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?


1 Answer 1


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. $\endgroup$ May 8, 2018 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, 2018 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, 2018 at 16:13

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