Pricing a Power Option with Black–Scholes

# Price of the form $v(t,x)=\phi(t,T)x^n$ for a power option


# Price of the form $v(t,x)=\phi(t,T)x^n$ for a power option












I'm trying to solve the next exercise:

Let $g(S_{T})=S_{T}^{n}$ be the pay-off of a power option. Show that it's price is given by $v(t,x)=\phi(t,T)x^{n}.$ Find the function $\phi(t,T)$ using risk-neutral value and PDE of Black-Scholes to find $\phi(t,T)$ solving a ODE.

I've done both computations but my answers do not match. I'm not sure if I'm understanding right both methods.

Using first method gives me $e^{(n-1)r(T-t)-\frac{n\sigma^2(T-t)(1-n(T-t))}{n}},$ by the other hand $e^{r(1-n)-\frac{r(1-n)-n(n-1)\sigma^2t}{2}}$

How am I doing wrong?

## Answer by Quantuple (score 2, accepted)

https://quant.stackexchange.com/a/36561

From Black-Scholes PDE

The option price $V(t,S)$ satisfies the Black-Scholes PDE: $$ \frac{\partial V}{\partial t} + rS \frac{\partial V}{\partial S} + \frac{1}{2} \sigma^2 S^2 \frac{\partial^2 V}{\partial S^2} - r V = 0 $$ with boundary condition $V(T,S) = g(S) = S^n $.

Plugging in the fact that $V(t,S) := \phi(t,T) S^n$ yields the following ODE for the function $\phi(t,T)$ $$ \phi' + \left( r n + \frac{1}{2} \sigma^2 n (n-1) - r \right) \phi = 0 \tag{1} $$ The solution of writes $$ \phi(t,T) = A e^{\mu t} $$ for some constant $A \in \Bbb{R}$ yet to be determined and $\mu$ the root of the characteristic polynomial of $(1)$ $$ \mu = - (n-1) \left( r + \frac{1}{2} \sigma^2 n \right) $$ Now applying the boundary condition $$V(T,S) = S^n \iff \phi(T,T) = 1$$ allows to determine the constant $A$ and we eventually find $$ \phi(t,T) = \exp\left( (n-1) \left( r + \frac{1}{2} \sigma^2 n \right) (T-t) \right) $$

From martingale pricing

In a Black-Scholes framework (GBM diffusion setting) and conditionally on the information available at $t$ we have that under the measure associated to the money market account numéraire (risk-neutral measure) $$ S_T = S_t \exp\left( \left( r - \frac{1}{2}\sigma^2\right)(T-t) + \sigma\left(W_T-W_t\right) \right) $$ such that $$ g(S_T) = S_T^n = S_t^n \exp\left( n\left( r - \frac{1}{2}\sigma^2\right)(T-t) + n\sigma\left(W_T-W_t\right) \right) $$ Adding and subtracting half of the quadratic variation of the process $n \sigma (W_T - W_t)$ in the argument of the exponential allows us to rewrite this as \begin{align} S_T^n &= S_t^n \exp\left( n\left( r - \frac{1}{2}\sigma^2\right)(T-t) + \frac{n^2 \sigma^2}{2}(T-t)- \frac{n^2 \sigma^2}{2}(T-t) + n\sigma\left(W_T-W_t\right) \right) \\ &= S_t^n \exp\left( \left( nr+ n(n-1)\frac{1}{2}\sigma^2\right) (T-t) \right) \mathcal{E}\left[ n\sigma(W_T-W_t) \right] \end{align} where $\mathcal{E}[\cdot]$ is a stochastic exponential of unit expectation (Doléans-Dade exponential). Now, since we're working in the measure associated to the money market account numéraire the price of the power option writes: \begin{align} V_t &= \Bbb{E}_t \left[ \exp(-r(T-t)) g(S_T) \right] \\ &= \exp(-r(T-t)) \cdot S_t^n \exp\left( \left( nr+ n(n-1)\frac{1}{2}\sigma^2\right) (T-t) \right) \cdot 1 \\ &= \underbrace{\exp\left( (n-1) \left( r+ n \frac{1}{2}\sigma^2\right) (T-t) \right)}_{:= \phi(t,T)} S_t^n \end{align}