I'm working through Shreve II, and on question 4.6, you are asked to compute

$d(S_t^p)$ where $S_t$ = $S_0e^{\sigma W_t + (\alpha - \frac{1}{2}\sigma^2)t}$

I get the answer $pS_t^p[\sigma dW_t + (\alpha + \frac{1}{2}\sigma^2p - \sigma^2)dt]$

whereas the online solutions manual gets $pS_t^p[\sigma dW_t + (\alpha + \frac{1}{2}\sigma^2(p - 1)dt]$

Clarifying that either I or the author of the solution has made an error. Going through their solution, I believe they drop an expression containing $\sigma$ where they should not drop it.


$$d(S^p) = pS^p (\alpha +\sigma dW) + \frac{1}{2}p(p-1)S^p\sigma^2 dt $$

$$ = pS^p \left[ \left(\alpha +\frac{1}{2}\sigma^2(p-1)\right)dt + \sigma dW \right]$$

  • $\begingroup$ Well, what do you say about Mr. Rodriguez? $\endgroup$ – user7348 Nov 21 '14 at 21:21
  • $\begingroup$ You said I'm right, but then you wrote down Shreve's solution. $\endgroup$ – user7348 Nov 21 '14 at 21:25
  • $\begingroup$ It’s a matter of using Ito’s chain rule. $$df = \frac{\partial f}{\partial t} dt + \frac{\partial f}{\partial X} dX + \frac{1}{2}\frac{\partial^2 f}{\partial X^2} dX^2$$ The second derivative of $S^p$ should be $p(p-1)S^{p-1}$ and everything follows. $\endgroup$ – fni Nov 21 '14 at 21:25
  • $\begingroup$ OH, I just realized that the dt term is non-zero. It is so often 0, that I just overlooked it on this function. Thanks. $\endgroup$ – user7348 Nov 21 '14 at 21:27

Shreve's answer is the correct one:

The drift term of $\frac{dS^p}{S^p}$ has two parts:

  • $p \left(\alpha - \frac{1}{2} \sigma^2 \right)$ from regular differentiation
  • $\frac{1}{2} p^2 \sigma^2 $ is the Ito term.

When you sum them up you get $p \left(\alpha + \frac{1}{2} (p-1) \sigma^2 \right)$


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