Why Integrating a Differential of a Stochastic Process Gives Its Increment

# Why this stochastic integral is calculated with Riemann integral


# Why this stochastic integral is calculated with Riemann integral












This picture is from Neftci's textbook, 'An Introduction to the Mathematics of Financial Derivatives, Third Edition'

What makes me uncomfortable is equation [10.61] In above picture. In this equation,$sW_s$ in the $d[sW_s]$ is definitively stochastic term.

So I think It is not rigorous to apply Riemann integral in this term. I mean, $∫^t_0dX = [X]^t_0 = t - 0... So ∫^t_0d[sW_s] = [sW_s]^t_0 = tW_t - 0*W_0. (because of W_0 = 0)$ this fundamental relationship should not be applied, because it has stochastic term.

But In the above picture, equation [10.61] seems to apply Riemann integral property not Ito integral property.

I want to know the reason why.

Thanks.

## Answer by Kurt G. (score 5, accepted)

https://quant.stackexchange.com/a/68369

For any semi martingale $X$ (in particular for $X_t=W_t$ or for $X_t=t$) we have $$\tag{1} \int_0^t dX_s=X_t-X_0\,. $$ You are correct that the Ito integral uses the limit procedure $$\tag{2} \int_0^tf(s)\,dX_s=\lim_{\max|t_i-t_{i-1}|\to 0}\sum_{i=1}^nf(t_{i-1})(X_{t_i}-X_{t_{i-1}}) $$ by which the integrand $f$ must be evaluated at the left endpoint of the interval $[t_{i-1},t_{i}]\,.$ In contrast, in the Riemann integral (or more generally the Stieltjes integral) the integrand can be evaluated at any point of that interval.

However, when the integrand is constant this difference does clearly not matter. In other words, we can interpret (1) as an Ito integral and a Riemann-Stieltjes integral at the same time. Also, when instead of (2) you define $$ \int_0^tf(s)\circ\,dX_s=\lim_{\max|t_i-t_{i-1}|\to 0}\sum_{i=1}^n\frac{f(t_i)+f(t_{i-1})}{2}(X_{t_i}-X_{t_{i-1}}) $$ you will obtain the Stratonovich integral. Clearly, (1) can be interpreted as Stratonovich integral as well.

To make along story short: (1) holds for all known types of integrals.