# What is the price of the European option with the payoff of $\max(S^a-K,0)$?

I interpret such an option as a power option but I do not find any literatures or existing methods to price it. Can it be priced with Black-Scholes with simple changes?

Under the Black-Scholes framework the dynamics of $$S$$ is a GBM ($$dS_t = \alpha S_t dt + \sigma S_t dW_t$$).

Introduce a new variable $$Y_t:= S_t^a$$ for $$a$$ being a real valued constant. Then by Ito the dynamics of $$Y$$ is given by: $$dY=aS^{a-1}dS+\frac{1}{2}a(a-1)S^{a-2}(dS)^2 \\ = (a\alpha+\frac{1}{2}a(a-1)\sigma^2)S^adt+a\sigma S^a dW \\ = (a\alpha+\frac{1}{2}a(a-1)\sigma^2)Ydt+a\sigma Y dW$$ Let $$\mu :=a\alpha+\frac{1}{2}a(a-1)\sigma^2$$ and $$\gamma:=a\sigma$$ then $$Y_t$$ is a GBM ($$dY_t = \mu Y_t dt + \gamma Y_tdW_t$$) with drift $$\mu$$ and volatility $$\gamma$$.

Can it be priced with Black-Scholes with simple changes?

Yes. $$\max (S^a_T-K,0)=\max (Y_T-K,0)$$. $$X:=\max (Y_T-K,0)$$ is a payout equivalent to a European Call option with the underlying having a price process $$Y_t$$.You can use Black-Scholes formula to find the value of X at any time $$t \in [0,T]$$

• GBM = Geometric Brownian motion Mar 18 '19 at 16:21
• To complete this excellent answer and using @Sanjay notation, note that the BS formula for a call option, if the asset follows a GBM with drift $\mu$ and vol $\gamma$, is: $C(Y_t,t)=e^{-r(T-t)}(Y_te^{\mu(T-t)}\Phi(d_1)-K\Phi(d_2))$ where $d_1=\frac{1}{\gamma\sqrt{T-t}}(\ln\frac{Y_t}{K}+(\mu+\frac{1}{2}\gamma^2)(T-t))$ and $d_1=d_1-\gamma\sqrt{T-t}$ so as long as your option is written on a price process whose dynamics can be expressed as a GBM, then you can reuse the BS formula. Mar 18 '19 at 19:42
• @DaneelOlivaw I was to lazy to write down the formula :) I figured it was trivial and OP might already be familiar with the formula due to the phrasing of the question. But it does indeed complete my answer Mar 19 '19 at 15:04

Yes indeed. Bearing in mind that for the option to be in the money, the underlying needs to fulfill the following inequality to be in-the-money: $$S^a\geq K$$, i.e. $$S\geq K^{1/a}$$. Calling the terminal pdf of underlying prices $$\rho(\xi)$$, the price can be computed by evaluating the following integral:

$$C_a(K)=\int_{K^{1/a}}^{\infty}(\xi^a-K)\rho(\xi)d\xi$$