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,



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 '20 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 '20 at 13:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.