2
$\begingroup$

I have a question regarding how to solve the NA price for a slightly modified call option.

Say that I have a money account $B(T)=e^{r(T-t)}$ and a stock dynamic $\frac{dS(t)}{S(t)}=(r-\delta)dt+\sigma dW(t)$ where r is the riskless return, $\delta$ the continous dividend yield and $W$ a brownian motion. By Itô's lemma we can easily derive $S(T)=S(t)e^{r-\delta-\frac{\sigma^2}{2}+\sigma(W(T)-W(t))}$.

I want now to compute the NA price for the T-claim $X=max(B(T),S(T))$. My solution so far is as follows below:

$\Pi(t;X)=\frac{1}{B(T)}E^Q_t[max(B(T),S(T))]\\ =\frac{1}{B(T)}E^Q_t[B(T)+max(0,S(T)-B(T))]\\ =1+\frac{1}{B(T)}E^Q_t[max(0,S(T)-B(T))]$

Last expression is a call option with strike $B(T)$. Can I then continoue to apply the Black-Scholes formula and write the NA price as following below?

$\Pi(t;X)=1+N(d_1)\frac{S(T)}{B(T)}-N(d_2)$

Or am I missing something? $d_1$ and $d_2$ is defined in the Black Scholes formula as,

https://en.wikipedia.org/wiki/Black%E2%80%93Scholes_model

$\endgroup$
4
$\begingroup$

Yes you can use the Black-Scholes Model with $K=B(T)$ because $B(T)$ is deterministic (a constant like $K$, since $T$ is constant).

However your current solution is incorrect as the Black-Scholes call price is already discounted ($C:=e^{-rT}E^Q[(S_T-K)^+])$ and based on the current stock price $S_t$ (not $S_T$). Further note that $e^{-r(T-t)}B(T)=1$ and $1-N(x)=N(-x)$:

\begin{align*}\Pi(t;X)&=1+N(d_1)S_te^{-\delta(T-t)}-N(d_2)e^{-r(T-t)}B(T)\\ &=N(d_1)S_te^{-\delta(T-t)}+N(-d_2) \end{align*}

$\endgroup$
0
2
$\begingroup$

Your solution is correct. Rewriting your modified payoff in terms of the payoff of a call option is a common technique. Note however that you have a little typo: you need $S(t)e^{-\delta(T-t)}$ instead of $S(T)$ in the last line, i.e. \begin{align*} \Pi(t,X) &= 1+S_te^{-\delta(T-t)}N(d_1)-N(d_2) \\ &= S_te^{-\delta(T-t)}N(d_1) +N(-d_2), \end{align*} since $1-N(d_2)=N(-d_2)$. In fact, we have $\mathbb{E}^{\mathbb{Q}}_t[S(T)]=S(t)e^{-\delta(T-t)}$. Furthermore, you can simplify $d_1$ and write \begin{align*} d_1 &= \frac{\ln\left(\frac{S(t)}{B(T)}\right)+\left(r-\delta+\frac{1}{2}\sigma^2\right)(T-t)}{\sigma\sqrt{T-t}} \\ &= \frac{\ln\left(S(t)\right)-r(T-t)+\left(r-\delta+\frac{1}{2}\sigma^2\right)(T-t)}{\sigma\sqrt{T-t}} \\ &= \frac{\ln\left(S(t)\right)-\left(\delta-\frac{1}{2}\sigma^2\right)(T-t)}{\sigma\sqrt{T-t}} \end{align*}

$\endgroup$

This site is temporarily in read only mode and not accepting new answers.

Not the answer you're looking for? Browse other questions tagged .