I'm reading Shreve's Stochastic Calculus for Finance II. On page 191, Exercise 4.6, we are given the problem

Exercise 4.6. Let $S(t)=S(0)\exp\Big \{\sigma W(t)+(\alpha-\frac{1}{2}\sigma^2)t\Big\}$ be a geometric Brownian motion. Let $p$ be a positive constant. Compute $\mathrm{d}(\{S(t)\}^p)$, the differential of $S(t)$ raised to the power of $p$.

The details of how I solve this problem aren't too relevant to the question, so a reader can skip to The Question.

Solving the Problem

I can solve this problem with a direct application of the Itô-Doeblin formula (page 138). I reproduce that formula here for completeness.

Theorem 4.4.1 (Itô-Doeblin formula for Brownian motion). Let $f(t, x)$ be a function for which the partial derivatives $f_t(t, x)$, $f_x(t, x)$, and $f_{xx}(t, x)$ are defined and continuous, and let $W(t)$ be a Brownian motion. Then, for every $T\geq0$,

$$f(T, W(T)) = f(0,W(0)) + \int^{T}_0 f_t(t, W(t)) \mathrm{d}t + \int^{T}_0 f_x(t, W(t))\mathrm{d}W(t)+\frac{1}{2}\int^T_0f_{xx}(t, W(t))\mathrm{d}t\text{.}\tag{1}$$

We define

$$f(t, x) = \Big( S(0)e^{\sigma x + (\alpha-\frac{1}{2}\sigma^2)t}\Big)^p\text{.}\tag{2}$$

We then directly apply Theorem 4.4.1 to get \begin{align} \mathrm{d}f(t, W(t)) = \mathrm{d}(\{S(t)\}^p) &= pS(0)^p(\alpha+(\frac{p-1}{2})\sigma^2) e^{p(\sigma W(t)+(\alpha-\frac{1}{2}\sigma^2)t)}\mathrm{d}t + pS(0)^p(\sigma)e^{p(\sigma W(t)+(\alpha-\frac{1}{2}\sigma^2)t)}\mathrm{d}W(t) \\ &= pS(t)^p\Big[\sigma \mathrm{d}W(t) + (\alpha + \frac{p-1}{2})\mathrm{d}t \Big]\tag{3}\end{align}

or, to put it in integral form,

$$\int_0^T (S(t))^p \mathrm{d}t = \int_0^T pS(t)^p\sigma \mathrm{d}W(t) + \int_0^T pS(t)^p(\alpha + \frac{p-1}{2})\mathrm{d}t\text{.}\tag{4}$$

The Question

The math involved in the solution above makes sense, and I believe this solution matches the author's intent. However, I don't see why the above work has been productive. When I "compute" something, I imagine that you execute an algorithm to solve a problem (perhaps just applying the Itô-Doeblin formula was the "computation"), or, in most cases, put a problem in a "closed-form" where known algorithms can then solve it. Often you simplify a formula to remove complicating parts.

I don't see why I am done when I write line $(4)$. I am not sure how you would compute $(4)$ or ever use $(4)$. I am not sure why $(4)$ is a better form than just writing $\mathrm{d}\{S(t)\}^p)$. Assuming you had some interest in $\mathrm{d}\{S(t)\}^p)$, how would you ever use $(4)$? What would be the inputs? What would be the outputs? Would you use it with a computer, or just paper and pencil? Is there any use of this formula that isn't just getting a distribution of $\mathrm{d}\{S(t)\}^p)$ for time $t$ when you have all of the information up to $t-1$? Also, the book suggests integrals w.r.t. $\mathrm{d}t$ are "just Lebesgue" integrals, but when the integrand involves Brownian Motion, I don't see why that is helpful.

  • 2
    $\begingroup$ If Brownian motion is in the integrand but is not the integrator, then the integral can be interpreted in the pathwise sense. $\endgroup$ Commented Apr 21, 2021 at 17:47

1 Answer 1


Note that SDE (4) does have a "closed-form" representation.

Let $X$ be $$X := S^p, $$

so (4) is a geometric Brownian motion SDE $$dX = (p\alpha + 2^{-1}p(p-1) \sigma^2) X dt + p \sigma X dW, $$

which, again due to Ito Lemma, is equivalent to

$$ d \ln X = (p\alpha + 2^{-1}p(p-1) \sigma^2 - 2^{-1}p^2 \sigma^2) dt + p \sigma dW $$


$$ d \ln X = (p\alpha - 2^{-1} p\sigma^2) dt + p \sigma dW $$

and, finally:

$$ X_t = X_0 \exp (p \sigma W_t + p(\alpha - 2^{-1} \sigma^2)t). $$

Of course, this was expected from the beginning, as $S$ was given in "closed-form":

$$ \left( S_0 \exp (\sigma W_t + (\alpha - 2^{-1} \sigma^2)t) \right)^p = S_0^p \exp (p \sigma W_t + p(\alpha - 2^{-1} \sigma^2)t)$$


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