Let $(\Omega,\mathcal A,\operatorname P)$ be a probability space and $(X_t)_{t\ge0}$ be an $[0,\infty)$-valued process on $(\Omega,\mathcal A)$. Assume $X:\Omega\times[0,\infty)\to[0,\infty)$ is $(\mathcal A\otimes\mathcal B([0,\infty)),\mathcal B([0,\infty)))$-measurable and $X(\omega):[0,\infty)\to[0,\infty)$ is (right-)continuous at $0$ for $\operatorname P$-almost all $\omega\in\Omega$.

Let $$Y_t:=\int_0^tX_s\:{\rm d}s\in[0,\infty]\;\;\;\text{for }t\ge0$$ and $$Z_t:=\operatorname E[Y_t]\;\;\;\text{for }t\ge0.$$

Are we able to show that $$\frac{Z_t}t\xrightarrow{t\to0+}\operatorname E[X_0]\tag1?$$

Note that we can easily show that $$\frac{Y_t(\omega)}t\xrightarrow{t\to0+}X_0(\omega)\tag2$$ for $\operatorname P$-almost all $\omega\in\Omega$. If $$\sup_{(\omega,\:t)\in[0,\:\infty)}X_t(\omega)<\infty,$$ then we can easily conclude $(1)$ by the dominated convergence theorem.

For the general case, let $X^n:=X\wedge n$, $$Y^n_t:=\int_0^tX_s^n\:{\rm d}s\;\;\;\text{for }t\ge0$$ and $$Z^n_t:=\operatorname E[Y^n_t]\;\;\;\text{for }t\ge0$$ for $n\in\mathbb N$. By the former case, $$\frac{Z^n_t}t\xrightarrow{n\to\infty}\operatorname E[X^n_0]\tag3$$ for all $n\in\mathbb N$.

Now the idea is to conclude using the monotone convergence theorem. However, we need to let $n\to\infty$ and $t\to0+$ simultaneously ... Are we able to show that this is actually possible?


1 Answer 1


Here is a counter-example to show that $\lim_{t\rightarrow\infty} E[Y_t/t]$ may not be $E[X_0]$:

Let $V$ be any nonnegative random variable with $E[V]=\infty$. Define $X_s=sV$ for $s\geq 0$. Then for each $\omega \in \Omega$, we have $X_s(\omega)=sV(\omega)$ is continuous at all points $s\geq 0$, and $X_0(\omega)=0$ for all $\omega \in \Omega$. However

$$Y_t = \int_0^t X_s ds = Vt^2/2 \quad \forall t \geq 0$$

So $Y_t/t = Vt/2$ and $E[Y_t/t]=\infty$ for all $t>0$. $\Box$

On the other hand, by Fatou's lemma and your equation (2) we get $$ E[X_0] = E[\liminf_{t\rightarrow\infty} Y_t/t] \leq \liminf_{t\rightarrow\infty} E[Y_t/t]$$ I suppose we could also conclude this from your equation (3). The counter-example shows the inequality can be strict.

  • $\begingroup$ Thank you for answer. Do you think the situation is different in the following case: $X=f\circ W$, where $E$ is a topological space, $(W_t)_{t\ge0}$ is an $E$-valued càdlàg process on $(\Omega,\mathcal A)$ and $f:E\to[0,\infty)$ is Borel measurable and locally bounded? $\endgroup$
    – 0xbadf00d
    Commented Jul 18, 2022 at 18:37
  • $\begingroup$ Would my same example with $E=[0, \infty)$, $W_t=tV$, and $f(x)=x$ fit that situation? $\endgroup$
    – Michael
    Commented Jul 18, 2022 at 18:47
  • $\begingroup$ It obviously does ... Sorry. Do you think there is any assumption milder than boundedness on $f$ which would turn the claim to be true? $\endgroup$
    – 0xbadf00d
    Commented Jul 18, 2022 at 18:50
  • $\begingroup$ What I think could be an interesting additional assumption is that $\operatorname P$ be replaced by a probability measure $\operatorname P_x$ with $W_t=x$ $\operatorname P_x$-almost surely and $\operatorname P_x[X_s\in U\text{ for all }s\in[0,t]]\to1$ as $t\to0+$ for all open neighborhoods $U\subseteq E$ of $x$. $\endgroup$
    – 0xbadf00d
    Commented Jul 18, 2022 at 18:52
  • $\begingroup$ Please see math.stackexchange.com/q/4495504/47771. $\endgroup$
    – 0xbadf00d
    Commented Jul 18, 2022 at 19:25

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