# Mean Value Theorem for stochastic processes

In Lemma 3.1 and Theorem 3.2 from the article Numerical solution of random differential equations: A mean square approach, it is stated and proved a Mean Value Theorem for stochastic integrals and derivatives in the mean square sense. See Chapter 4 in the book Random differential equations in science and engineering for an introduction to mean square theory.

My question is about the proof of Lemma 3.1.

Statement of Lemma 3.1: Let $Y(t)$ be a mean square continuous process with finite second order moments on $T=[t_0,t_1]$. Then, there exists $\xi\in [t_0,t_1]$ such that $\int_{t_0}^t Y(s)ds=Y(\xi)(t-t_0)$, $t_0<t<t_1$.

Question: The first thing that surprises me is the fact that $\xi$ is the same for every $t_0<t<t_1$. Is this correct?

Proof of Lemma 3.1 in the article: By page 90 from the book, $\Gamma_Y(r,s)=E[Y(r)Y(s)]$ is continuous on $T\times T$. Then $\Gamma_Y(r,\cdot)$ is continuous on $T$, for each $r\in T$. By the Mean Value Theorem for Riemann integrals, $\int_{t_0}^t \Gamma_Y(r,s)ds=\Gamma_Y(r,\xi)(t-t_0)$, $\xi\in[t_0,t]$. From here, $\xi$ is considered constant.

Question: I think $\xi$ depends on $r$ and $t$, so this invalidates the proof. Is this correct?

Question: In case Lemma 3.1 is correct, for Theorem 3.2 I think it misses the hypothesis $\dot{X}(t)$ be mean square integrable, by page 104 from the book. Is this correct?

Of course, $\xi$ in formula (3.2) must in general depend on $r$ and $t$. Also, there is no reason for the last sentence in the "proof" of Lemma 3.1 in the paper to be true, and I think one can easily construct a relevant counterexample.

It is also easy to construct examples showing that $\xi$ in the equality $\int_{t_0}^t Y(s)ds=Y(\xi)(t-t_0)$ in Lemma 3.1 in the paper must in general depend, not only on $t$, but also on the path/realization of the process $Y(\cdot)$.

For instance, for $t_0=0$ consider a process $Y(\cdot)$ on $[t_0,t_1]=[0,1]$ with only two possible paths, say $s\mapsto y_1(s):=s$ and $s\mapsto y_2(s):=s^2$, each of them having the same probability, $1/2$. Then for real $t>0$
$$\text{on the event \{Y=y_1\} one has \int_{t_0}^t Y(s)ds=\frac{t^2}2=Y\Big(\frac t2\Big)(t-t_0) and hence \xi=\frac t2}$$ and $$\text{on the event \{Y=y_2\} one has \int_{t_0}^t Y(s)ds=\frac{t^3}3=Y\Big(\frac t{\sqrt3}\Big)(t-t_0) and hence \xi=\frac t{\sqrt3}.}$$ So, we see that $\xi$ depends on the path and on $t$.

Thus, Lemma 3.1 in the paper is not correct if $\xi$ is thought of as not depending on $t$ and/or the path. The proof of that lemma is incorrect anyway.

However, if $\xi$ is allowed to depend on $t$ and on the path and if $Y(\cdot)$ is, not only mean-square continuous, but also path-wise continuous, then the path-wise mean value theorem obviously holds, as it immediately follows from the standard, "non-random" mean value theorem of calculus.

On the other hand, if $Y(\cdot)$ is only assumed to be mean-square continuous but not path-wise continuous, then $\xi$ may not exist at all. E.g., let $Y(\cdot)$ be the stochastic process on the interval $[0,1]$ defined by the formula $Y(t)=I\{t>U\}$, where $U$ is a random variable uniformly distributed on $[0,1]$ and $I$ denotes the indicator. Then $Y(\cdot)$ is bounded and hence has finite second order moments. Moreover, $Y(\cdot)$ is mean square continuous, since $E(Y(t)-Y(s))^2=t-s$ if $0\le s\le t\le1$. However, for any $t\in(0,1]$, on the event $\{0<U<t\}$ we have $$\int_0^t Y(s)ds=\int_U^t Y(s)ds=t-U\ne Y(\xi)t$$ for any $\xi\in[0,1]$, since $Y(\cdot)$ can only take values $0$ and $1$.

• So those results from the paper are not correct, right? Is the conclusion that there is no Mean Value Theorem in the mean square sense for stochastic processes correct? Mar 9, 2018 at 21:05
• @stoch : Indeed, Lemma 3.1 in the paper is not correct if $\xi$ is thought of as not depending on $t$ and/or the path. The proof of the lemma is incorrect anyway. However, if $\xi$ is allowed to depend on $t$ and the path and if $Y(\cdot)$ is, not only mean-square continuous but also has continuous paths, then the path-wise mean value theorem obviously holds. On the other hand, if $Y(\cdot)$ is only assumed to be mean-square continuous but not path-wise continuous, then $\xi$ may not exist at all. I have added the relevant details to my answer. Mar 11, 2018 at 3:06
• Thank you for your answer, I understood it completely. It's surprising that a paper with a clear error got published in a good journal. Mar 11, 2018 at 18:36