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Valuing a Power Payoff with a Lognormal Asset Price

Article Quant Q&A · Author: Jase

Summary

The document derives a risk-neutral value for a derivative that pays a power of the underlying asset price divided by a fixed strike at maturity. It starts with the discounted conditional expectation of the payoff, then assumes the asset follows a constant-rate, constant-volatility geometric Brownian motion. The terminal price is lognormally distributed, so its power expectation can be evaluated using the exponential moment of a normal random variable.

The resulting expression depends on the current asset price raised to the payoff power, the strike, discounting, the risk-free rate, volatility, and time to maturity. This illustrates how higher powers alter the volatility contribution to expected payoff. The derivation is conditional on the stated risk-neutral model and constant parameters; it does not address calibration, dividends, market frictions, or whether the assumed dynamics fit observed prices.

Key ideas

  • Risk-neutral valuation discounts the conditional expected payoff back from maturity.
  • Under geometric Brownian motion, the terminal asset price has a lognormal distribution.
  • The expectation of a power payoff follows from the exponential moment of a normal random variable.
  • The payoff exponent changes both the drift and volatility terms in the expected value.
  • The derivation relies on constant rates and volatility and on the assumed risk-neutral price process.

Tags

Full text
# Pricing a Power Contract derivative security


# Pricing a Power Contract derivative security












I'm trying to price a "power contract" and would appreciate guidance on the next step. The payoff at time $T$ is $(S(T)/K)^\alpha$, where $K > 0$, $\alpha \in \mathbb{N}$, $T > 0$. $S$ is adapted to $\mathscr{F}$, and we are currently at time $t \in [0,T)$. Let $Q$ denote the risk-neutral measure and $\beta(t) = e^{\int_0^t r(s)ds}$ be the domestic savings account/discount factor. Also, $W(t)$ is standard Brownian Motion.

Here's my progress:

$\displaystyle \ \ \text{value}_t = E^Q[\frac{\beta(t)}{\beta(T)}(S(T)/K)^\alpha \big|\mathscr{F}_t]$

$\displaystyle \ \ = \frac{\beta(t)}{\beta(T)K^\alpha}E^Q[S(T)^\alpha \big|\mathscr{F}_t]$

We take $\displaystyle \ \ S(T)^\alpha = S(t)^\alpha \exp{\{\bigg[ (r-\frac12 \sigma^2)(T-t)+\sigma(W(T)-W(t))\bigg]\alpha \}}$.

Therefore:

$\displaystyle \ \ \text{value}_t = \frac{\beta(t)S(t)^\alpha}{\beta(T)K^\alpha}\exp{\{ \alpha(r-\frac12 \sigma^2)(T-t) + \frac12 \alpha^2 \sigma^2(T-t) \}}$

by the fact that $E[e^z] = e^{\mu + \frac12 \sigma^2}$ when $z \sim \mathscr{N}(\mu,\sigma^2)$.

This is homework but is not graded.

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.