Pricing a Squared-Spot European Payoff Under Geometric Brownian Motion
Summary
The document prices a European payoff equal to the square of the underlying asset price under a risk-neutral geometric Brownian motion. It writes the terminal asset value using a normally distributed Brownian increment, discounts the expected payoff, and reduces the calculation to an integral over a standard normal variable. Because the payoff is positive for every terminal price, there is no strike-based integration region to isolate as in a call option.
The answer evaluates the full Gaussian integral by completing the square, using the normal density’s exponential moment. This yields the discounted second moment of the asset price, which can be expressed as the current price squared multiplied by an exponential term involving the risk-free rate, volatility, and time to maturity. The worked step addresses the integral in the question; it assumes the stated constant-rate, constant-volatility model and does not discuss dividends or other payoff conventions.
Key ideas
- Under risk-neutral geometric Brownian motion, the terminal asset price is lognormally distributed.
- A squared-price payoff requires the second moment of the terminal price rather than a strike-truncated integral.
- The Gaussian integral is evaluated by completing the square or applying the normal exponential moment.
- Discounting the second moment gives a value proportional to the current asset price squared.
- The calculation assumes the stated constant-rate, constant-volatility setting.
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Full text
# European option with payoff $X_T^2$
# European option with payoff $X_T^2$
I have been ask to price a European option with payoff $H(X_T,T) = X_T^2$ using the equivalent martingale measure (EMM).
For this I used the process:
\begin{equation} dX_t = r X_t dt + \sigma X_t d\tilde{B}_t \end{equation}
which is a geometric brownian motion in the EMM.
For the pricing we have:
\begin{equation} E[e^{-r(T-t)}f(X_T)|\mathfrak{F}_s] = e^{-r(T-t)}E\left[\left(X_t e^{(r - \frac{1}{2}\sigma^2)(T-t) + \sigma \omega \sqrt{T-t}}\right)^2|\mathfrak{F}_t\right] \end{equation}
where $\tilde{B}_{T-t}$ was replace by $\omega \sqrt{T-t}$ as $\tilde{B}_{T-t} \sim \mathcal{N}(0,T-t)$.
The condition of the expected value can be taken of as the process is a martingale under measure $Q$ and there fore the expected value can be computed, and this is where I get stuck. I have the following.
\begin{equation} \begin{aligned} e^{-r(T-t)}E\left[\left(X_t e^{(r - \frac{1}{2}\sigma^2)(T-t) + \sigma \omega \sqrt{T-t}}\right)^2|\mathfrak{F}_t\right] = e^{-r(T-t)}E\left[\left(X_t e^{(r - \frac{1}{2}\sigma^2)(T-t) + \sigma \omega \sqrt{T-t}}\right)^2\right] \\ = e^{-r(T-t)} \frac{1}{\sqrt{2\pi}}\int_{R} \left(X_t e^{(r - \frac{1}{2}\sigma^2)(T-t) + \sigma \omega \sqrt{T-t}}\right)^2 e^{-\frac{1}{2} \omega^2}d\omega \\ = e^{-r(T-t)} \frac{1}{\sqrt{2\pi}}\int_{R} X_t^2 e^{(2r - \sigma^2)(T-t) + 2\sigma \omega\sqrt{T-t}-\frac{1}{2}\omega^2}d\omega \\ = e^{r(T-t)}X_t^2 \frac{1}{\sqrt{2\pi}}\int_R e^{-\sigma^2(T-t) + 2\sigma\omega\sqrt{T-t} - \frac{1}{2}\omega^2} d\omega \end{aligned} \end{equation}
In a normal payoff like $(X_T - K)^+$ we would the $\omega$ for which the integral is positive and then factor everything in the exponent. But in this case I'm not sure how to continue as I'm not seeing an easy solution to the integral.
## Answer by Kurt G. (score 2, accepted)
https://quant.stackexchange.com/a/71097
To long for a comment
This question should be closed as it is a basic financial question.
With the hint given in my first comment the integral \begin{align} \frac{1}{\sqrt{2\pi}}\int_\mathbb R e^{-\sigma^2(T-t) + 2\sigma\omega\sqrt{T-t} - \frac{1}{2}\omega^2} d\omega \end{align} is --setting $a=2\sigma\sqrt{T-t}$ and $\frac{a^2}{2}=2\sigma^2(T-t)$-- $$ \boxed{e^{-\sigma^2(T-t)+2\sigma^2(T-t)}\underbrace{ \frac{1}{\sqrt{2\pi}}\int_\mathbb R e^{a\omega-\frac{a^2}{2} - \frac{1}{2}\omega^2} d\omega}_{=1}=e^{\sigma^2(T-t)}.} $$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.