0
$\begingroup$

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.

Directly from the book

$\endgroup$

1 Answer 1

0
$\begingroup$

Too long for a comment. I'd offer even a third derivation: $$ \begin{align} dS/S&=(\mu-d)dt+\sigma dW\\ X&\equiv e^{-d(T-t)}S_t\\ \Rightarrow dX&=\frac{\partial X}{\partial t}dt+\frac{\partial X}{\partial S }dS\\ &=de^{-d(T-t)}S_tdt+e^{-d(T-t)}dS\\ &=dX_tdt+e^{-d(T-t)}S\left[(\mu-d)dt+\sigma dW\right]\\ &=dX_tdt+X_t(\mu-d)dt+X_t\sigma dW\\ &=X_t\mu dt+X_t\sigma dW\\ &=X_t(\mu dt+\sigma dW) \end{align} $$

$\endgroup$
7
  • $\begingroup$ 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). $\endgroup$
    – Giogre
    Jun 3, 2022 at 12:49
  • 1
    $\begingroup$ @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 $$ S(t) = S_0 e^{(\mu - d - \frac{\sigma^2}{2})\color{red}{t} + \sigma \color{red}{W_t}}. $$ $\endgroup$
    – Kurt G.
    Jun 3, 2022 at 14:08
  • $\begingroup$ @KurtG. question edited to include your correction $\endgroup$
    – Giogre
    Jun 3, 2022 at 14:22
  • $\begingroup$ With Kermittfrog's answer I don't see a problem. Do you ? $\endgroup$
    – Kurt G.
    Jun 3, 2022 at 14:53
  • $\begingroup$ @KurtG. I have included a snapshot of the 'incriminated' excerpt from the book, so you can see with your own eyes $\endgroup$
    – Giogre
    Jun 3, 2022 at 16:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.