Skip to content
All library documents

Pricing Options on a Power of the Underlying in the Black Model

Article Quant Q&A · Author: Wolfy

Summary

The exercise derives a Black-model approach for a call whose payoff depends on a power of the terminal asset price. Raising a lognormal underlying to a power preserves lognormality: the transformed variable has a shifted expected value and a log variance scaled by the square of the power. The resulting option can therefore be priced using a Black formula with the transformed forward and volatility adjusted by the power.

The exponential is split to isolate the mean adjustment from a mean-normalized lognormal factor, which identifies the appropriate forward input. The same distributional reasoning applies to the put payoff, using the Black put formula or put–call parity. The source gives the method but does not work through a numerical example; it presumes the Black framework and the stated lognormal dynamics for the underlying.

Key ideas

  • A power of a lognormal asset remains lognormal.
  • The transformed forward includes an adjustment to account for the changed expected value.
  • The transformed volatility equals the original volatility scaled by the power exponent.
  • The put version can be priced with the Black put formula or put–call parity.

Tags

Full text
# Mark Joshi, The concepts and practice of mathematical finance chapter 6 exercise 20,21


# Mark Joshi, The concepts and practice of mathematical finance chapter 6 exercise 20,21












> 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.

## Answer by Freelunch (score 3, accepted)

https://quant.stackexchange.com/a/37861

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.

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

This summary was written by Stratmill's research agent from the original; it is not a copy of the source.