2. Mean Square Continuity and Differentiability of a Stochastic Process

Having characterized a process through its autocovariance function \(R(\tau)\) in 1. Covariance Stationary Stochastic Processes, we now ask how the smoothness of \(R\) at the origin translates into the continuity and differentiability of \(x(t)\) itself. We begin by defining two notions of stochastic continuity. First, a stronger notion of continuity than we shall need is “continuity almost everywhere.”

Definition 3

The stochastic process \(x_t = x(t,w)\) is said to be continuous almost everywhere at the point \(t\) if

\[ \text{Prob}\, \left\{w: \qquad \lim_{\varepsilon \to 0}\, x(t + \varepsilon, w) = x(t, w) \right\} = 1. \]

Definition 3 states that the realizations of \(x(t,w)\) that are continuous at \(t\) have probability 1.

The weaker concept of continuity that we shall use is “mean square continuity.”

Definition 4

A stochastic process \(x_t = x(t,w)\) is said to be mean square continuous at the point \(t\) if

\[ \lim_{\varepsilon \to 0} \int \big(x(t + \varepsilon, w) - x(t, w)\big)^2 dP(w) = 0. \]

We immediately have the following theorem:

Theorem 1

The process \(x(t,w)\) is mean square continuous at \(t\) if and only if \(R(t_1, t_2)\) is continuous in \(t_1\) and \(t_2\) at \(t_1 = t_2 = t\).

Proof. Note that

\[ E\big(x(t+\tau) - x(t)\big)^2 = E x(t + \tau)^2 + E x(t)^2 - 2 E x(t + \tau) x(t) \]

or

\[ E\big(x(t + \tau) - x(t)\big)^2 = R(t + \tau, t + \tau) + R(t,t) - 2R(t + \tau, t). \]

Taking limits of both sides as \(\tau \to 0\) proves the theorem.

Next we have the definition:

Definition 5

A stochastic process \(x_t = x(t,w)\) is said to be mean square continuous if it is mean square continuous at each point \(t \in T\).

We immediately have:

Theorem 2

If a stochastic process \(x(t,w)\) is mean square continuous, then \(E x(t) = \mu(t)\) is continuous.

Proof. For any random variable \(z\), \(E z^2 =\) variance \(z + (Ez)^2 \geq (Ez)^2\). It follows that

\[ E\,\big\{x(t + \tau, w) - x(t, w)\big\}^2 \geq \big\{E\,\big(x(t + \tau, w) - x(t, w)\big)\big\}^2. \]

Taking limits as \(\tau \to 0\) proves the theorem.

We also have the following theorem:

Theorem 3

A covariance stationary stochastic process \(x(t,w)\) is mean square continuous if and only if its autocorrelation function \(R(\tau) = E x(t) x(t-\tau)\) is continuous for \(\tau = 0\).

Proof. Theorem 3 is implied by the proof of Theorem 1.

Next we turn to a concept of stochastic differentiation. It would be convenient to have a concept of differentiation that rationalized the following interchange of orders of integration and differentiation:

\[\begin{split} \begin{aligned} \frac{\partial^2}{\partial t_1\,\partial t_2}\ \ R(t_1, t_2) &= \frac{\partial^2}{\partial t_1\,\partial t_2}\ \int x(t_1, w)\, x(t_2, w)\, dP(w) \\ &= \int\ \frac{d}{dt}\ x(t_1, w)\ \frac{d}{dt}\ x(t_2, w)\, dP(w) \\ &= E\,\big\{\tfrac{d}{dt}\ x(t_1)\ \tfrac{d}{dt}\ x(t_2)\big\}. \end{aligned} \end{split}\]

or

\[ \frac{\partial^2}{\partial t_1\,\partial t_2}\ \ R(t_1, t_2) = E\,\big\{\tfrac{d}{dt}\ x(t_1)\ \tfrac{d}{dt}\ x(t_2)\big\}. \]

We shall see that the concept of mean square differentiability has this property.

