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Pricing a Power Claim in Black–Scholes with a Lognormal Moment

Article Quant Q&A · Author: J. D.

Summary

The document works through pricing a terminal claim equal to a power of the underlying stock price in the standard Black–Scholes model. It sets up the pricing partial differential equation, describes the risk-neutral stock process, and uses the lognormal distribution of the terminal stock-to-current-price ratio to evaluate the required power moment. The resulting value is the discounted expected payoff, expressed as a function of spot, maturity, volatility, interest rate, and the power exponent.

An accepted answer supports the calculation with the general moment formula for a lognormally distributed variable and maps its parameters to the Black–Scholes setting. The post also asks whether the price formula is itself the arbitrage-free process; conditional risk-neutral valuation provides the price at each time and spot. The discussion is an exercise-level derivation, with no market data or numerical validation, and assumes the standard model and its constant parameters.

Key ideas

  • A power payoff can be priced by taking its discounted risk-neutral expectation.
  • The terminal stock-to-current-price ratio is lognormally distributed in the Black–Scholes model.
  • The moment-generating property of a lognormal variable gives the expected power payoff.
  • The resulting pricing function gives the claim value for each time and underlying spot price.

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Full text
# Exercise on arbitrage-free process


# Exercise on arbitrage-free process












Consider the following problem, from Bjork's Arbitrage Theory in Continuous Time:

> Consider the standard Black-Scholes model. Derive the arbitrage free price process for the $T$-claim $\mathcal{X}$ where $\mathcal{X}$ is given by $\mathcal{X}=\{S(T)\}^\beta$. Here $\beta$ is a known constant.

My approach.

Let $F(t,s)$ be the price of the claim $\mathcal{X}$ at time $t$, when the underlying spot price is $s$.

The Black-Scholes equation for $F$ is: $$ \begin{align} F_t + rsF_s + \frac12 \sigma^2s^2 F_{ss} - rF &= 0 \\ F(T, S(T)) &= S(T)^\beta. \end{align} $$

It is convenient to make a change of variables of the form $\tilde{F}(t,s) = e^{-rt}F(t,s)$, so that the associated stochastic process is: $$ \begin{align} dX &= rX dt + \sigma X dW \\ X(t) &= s. \end{align} $$

After changing variable to $Y = \log X$ and integrating, I find $$ X(T) = s \exp\left((r-\frac12 \sigma^2)(T-t) + \sigma(W(T) - W(t))\right). $$ So by the Feynman-Kac formula I have: $$ \begin{align} F(t,s) &= e^{-r(T-t)}\mathbb{E}\left[X(T)^\beta\right] \\ &= e^{-r(T-t)}\int_{-\infty}^{\infty}s^\beta e^{\beta z} \exp\left(-\frac12 \frac{(z - (r-\frac12\sigma^2)(T-t))^2}{\sigma^2(T-t)}\right) dz, \end{align} $$ which after some computation gives, if I did not make any mistake: $$ F(t,s)=e^{-r(T-t)}s^\beta \exp\left(\frac12\sigma^2\beta^2(T-t) + (r-\frac12\sigma^2)\beta(T-t)\right). $$

Does it sound right?

Also, regardless of whether the pricing formula is correct, I am not sure if what I found is really the arbitrage free stochastic process for $\mathcal{X}$.

## Answer by user34971 (score 1, accepted)

https://quant.stackexchange.com/a/48601

If $X$ is a lognormally distributed variable, $X = e^{\mu + \nu Z}$, so $\ln X$ has mean $\mu$ and variance $\nu^2$, and $Z$ is normally distributed, then $$ E\left[ X^n \right] = e^{n\mu + \frac{1}{2} n^2\nu^2} $$ This solves your question with $X = S_T/S_t$, $n = \beta$, $\mu = (r -\frac{1}{2} \sigma^2) (T-t)$ and $\nu = \sigma \sqrt{T-t}$ in the Black Scholes world.

See also the following wiki page for other properties of lognormal distribution:

https://en.wikipedia.org/wiki/Log-normal_distribution

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.