3
$\begingroup$

Find the Black-Scholes price of an option paying $$(S_T^{\alpha} - K)_{+}$$ at time $T$.

Solution - The forward price is given by

$$F_T(t) = e^{r(T-t)}S_t$$

So,

$$F_T(0) = e^{rT}S_0$$

and

$$F_T(T) = S_T = F_T(0)e^{-\frac{1}{2}\sigma^2 T + \sigma\sqrt{T}N(0,1)}$$

So,

\begin{align*} F_T(T)^{\alpha} &= F_T(0)^{\alpha}e^{-\frac{1}{2}\sigma^2 T \alpha + \sigma\alpha\sqrt{T}N(0,1)}\\ &= F_T(0)^{\alpha}e^{-\frac{1}{2}\sigma^2 T \alpha + \frac{\sigma^2 \alpha^2}{2}}e^{- \frac{\sigma^2 \alpha^2}{2} + \sigma\alpha\sqrt{T}N(0,1)} \end{align*}

Then use the Black formula for a call option with forward price

$$F_T(0)^{\alpha}e^{-\frac{1}{2}\sigma^2 T \alpha + \frac{\sigma^2 \alpha^2}{2}}$$

and volatility $\alpha\sigma$.

Question:

I do not understand why Joshi splits up the exponential in this part of the solution:

\begin{align*} F_T(T)^{\alpha} &= F_T(0)^{\alpha}e^{-\frac{1}{2}\sigma^2 T \alpha + \sigma\alpha\sqrt{T}N(0,1)}\\ &= F_T(0)^{\alpha}e^{-\frac{1}{2}\sigma^2 T \alpha + \frac{\sigma^2 \alpha^2}{2}}e^{- \frac{\sigma^2 \alpha^2}{2} + \sigma\alpha\sqrt{T}N(0,1)} \end{align*}

I do not understand the logic of then concluding that we use the Black formula for a call option with forward price

$$F_T(0)^{\alpha}e^{-\frac{1}{2}\sigma^2 T \alpha + \frac{\sigma^2 \alpha^2}{2}}$$

and volatility $\alpha \sigma$.

Lastly, in exercise 21 we are asked to price the put $(K - S_T^{\alpha})_{+}$. The steps are exactly the same exact the volatility term is $\alpha\sigma \sqrt{T}$, which does not make sense to me. Any suggestions on these points are greatly appreciated.

$\endgroup$

1 Answer 1

3
$\begingroup$

Note that \begin{equation} E\big[e^{\sigma \alpha \sqrt{T} N(0,1)}\big] = e^{\frac{\sigma^2 \alpha^2}{2}T} \end{equation}

Hence $F_T(T)^\alpha$ will be a lognormal variable with expected value $F_T(0)^\alpha e^{-\frac{1}{2}\sigma^2T \alpha + \frac{1}{2}\sigma^2 \alpha^2T}$ and log-variance $\sigma^2 \alpha^2 T$. Compare this to the Black formula for computing the price of a call option where you also have a lognormal variable but the expected value is the current price of the forward and the variance is $\sigma^2T$. The case with a put option is analogous, where you may use the Black formula for put options or use the previous answer and put–call parity.

$\endgroup$
4
  • $\begingroup$ I still don't understand where the expectation comes from $\endgroup$
    – Wolfy
    Jan 23, 2018 at 22:56
  • $\begingroup$ It's the expectation of a log-normal variable. $\endgroup$
    – Freelunch
    Jan 24, 2018 at 7:17
  • $\begingroup$ I know but I just don’t understand why we take the expectation to show what you show. I’m just having a hard time with this problem $\endgroup$
    – Wolfy
    Jan 24, 2018 at 9:02
  • $\begingroup$ The Black formula lets you calculate the expected value $E^\mathbb{Q}[(X_T-K)^+]$ given that you know that $X_T$ is a log-normal variable defined by its expected value and log-variance. Those are the only two relevant parameters for you to calculate for any general $X_T$. $\endgroup$
    – Freelunch
    Jan 24, 2018 at 10:34

Your Answer

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

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