Definition 6

A stochastic process \(x(t, w)\) is said to have a mean square derivative \(x'(t)\) at \(t\) if there is a random variable \(x'(t)\) such that

\[ \lim_{\epsilon \to 0}\ E\, \biggl\{ \Big( \frac{x(t + \epsilon) - x(t)}{\epsilon}\ - x'(t) \Big)^2\biggr\} = 0. \]

If \(x(t, w)\) has a mean square derivative, we say that \(x(t, w)\) is mean square differentiable.

We have the following Cauchy criterion for the mean square differentiability of \(x(t,w)\) at \(t\):

Criterion 1 (Cauchy criterion)

A process \(x(t,w)\) is mean square differentiable at \(t\) if and only if

\[ \lim_{\epsilon_1 \to 0,\ \epsilon_2 \to 0}\ E\biggl\{ \Big[ \frac{x(t + \epsilon_1) - x(t)}{\epsilon_1}\ - \ \frac{x(t + \epsilon_2) - x(t)}{\epsilon_2}\Big]^2\biggr\} = 0. \]

We have the following theorem:

Theorem 4

Let \(x(t,w)\) be a covariance stationary process with autocorrelation function \(R(\tau)\). If \(x'(t)\) exists in the mean square sense, then \(R''(0)\) exists.

Proof. First notice that

(4)\[\begin{split}\begin{aligned} E\, \biggl[ \Big( \frac{x(t + \epsilon) - x(t)}{\epsilon}\Big)^2 \biggr]\ &= \frac{2R(0) - 2R(\epsilon)}{\epsilon^2} \\ &=\ - \ \Big[ \frac{R(\epsilon) - 2R(0) + R(-\epsilon)}{\epsilon^2}\Big]. \end{aligned}\end{split}\]

Taking limits as \(\epsilon \to 0\), it follows that

(5)\[\lim_{\epsilon \to 0}\ E\, \Big( \frac{x(t + \epsilon) - x(t)}{\epsilon}\Big)^2\ = \ -R''(0),\]

for the limit on the left exists by the assumption of mean square differentiability.

We also have a converse of the above theorem:

Theorem 5

Let \(x(t,w)\) be a covariance stationary process with autocorrelation function \(R(\tau)\). If \(R''(0)\) exists, then \(x(t,w)\) is mean square differentiable.

Theorem 4 and Theorem 5 together say that, for a covariance stationary process, mean square differentiability is equivalent to the existence of \(R''(0)\). Nothing more than the second derivative at the origin is required; the proof below uses only that.

Proof. To apply the Cauchy criterion, we shall need to evaluate

\[\begin{split} \begin{aligned} E\, &\big\{ \frac{x(t + \epsilon_1) - x(t)}{\epsilon_1}\ \cdot\ \frac{x(t + \epsilon_2) - x(t)}{\epsilon_2}\big\} \\ &= \frac{R(\epsilon_1 - \epsilon_2) - R(\epsilon_1) - R(-\epsilon_2) + R(0)}{\epsilon_1\, \epsilon_2} \\ &= \frac{ \frac{R(\epsilon_1 - \epsilon_2) - R(\epsilon_1)}{\epsilon_2} \ + \ \frac{R(0) - R(-\epsilon_2)}{\epsilon_2} }{\epsilon_1}. \end{aligned} \end{split}\]

Taking limits first as \(\epsilon_2 \to 0\), then as \(\epsilon_1 \to 0\) gives

(6)\[\begin{split}\begin{aligned} \lim_{\epsilon_1,\, \epsilon_2 \to 0}\ &E\big\{ \frac{x(t + \epsilon_1) - x(t)}{\epsilon_1}\ \cdot\ \frac{x(t + \epsilon_2) - x(t)}{\epsilon_2}\big\} = \\ \lim_{\epsilon_1 \to 0}\ & -\ \frac{R'(\epsilon_1) - R'(0)}{\epsilon_1}\ =\ - R''(0). \end{aligned}\end{split}\]

Using (5) and (6), we find that

\[\begin{split} \begin{aligned} \lim_{\epsilon_1,\, \epsilon_2 \to 0}\ &E\biggl\{ \Big[ \frac{x(t + \epsilon_1) - x(t)}{\epsilon_1}\ - \ \frac{x(t + \epsilon_2) - x(t)}{\epsilon_2}\Big]^2\biggr\} \\ &= \ - 2R''(0) + 2R''(0) = 0. \end{aligned} \end{split}\]

Thus, if \(R''(0)\) exists, then \(x(t)\) is mean square differentiable.

