Let's take the case where the underlying stock has the continuous dividend yield $\delta$. Then, in the risk-neutral world, $\frac{dS}{S}=(r-\delta)dt+\sigma dW^Q$. Suppose we want to price a derivative on the underlying stock. The standard way to go about it is to create a risk-less portfolio by using a combination of the derivative and the stock, applying Itô to calculate the small change in said portfolio, and equating its growth rate to the risk-free rate.

$\begin{align*}\pi=C-\Delta S\implies d\pi=dC-\Delta dS-(\delta\Delta) Sdt\\=C_tdt+C_{S}dS+\frac{C_{SS}}{2}\sigma^2S^2dt-\delta\Delta Sdt-\Delta dS\end{align*}$

If we choose $\Delta=C_S$, then all the stochastic terms would go away, and we would get-

$d\pi=C_tdt+\frac{C_{SS}}{2}\sigma^2S^2dt-\delta\Delta Sdt=\pi rdt$. Cancelling out $dt$ on both sides, and re-arranging we get the Black-Scholes equation in the case of dividends-


My question is regarding the expansion of $dS$ in the first step. If we hold $-\Delta$ of the stock, then shouldn't we be using the total return process of the stock, i.e. $S'=Se^{\delta t}$? Using Itô, $\frac{dS'}{S'}=rdt+\sigma dW^Q$ and $d\pi=dC-\Delta dS'$ and $dS'\neq dS+\delta Sdt$.

Where am I going wrong here?


Holding a quantity $\Delta$ of the stock over an infinitesimal period $dt$ gives you:

  • $\Delta \times dS$ : return of the stock over $dt$
  • $\Delta S \times \delta \times dt$: continuous dividends collected over $dt$

The hedge is to hold a quantity $\Delta$ of stocks, not $\Delta$ of the total return process. If you want to consider the total return process, the hedge ratio will be different, it will be equal to $\Delta e^{-\delta t}$:

$\frac{\partial C}{\partial S^{'}} = \frac{\partial C}{\partial S} \times \frac{\partial S}{\partial S^{'}} = \Delta \times e^{-\delta t}$

Using either $S$ or $Se^{\delta t}$ will give the same result, same PDE, same hedge ratio, up to a change of variable $S \leftrightarrow S'$

  • $\begingroup$ If we use the total return process S', then $d\pi=dC-\Delta dS'$, i.e. without the $\delta$ term. This would then result in the Black-Scholes PDE being derived with $r$ instead of the $r-\delta$ term. This follows from the fact that $C_SS=C_{S'}S'$ and $C_{SS}S^2=C_{S'S'}(S')^2$ as you've shown above. $\endgroup$ – Amrit Prasad May 8 '18 at 5:21
  • $\begingroup$ With $S^{'}$ you get the PDE without $\delta$ and with a hedge ratio $\Delta e^{-\delta t}$ of $S^{'}$. This PDE is satisfied by $S^{'}$ and not $S$. Now, if you change the variable to $S$, you get the PDE satisfied by $S$, it contains the $\delta$ term, and the hedge ratio of stocks to hold is $\Delta$. $\endgroup$ – byouness May 8 '18 at 7:44
  • $\begingroup$ Hello @AmritPrasad. Is everything clear now? If so, could you accept the answer please? Otherwise, could you explain what is not clear so that I can further help? $\endgroup$ – byouness Jun 5 '18 at 16:13

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.