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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?

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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:

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

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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 $t \in [0,T]$

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  • $\begingroup$ GBM = Geometric Brownian motion $\endgroup$ – Sanjay Mar 18 at 16:21
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    $\begingroup$ 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. $\endgroup$ – Daneel Olivaw Mar 18 at 19:42
  • $\begingroup$ @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 $\endgroup$ – Sanjay Mar 19 at 15:04

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