Since \(R(\tau) = R(-\tau)\), it follows from the fact that \(R''(0)\) exists for a mean square differentiable process that \(R'(0) = 0\).

For a nonstationary stochastic process, the counterpart of the two preceding theorems is the following theorem, which we present without proof

Theorem 6

A nonstationary stochastic process \(x(t, w)\) is mean square differentiable if

\[ \frac{\partial^2 R(t_1, t_2)}{\partial t_1\,\partial t_2} \]

exists for \(t_1 = t_2\).

For a mean square differentiable process \(x(t, w)\), the following interchange of order of integration (expectation) and differentiation is appropriate:

\[\begin{split} \begin{aligned} E x'(t) &= E\ \lim_{\epsilon \to 0}\ \frac{x(t + \epsilon) - x(t)}{\epsilon} \\ &= \ \lim_{\epsilon \to 0}\ \frac{E x(t + \epsilon) - E x(t)}{\epsilon} \ = \ \frac{d}{dt}\ E x(t). \end{aligned} \end{split}\]

Thus,

\[ E x'(t) = \frac{d}{dt}\ \mu(t). \]

Thus, we obtain the mean function of \(x'(t)\) by once differentiating the mean function \(\mu(t)\) of \(x(t)\).

We can derive the autocorrelation function of \(x'(t)\) by twice differentiating the autocorrelation function of \(x(t)\). To establish this, we need some additional notation. We define the autocorrelation function of \(x(t)\) as

\[ E x(t_1)\, x(t_2) = R_{xx}(t_1, t_2). \]

We also define

\[ E x(t_1)\, x'(t_2) = R_{xx'}(t_1, t_2), \]

and

\[ E x'(t_1)\, x'(t_2) = R_{x'x'}(t_1, t_2). \]

First, we shall show that

(7)\[\frac{\partial R_{xx}(t_1, t_2)}{\partial t_2} = R_{xx'}(t_1, t_2).\]

To show this, note that

\[\begin{split} \begin{aligned} E &\ \big\{x(t_1) \frac{x(t_2 + \epsilon) - x(t_2)}{\epsilon} \big\} \\ &= \frac{R_{xx}(t_1, t_2 + \epsilon) - R_{xx}(t_1, t_2)}{\epsilon} \end{aligned} \end{split}\]

Taking limits as \(\epsilon \to 0\) gives the desired results (7). Similarly,

\[ R_{x'x'}(t_1, t_2) = \lim\, E\, \big\{ \frac{x(t_1 + \epsilon) - x(t_1)}{\epsilon}\ x'(t_2) \big\}\ = \frac{\partial R_{xx'}(t_1, t_2)}{\partial t_1} \]

Therefore, we have proved the desired result, which we state in the following theorem:

Theorem 7

If the stochastic process \(x(t,w)\) has mean square derivative \(x'(t)\), then the autocorrelation function of \(x'(t)\) is given by

\[ R_{x'x'}(t_1, t_2) = \frac{\partial^2 R(t_1, t_2)}{\partial t_1\,\partial t_2}. \]

For covariance stationary processes, we have the immediate corollary.

Corollary 1

If the covariance stationary stochastic process \(x(t, w)\) is mean square differentiable, the autocorrelation function of \(x'(t)\) is given by

\[ R_{x'}(\tau) = - R''(\tau) \]

where \(R(\tau)\) is the autocorrelation function of \(x(t)\).

By successively applying the preceding reasoning to \(x'(t)\) and each of its derivatives in an evident way, we can prove the following theorem:

Theorem 8

Let \(x(t, w)\) be a stochastic process with autocorrelation function \(R(t_1, t_2)\). The process is \(n\) times mean square differentiable if

\[ \frac{\partial^{2n} R(t_1, t_2)}{\partial t_1^n \partial t_2^n} \]

exists. The autocorrelation of the \(n^{th}\) mean square derivative process \(x^{(n)}(t)\) equals \(\partial^{2n} R(t_1, t_2)/ \partial t_1^n \partial t_2^n\).

As an example, consider a stochastic process for which

\[ R(\tau) = e^{\lambda |\tau|},\ \lambda < 0 \]

This process is mean square continuous, but not mean square differentiable. (Why?) Next, consider a process for which

\[ R(\tau) = k_1 e^{\lambda_1 |\tau|} + k_2 e^{\lambda_2 |\tau|};\ \lambda_1,\, \lambda_2\, < 0. \]

The process is mean square continuous, but is mean square differentiable only if \(\lambda_1\, k_1 + \lambda_2\, k_2 = 0\). (Why?)

The following construction demonstrates a link between the existence of mean square derivatives of arbitrarily high orders, and the predictability of a series. Let \(x(t, w)\) be a covariance stationary process with autocorrelation function \(R(\tau)\). We say that \(R(\tau)\) is analytic if its derivatives of all orders exist for all \(\tau\), and if \(R(\tau)\) has the Taylor (Maclaurin) series representation:

(8)\[ R(\tau) = \sum_{n=0}^{\infty} \ R^{(n)}\ (0)\ \frac{\tau^n}{n!} \]

We note that if \(R(\tau)\) is analytic, then for all integer \(n > 0\), the \(n^{th}\) mean square derivative \(x^{(n)}\,(t)\) exists; we can now state the following theorem.

Theorem 9

Let \(x(t, w)\) be a covariance stationary stochastic process with analytic autocorrelation function \(R(\tau)\). Then \(x(t)\) can be expanded in a Taylor series, i.e.,

\[ x(t + \tau) = \sum_{n=0}^{\infty} x^{(n)}(t) \frac{\tau^n}{n!},\ \text{ for all } \tau > 0. \]

Proof. We have to show that

(9)\[\begin{split} \begin{aligned} E\, \big\{ x(t + \tau) - \hat x(t + \tau) \big\}^2 &= E\, \big\{ [ x(t + \tau) - \hat x(t + \tau)] x(t + \tau) \big\} \\ &- E\, \big\{ [x(t + \tau) - \hat x(t + \tau) ] \hat x(t + \tau) \big\} = 0, \end{aligned} \end{split}\]

where

(10)\[ \hat x(t + \tau) = \sum_{n=0}^{\infty} x^{(n)}(t)\ \frac{\tau^n}{n!}. \]

From the analytic nature of \(R(\tau)\) it follows from (8) that

(11)\[ R^{(m)}(\tau) = \sum_{n=m}^{\infty} R^{(n)}(0)\ \frac{\tau^{n-m}}{(n-m)!}\, ,\ \text{ for } m \geq 1. \]

It also follows from a Taylor series of \(R(\tau + \lambda)\) about \(\lambda = 0\) that

(12)\[ R(0) = \sum_{n=0}^{\infty} R^{(n)}(\tau)\ \frac{(-\tau)^n}{n!}. \]

Substituting the right side of (10) into (9), noting by the reasoning that led to Theorem 8 that \(E x^{(n)}(t) x^{(m)}(t - \tau) = (-1)^m R^{(n + m)}(\tau)\), and using (11) and (12) to evaluate the two terms in braces in (9) gives the desired results.

The preceding states that if \(R(\tau)\) is analytic, then the stochastic process \(x(t)\) is differentiable an arbitrarily large number of times, and that \(x(t)\) is perfectly forecastable arbitrarily far into the future from values of \(x\) and its mean square derivatives at time \(t\). In our work, we shall usually want to deal with stochastic processes that are only imperfectly forecastable from knowledge of the past. This means that we shall usually deal with processes for which the autocorrelation function \(R(\tau)\) is not analytic. The differentiability criteria of this chapter take a sharper, more usable form once a process is written in its moving-average (Wold) representation: 9. Characterizations of Mean Square Differentiability and Mean Square Continuity shows that \(x(t)\) is mean square differentiable precisely when the moving-average kernel satisfies \(p(0) = 0\), and 13. Locally Unpredictable Stochastic Processes ties the failure of that condition to local unpredictability.