displaymath}\lim_{t \rightarrow t_0} E[Y_t X] = E[ZX].

\item $Z = Y_t + Z-Y_t$. Then since $(a+b)^2 \...
...t-Z)^2]E[X^2] \\
&= 0.

Properties of the derivative

Suppose $\{X_t, t\in T\}$ is mean-square differentiable with mean-square derivative $\{X_t',t \in T\}$ . Suppose that $\{X_t\}$ is second order. Then:

  1. $\{X_t'\}$ is also second order. (This follows from the first fact above.)
  2. $\partiald{}{t} R_X(t,s)$ exists, and equals $R_{X',X}(t,s)$ .
R_{X'X}(t,s) &= E[X_{t}'X_s] =...
...{q-t} \\
&= \partiald{}{t} R_X(t,s).
\end{aligned}\end{displaymath} \end{proof}
  3. $\partiald{}{s}\partiald{}{t} R_X(t,s)$ exists and is equal to $R_{X'}(t,s)$ for all $t,s \in T$ .
R_{X'}(t,s) &= E[X_t'X_s'] = E[...
...tiald{}{s} \partiald{}{t} R_X(t,s).
  4. $\partiald{}{t} \mu_X(t)$ exists and is equal to $\mu_{X'}(t)$ .
  5. Suppose $T=\Rbb$ and $X_t$ is also W.S.S. Then $\{X_t'\}$ is also W.S.S. Also, $\{X_t\}$ and $\{X_t'\}$ are jointly W.S.S. Also,
    1. $\mu_x' = 0$ .
    2. $R_{X'X}(\tau) = \frac{d}{d\tau} R_X(\tau).$
    3. $R_{X'}(\tau) = -\frac{d^2}{d\tau^2} R_X(\tau)$ .

On the existence of the mean-square derivative

It can be shown that a sufficient condition for the existence of $X_{t_0}'$ is the existence of $\partialw{}{t}{s} R_X(t,s)$ at $(t,s)
= (t_0,t_0)$ . A necessary condition is the existence and equality of the mixed partials

\begin{displaymath}\partiald{}{t} \left[ \partiald{}{s} R_X(t,s)\right]_{(t=s=t_...
...rtiald{}{s} \left[ \partiald{}{t} R_X(t,s)\right]_{(t=s=t_0)}.

If $X_t$ is W.S.S., then these two conditions are the same. So $X_{t_0}'$ exists if and only if

\begin{displaymath}\left\vert \frac{d^2}{d\tau^2} R_X(\tau)\right\vert _{\tau=0} < \infty.

By a result we will show later, we will find that

\begin{displaymath}R_{X'}(0) = -R_X(0) \int_{-\infty}^\infty \omega^2 S_X(\omega).

So the existence of $X_t'$ for a WSS process $X_t$ is equivalent to the condition that the second moment of the PSD of $X_t$ is finite.
Let $X_t$ be a process having
\begin{displaymath}S_X(\omega) =...
...isplaymath}so the process is {\em not} mean-square differentiable.
Copyright 2008, by the Contributing Authors. Cite/attribute Resource . admin. (2006, June 07). Analytic Properties of Random Processes. Retrieved January 07, 2011, from Free Online Course Materials — USU OpenCourseWare Web site: This work is licensed under a Creative Commons License Creative Commons License