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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.

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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.

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  • $\begingroup$ I still don't understand where the expectation comes from $\endgroup$ – Wolfy Jan 23 '18 at 22:56
  • $\begingroup$ It's the expectation of a log-normal variable. $\endgroup$ – Freelunch Jan 24 '18 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 '18 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 '18 at 10:34

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