Pricing a Squared-Threshold Digital Call under Black–Scholes

# Pricing call option using risk-neutral martingale approach with squared stock price boundary?


# Pricing call option using risk-neutral martingale approach with squared stock price boundary?












I have to use the risk-neutral martingale 5 step approach under BS pricing framework to price the following call option at time 0: $$X = \begin{cases}1, &{if} &S_T^2\geq K,\\0, & {otherwise}.\end{cases}$$ As the squared stock price in the boundary condition is unusual (and unlike anything I have seen in literature) I tried to calculate $dS_T^2$ under P to then calculate the equivalent martingale measure (under Q). By using Ito's lemma, I got: $$dS_T^2 = (2 \alpha +\sigma^2)S_t^2dt + 2 \sigma S_t^2 dW_t$$ I then tried to calculate this under Q by adding and subtracting $2rS_t^2dt$ (suggested in lecture material) and then rearranging: $$dS_T^2 = (2r +\sigma^2)S_t^2dt + 2 \sigma S_t^2 d\tilde W_t$$ $$d\tilde W_t = \frac{\alpha - r}{\sigma} dt + dW_t$$ I planned on using this to then compute the closed-form expression for the fair price at t=0 for a digital call using the remainder of the martingale approach, however when calculating $d(S_T^2)^*$ I obtained: $$d(S_T^2)^*=[(2\alpha + \sigma^2)-2r] (S_t^2)^*dt + 2\sigma (S_t^2)^* dW_t$$ Which under Q (relacing $(2\alpha + \sigma^2)$ with $(2r +\sigma^2)$) is not a martingale so I cannot use it for the remainder of the steps. Is this method correct? If so, where have I gone wrong? If not, what approach should I take to the squared stock price in the boundary?

Sorry about any format issues, I am very new to MathJax! Thanks!

## Answer by Kevin (score 6, accepted)

https://quant.stackexchange.com/a/51658

You do not really need the dynamics of $S_t^2$. You can simply apply your standard technique from risk-neutral pricing. The time zero price of a European-style contract with payoff $X$ is given by $$V_0=e^{-rT}\mathbb{E}^\mathbb{Q}[X\mid\mathcal{F}_0].$$ Thus, \begin{align*} V_0 &= e^{-rT}\mathbb{E}^\mathbb{Q}[\mathbb{1}_{\{S_T^2\geq K\}}] \\ &= e^{-rT}\mathbb{E}^\mathbb{Q}[\mathbb{1}_{\{S_T\geq \sqrt{K}\}}] \\ &= e^{-rT}\mathbb{Q}[\{S_T\geq \sqrt{K}\}] \\ &= e^{-rT}\mathbb{Q}\left[\left\{\left(r-q-\frac{1}{2}\sigma^2\right)T+\sigma W_T \geq \ln\left(\frac{\sqrt{K}}{S_0}\right)\right\}\right] \\ &= e^{-rT}\mathbb{Q}\left[\left\{Z \geq \frac{\ln\left(\frac{\sqrt{K}}{S_0}\right)-\left(r-q-\frac{1}{2}\sigma^2\right)T}{\sigma \sqrt{T}}\right\}\right] \\ &= e^{-rT}\left(1-\Phi\left(\frac{\ln\left(\frac{\sqrt{K}}{S_0}\right)-\left(r-q-\frac{1}{2}\sigma^2\right)T}{\sigma \sqrt{T}}\right)\right) \\ &= e^{-rT}\Phi\left(\frac{\ln\left(\frac{S_0}{\sqrt{K}}\right)+\left(r-q-\frac{1}{2}\sigma^2\right)T}{\sigma \sqrt{T}}\right) \end{align*} where we used that $\Phi(-x)=1-\Phi(x)$ for all $x$ and $W_T\sim N(0,T)$.

This is now just the price of a digital (binary) cash-or-nothing call option with strike price $\sqrt{K}$ and maturity $T$.