Let $(X_t)_{t \in \mathbb{R}}$ be a square-integrable real-valued process with a continuous mean value function $\mu:\mathbb{R}\rightarrow\mathbb{R}$ and a continuous covariance function $\Sigma:~\mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}$.

I want to show, that this process is stochastic continuous ($\lim_{s \rightarrow t} \mathbb{P}(|X_t - X_s| > \epsilon)=0~ \forall t \in \mathbb{R})$.

My first idea was using the markov inequality, then I would get

$\mathbb{P}(|X_t-X_s|>\epsilon) \leq \dfrac{\mathbb{E}(|X_t-X_s|^2)}{\epsilon^2} = \dfrac{\mathbb{E}(X_t^2-2X_tX_s+X_s^2)}{\epsilon^2} = \dfrac{\mathbb{E}(X_t^2)-2\mathbb{E}(X_tX_s)+\mathbb{E}(X_s^2)}{\epsilon^2}$.

Is this a helpful beginning of the proof or is there a better way? And how can I proceed?

Thanks in advance.

  • 1
    $\begingroup$ You are almost done. Manipulate the numerator in your last equation to get something that looks like $Cov(X_t,X_t)-Cov(X_t,X_s)$ and $Cov(X_s,X_s)-Cov(X_t,X_s)$. $\endgroup$
    – Michael
    May 7, 2015 at 5:24
  • $\begingroup$ Thank you for your answer! And when I have this, I can argue that $\lim_{s \rightarrow t}Cov(X_s,X_t)=Cov(X_t,X_t)$ and $\lim_{s \rightarrow t}Cov(X_s,X_s)=Cov(X_t,X_t)$ since the covariance function is continuous? Or do I have to argue in a different way? $\endgroup$
    – Max93
    May 7, 2015 at 5:29
  • 1
    $\begingroup$ Looks good to me. Of course you will have some extra terms (that are just as easy to take care of) after your manipulation. $\endgroup$
    – Michael
    May 7, 2015 at 5:30
  • $\begingroup$ When I do this manipulation, I get $Cov(X_t,X_t)-2Cov(X_s,X_t)+Cov(X_s,X_s)+(\mathbb{E}(X_s))^2-2\mathbb{E}(X_s) \mathbb{E}(X_t)+(\mathbb{E}(X_t))^2$ and since both the mean value function and the covariance function are continuous, I can interchange lim and $\mathbb{E}$ resp. lim and Cov and so I get the desired result, right? $\endgroup$
    – Max93
    May 7, 2015 at 5:45
  • 1
    $\begingroup$ There is no need to exchange limits and expectations. The value $\lim_{s\rightarrow t} X_s$ is just weird and not computable for this problem (it may not even exist). Even if it exists, you would need additional assumptions to ensure you can pass the limit through the expectation. So, don't do that. If it helps, you can think of $E[X_t]$ as a continuous function $m(t)$. You can think of $Cov(X_t,X_s)$ as a continuous function $f(t,s)$. This notation might help you realize that you are essentially done. For example, your expression becomes $f(t,t) - 2f(s,t) + \cdots + m(t)^2$. $\endgroup$
    – Michael
    May 7, 2015 at 7:22

1 Answer 1


Define $m(t)$ and $f(t,s)$ as the mean and covariance functions: \begin{align} m(t) &= E[X_t]\\ f(t,s) &= E[X_tX_s] - E[X_t]E[X_s] \end{align} We are told that the functions $m(t)$ and $f(t,s)$ are continuous. In general, if a function $h(t)$ is continuous, we know that $\lim_{s\rightarrow t} h(s) = h(t)$. Thus: \begin{align} &\lim_{s\rightarrow t} m(s) = m(t)\\ &\lim_{s\rightarrow t} f(t,s) = f(t,t) \end{align}

Note that once you have the functions $m(t)$ and $f(t,s)$, there is no randomness involved (and we can take limits as usual). This is in contrast to $\lim_{s\rightarrow t} X_s$, which is the limit of a random process. The limit may not exist. If it does exist, the limit is itself a random variable. We are not told anything about continuity of the random process $X_t$. It could be discontinuous even if its mean and covariance functions are continuous. See example below.

Example 1: Here is an example of a discontinuous process $X_t$ with continuous mean and covariance functions. Let $Y$ be a random variable that is exponentially distributed with rate $\lambda$. Define $X_t$ for all $t \geq 0$ as follows: $$ X_t = \left\{ \begin{array}{ll} 1 &\mbox{ if $t < Y$} \\ 0 & \mbox{ if $t \geq Y$} \end{array} \right.$$ Thus, $X_t$ starts out with the value 1, but flips discontinuously to 0 when $t$ crosses the random threshold $Y$.

Then for $t \geq 0$ we have: \begin{align} m(t) &= E[X_t] \\ &= Pr[X_t=1]\\ &= Pr[t < Y] \\ &= e^{-\lambda t} \end{align} Thus, $m(t)$ is continuous over $t \geq 0$.

Now for $t \geq 0$ and $s \geq 0$ we have: \begin{align} f(t,s) &= E[X_tX_s] - m(t)m(s) \\ &= Pr[X_t=1, X_s=1] - m(t)m(s) \\ &= Pr[\max[t,s] < Y] - m(t)m(s) \\ &= e^{-\lambda\max[t,s]} - e^{-\lambda t}e^{-\lambda s} \end{align} Thus, $f(t,s)$ is continuous over $t \geq 0$ and $s \geq 0$.

Example 2: Here is an example of a process $X_t$ that satisfies $E[\lim_{t\rightarrow\infty} X_t] \neq \lim_{t\rightarrow\infty} E[X_t]$. Again let $Y$ be exponentially distributed with rate $\lambda$. Define $X_t$ for all $t \geq 0$ by:

$$ X_t = \left\{ \begin{array}{ll} e^{\lambda t} &\mbox{ if $t < Y$} \\ 0 & \mbox{ if $t \geq Y$} \end{array} \right.$$ Thus, $X_t$ increases exponentially until it crosses the random threshold $Y$, at which point it (discontinuously) jumps down to $0$ and stays there. Since $Y$ is finite with probability 1, we know with probability 1 that $\lim_{t\rightarrow\infty} X_t = 0$. Thus, $E[\lim_{t\rightarrow\infty} X_t] = 0$. However, for all $t\geq 0$ we have: $$ E[X_t] = e^{\lambda t} Pr[t < Y] = 1 $$ Thus, $\lim_{t\rightarrow\infty} E[X_t] = 1$. And $1 \neq 0$.

Example 3: Here is an example of a process $X_t$ such that $\lim_{t\rightarrow 0} X_t$ does not exist, but $\lim_{t\rightarrow 0} E[X_t] = 0$. Let $A$ be a random variable that takes values in the set $\{-1,1\}$ and satisfies $Pr[A=1]=Pr[A=-1]=1/2$ (and so $E[A]=0$). Define: $$ X_t = \left\{ \begin{array}{ll} A\cos(1/t) &\mbox{ if $t \neq 0$} \\ 0 & \mbox{ if $t =0$} \end{array} \right.$$ Then $E[X_t]=0$ for all $t$, so $\lim_{t\rightarrow 0} E[X_t]=0$. But $\lim_{t\rightarrow 0} X_t$ does not exist because $X_t$ infinitely oscillates between $-1$ and $1$ as $t\rightarrow 0$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .