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?

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:

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

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: $$dY=aS^{a-1}dS+a(a-1)S^{a-2}(dS)^2 \\ = (a\alpha+\frac{1}{2}a(a-1)\sigma^2)S^adt+a\sigma S^a dW \\ = (a\alpha+\frac{1}{2}a(a-1)\sigma^2)Ydt+a\sigma Y dW$$ 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 – Sanjay Mar 18 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. – Daneel Olivaw Mar 18 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 – Sanjay Mar 19 at 15:04