# The continuity of the expectation of a continuous stochastic procees

Let $X_t$ be a continuous stochastic process on a filtered space $(\Omega, \mathcal F, \mathcal F_t, \mathbb P)$.

Is $\mathbb E[X_t]$ necessarily a continuous function?

My first answer would be no. For example if $X_t$ admits densities $f(t,x)$, the first equality in:

$$\lim_{t \rightarrow t_0} \int_{\mathbb R} x f(t, x) dx=\int_{\mathbb R} \lim_{t \rightarrow t_0} x f(t,x)=\int_{\mathbb R} x f(t_0,x)$$

requires $f$ to be continuous in $t$ uniformly in $x$ to hold.

Examples where $\mathbb E[X_t]$ is indeed continuous are abundant. Counterexamples where it is not? Thanks.

One counterexample is the following:

Let $$B_s$$ be a standard Brownian motion, and let $$S = \inf\{s : B_s = 1\}$$ be the first time it hits 1. Since Brownian motion is recurrent, $$S < \infty$$ almost surely. Let $$Y_s = B_{s \wedge S}$$; then $$Y_s$$ is a continuous martingale and $$\lim_{s \to \infty} Y_s = 1$$ almost surely. Finally, let $$X_t = \begin{cases} Y_{t/(1-t)}, & 0 \le t < 1 \\ 1, & t \ge 1.\end{cases}$$ $$X_t$$ is continuous since $$\lim_{t \to 1^-} X_t = \lim_{s \to +\infty} Y_s = 1$$. But for $$t < 1$$, we have $$E X_t = E Y_{t/(1-t)} = 0$$ since $$Y_s$$ is a martingale, and for $$t \ge 1$$, $$E X_t = 1$$. So $$E X_t$$ is discontinuous.

$$X_t$$ is a useful example to keep in mind; for instance, it is a local martingale with respect to its natural filtration, but not a martingale.

Edit: Hans asks in comments whether we can have an example where the second moments are finite but discontinuous. The answer is yes. Let $$X_t$$ be as above and set $$Z_t = \sqrt{1-X_t}$$. Note that $$X_t \le 1$$ everywhere so the square root makes sense, and $$Z_t$$ is a continuous process. Then we have $$E[Z_t^2] = \begin{cases} 1, & t < 1 \\ 0, & t \ge 1 \end{cases}$$ which is discontinuous. The second moments $$E[Z_t^2]$$ are not only finite but uniformly bounded.

Moreover, if we wish to consider variance (centered second moment) instead, notice that by continuity of $$Z_t$$, as $$t \uparrow 1$$ we have $$Z_t \to Z_1 = 0$$ almost surely. Since we showed $$E[Z_t^2]$$ is uniformly bounded, we have that $$\{Z_t\}$$ is uniformly integrable (see https://math.stackexchange.com/a/184484/822), hence $$E Z_t \to E Z_1 = 0$$. So as $$t \uparrow 1$$, $$\operatorname{Var}(Z_t) = E[Z_t^2] - (E Z_t)^2 \to 1$$, whereas $$\operatorname{Var}(Z_1) = 0$$.

• Thank you very much Nate. It is staggering how stopping/time reversing can produce so many counterexamples Nov 17, 2013 at 18:20
• What about the continuity of the second moment if the second moments are all finite?
– Hans
Feb 3, 2015 at 4:19
• @Hans: Consider $\sqrt{1-X_t}$. Feb 3, 2015 at 5:31
• Thank you, Nate. I actually meant the variance. What is a counterexample in this case?
– Hans
Feb 3, 2015 at 5:58
• @Hans: $Z_t = \sqrt{1-X_t}$ still works. Of course we have $E Z_t^2 = 1$ for $t < 1$ and $0$ for $t \ge 1$. But since $Z_t$ is $L^2$-bounded, it is uniformly integrable, so as $t \uparrow 1$ we have $E Z_t \to E Z_1 = 0$. So as $t \uparrow 1$ we have $\operatorname{Var}(Z_t) = E Z_t^2 - (E Z_t)^2 = 1 - (E Z_t)^2 \to 1$, while $\operatorname{Var}(Z_1) = 0$. Feb 3, 2015 at 7:26