# Mean Square Continuity of Random Process

Show that a stochastic process $X(t)$ is mean square continuous if and only if its autocorrelation function $R_X(t_1,t_2)$ is continous

$\Rightarrow$ Proof:

We have $E[(X(t)-X(t_0))^2]=R_X(t,t)-R_X(t_0,t)-R_X(t,t_0)+R_X(t_0,t_0)$, so if $t \rightarrow t_0$ the right side of the equation is equal to $0$, therefore $\Rightarrow$ implication is true.

$\Leftarrow$ Proof:

This is where I have trouble.

Is this implication true due to the fact that if $X(t)$ is continuous at $t_0$ in m.s then $\lim_{t \to t_0} m_X(t)=m_X(t_0)$ ??

• For the proof of the other direction, see here: arxiv.org/abs/1509.07526 Jan 21, 2023 at 15:50

The whole point of the concept of mean square continuity is to avoid strict assumptions about the continuity of every realization of $X(t)$.

The given proof simply shows that continuity of the autocorrelation function implies that the limit $\lim_{t\rightarrow t_0}E\{[X(t)-X(t_0)]^2\}$ exists and equals zero.

Note that continuity of the mean of $X(t)$ is a consequence of mean square continuity. Since

$$E\{[X(t)-X(t_0)]^2\}\ge E^2\{X(t)-X(t_0)\}\tag{1}$$

it follows from

$$\lim_{t\rightarrow t_0}E\{[X(t)-X(t_0)]^2\}=0$$

that

$$\lim_{t\rightarrow t_0}E\{X(t)-X(t_0)\}=0$$

must hold, i.e., the mean of $X(t)$ is continuous.

• Although this is a good answer, it doesn't answer the OP's question. Jan 21, 2023 at 15:52