# Deriving the stochastic process for a dividend-yielding stock (under Black-Scholes assumptions)

In order to derive the Black-Scholes equation for a stock $$S(t)$$ yielding dividends at the continuous rate $$d$$ $$S(t) = S_0 e^{(\mu - d - \frac{\sigma^2}{2})t + \sigma \sqrt{t} N(0,1)} \text{,}$$ M. Joshi in The concepts and practice of mathematical finance starts from the stochastic process for a delivery contract $$X(t) = e^{-d (T - t)} S(t)$$, equation (5.76):

$$dX_t = (\mu + d) X_t dt + \sigma X_t dW_t \qquad \qquad (1)$$

He defines a delivery contract $$X_t$$ as a contract where you pay for stock $$S_t$$ today, but it gets delivered to you at time $$T$$. He writes that for a non-dividend paying stock, $$X_t$$ at time $$T$$ has the same value of $$S_t$$ as both end up with you holding one $$S_t$$. Then he makes the case of a dividend paying stock (included in text snapshot below): at time $$T$$ you will have $$e^{d(T−t)}S_t$$ if you held the stock, while only $$S_t$$ if you held a delivery contract, so the latter's value at $$T$$ must be $$X_t=e^{−d(T−t)}S_t$$.

However equation (5.76), renamed (1) above is thrown there as is and not motivated by any derivation. I have tried deriving it from the $$X_t$$ and $$S_t$$ processes listed above, using the chain rule ($$=$$ Ito's lemma here because $$\dfrac{\partial^2 X_t}{\partial S^2} = 0$$):

\begin{align} dX_t(S_t, t) & =\\ &= \frac{\partial X_t}{\partial S_t} dS_t + \left[ \frac{\partial X_t}{\partial t} + \frac{\partial X_t}{\partial S_t} \frac{\partial S_t}{\partial t} \right] dt \\ &= e^{-d(T - t)} \left[ ( \mu - d) S_t dt + \sigma S_t dW_t \right] + \left[ e^{-d (T - t)} S_t d + e^{-d (T - t)} S_t \left(\mu - d - \frac{\sigma^2}{2} \right)\right] dt \\ &= X_t \left[ \left( 2 \mu - d - \frac{\sigma^2}{2} \right) dt + \sigma dW_t \right] \qquad \qquad (2) \end{align}

where I have used $$dS_t = (\mu - d) S_t dt + \sigma S_t dW_t$$.

Equations (1) and (2) differ, in that they have different deterministic components.

Can anyone enlighten me as where errors/incongruities are in the above?

Below, the passage from the book included as snapshot.

Too long for a comment. I'd offer even a third derivation: \begin{align} dS/S&=(\mu-d)dt+\sigma dW\\ X&\equiv e^{-d(T-t)}S_t\\ \Rightarrow dX&=\frac{\partial X}{\partial t}dt+\frac{\partial X}{\partial S }dS\\ &=de^{-d(T-t)}S_tdt+e^{-d(T-t)}dS\\ &=dX_tdt+e^{-d(T-t)}S\left[(\mu-d)dt+\sigma dW\right]\\ &=dX_tdt+X_t(\mu-d)dt+X_t\sigma dW\\ &=X_t\mu dt+X_t\sigma dW\\ &=X_t(\mu dt+\sigma dW) \end{align}
• Thank you, I'd agree with you, but on the book it is stated the derivation should lead to $d X_t = (\mu + d) X_t dt + \sigma X_t dW_t$ (eq 5.76, or eq.1 in my question above). Jun 3, 2022 at 12:49
• @Giogre : your equation $$S(t) = S_0 e^{(\mu - d - \frac{\sigma^2}{2})(T - t) + \sigma \sqrt{T - t} N(0,1)} \text{,}$$ looks wrong. The stock has no maturity $T$. The correct solution of the SDE for $S$ is $$S(t) = S_0 e^{(\mu - d - \frac{\sigma^2}{2})\color{red}{t} + \sigma \color{red}{W_t}}.$$ Jun 3, 2022 at 14:08