I have two questions relating stochastic integration which perhaps could be answered together.

First question:

First of all, I don't really understand why we can't use Riemann-Stieltjes integration when a Brownian motion is the integrator (has something to do with its infinite variation but I don't see how that affects the integral).

Second Question:

For the second question (I think the more general case), we first need to define the following spaces

$$ \begin{align} M_{0, loc}^{c} &:= \text{Space of all continuous local martingales } (M_{t})_{t \in [0, T]} \text{ with } M_{0} = 0 \\ FV_{0}^{c} &:= \text{Space of all adapted stochastic processes } (A_{t})_{t \in [0, T]} \text{ with } A_{0} = 0 \\& \hspace{0.6cm} \text{ and continuous sample paths of finite variation} \end{align} $$

Now, I have the following lemma:

Every continuous local martingale $(M_{t})_{t \in [0, T]}$ with sample paths of finite variation is constant. In particular, one has $M_{0, loc}^{c} \cap FV_{0}^{c} = \{0 \}.$

This lemma allegedly is responsible, that we cannot construct the integrals with respect to martingales based on classical Riemann-Stieltjes integration. I don't really see why this is the case either.

I hope you understand my questions and are able to answer them.

Best regards,



1 Answer 1


Let's take a standard Brownian motion $(B_t)$ and let's try to compute $\int_0^t B_s\mathrm{d}B_s$ in the Riemann-Stieltjes sense.

Let $0=t_0<t_1<...<t_n=t$ be a partition and let $y_i=t_{i-1}$ or $y=t_i$ for $i=1,...,n$ be two intermediate partitions. Thus, \begin{align*} S^1_n(t) &= \sum_{i=1}^n B_{t_{i-1}}(B_{t_i}-B_{t_{i-1}}), \\ S^2_n(t) &= \sum_{i=1}^n B_{t_{i}}(B_{t_i}-B_{t_{i-1}}), \end{align*} are Riemann-Stieltjes sums.

If the Riemann-Stieltjes integral exists, $S_n^1(t)-S_n^2(t)\to0$ as $\max\limits_{i=1,...,n}\{t_i-t_{i-1}\}\to0$. However, \begin{align*} S^2_n(t)- S^1_n(t)&= \sum_{i=1}^n (B_{t_i}-B_{t_{i-1}})^2 >0 \end{align*} and \begin{align*} \mathbb{E}[S^2_n(t)- S^1_n(t)]&= \sum_{i=1}^n (t_i-t_{i-1})=t \neq 0. \end{align*} Thus, the Riemann-Stieltjes integral does not exist for a Brownian motion as integrator.

In general, the Riemann-Stieltjes integral $\int_0^t f(s)\mathrm{d}g(s)$ exists if $f$ is piecewise continuous and $g$ has finite variation.* However, as you said, the sample paths of Brownian motion have infinite variation (yet finite quadratic variation). Your lemma states that every non-trivial continuous local martingale has infinite variation, as well. Thus, we have to use a new integral notion, Itô's integral. In fact, $\int_0^t B_s\mathrm{d}B_s=\frac{1}{2}(B_t^2-t)$ in the Itô sense.

*To prove this, we take a partition $0=t_0<t_1<...<t_n=t$ and choose $y_i^-$ such that $$f(y^-_i) = \begin{cases} \inf\limits_{t_{i-1}\leq y\leq t_i} f(y) &\mathrm{if}\; g(t_i)-g(t_{i-1})\geq0, \\ \sup\limits_{t_{i-1}\leq y\leq t_i} f(y) &\mathrm{if}\; g(t_i)-g(t_{i-1})<0, \end{cases} $$ and choose $y_i^+$ such that $$f(y^+_i) = \begin{cases} \sup\limits_{t_{i-1}\leq y\leq t_i} f(y) &\mathrm{if}\; g(t_i)-g(t_{i-1})\geq0, \\ \inf\limits_{t_{i-1}\leq y\leq t_i} f(y) &\mathrm{if}\; g(t_i)-g(t_{i-1})<0, \end{cases}.$$ Let \begin{align*} S^+_n(t) &= \sum_{i=1}^n f(y_i^+)(g(t_i)-g(t_{i-1})), \\ S^-_n(t) &= \sum_{i=1}^n f(y_i^-)(g(t_i)-g(t_{i-1})). \end{align*} Then, the Riemann-Stieltjes integral exists if $S^+_n(t)-S^-_n(t)\to0$ as $n\to\infty$.

However, if $\max\limits_{i=1,...,n} \{t_i-t_{i-1}\}\leq \delta$ for some $\delta>0$, then \begin{align*} S^+_n(t)-S^-_n(t) &\leq \sum_{i=1}^n |f(y_i^+)-f(y_i^-)||g(t_i)-g(t_{i-1})| \\ &\leq \sup\{|f(y)-f(y')| : y\geq0; y'\leq t,\;|y-y'|<\delta\} \sum_{i=1}^n |g(t_i)-g(t_{i-1})| \\ &\to 0, \end{align*} if $f$ is continuous (first term goes to zero) and $g$ has finite variation (the sum doesn't blow up). This, of course, also works if $f$ is piecewise continuous, we merely need to split up the integral domain.

This is the reason why we need finite variation for $g$! Otherwise, the Riemann-Stieltjes integral is simply not well-defined.

  • $\begingroup$ Thanks for the explanation. The lemma further states that every continuous adapted local martingale with finite variation and $M_{0} = 0$ is equal to zero right? Which means if we would use such an martingale as an integrator the integral would always be equal to zero or not? $\endgroup$
    – Peter
    May 7, 2020 at 13:29
  • $\begingroup$ Yeah, you're right. You know that the sample paths of such a process are almost surely constant and if $M_0=0$, then $M_t=0$ for all $t$ almost surely. These sample paths have, of course, finite variation and the Riemann–Stieltjes integral exists but if you ingrate any continous function with respect to a constant function, you obtain (as you said) zero. $\endgroup$
    – Kevin
    May 7, 2020 at 13:34

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