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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!

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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 $T$.

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    $\begingroup$ @Actstu That is correct. You simply use the equivalent martingale measure which makes the discounted stock price $e^{-rt}S_t$ a martingale and you can work with this one. You can use this measure for all terminal payoffs $f(S_T)$. $\endgroup$ – KeSchn Mar 16 at 9:24
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    $\begingroup$ Thank you so much! This makes perfect sense. $\endgroup$ – Actstu Mar 16 at 9:24
  • $\begingroup$ @KeSchn: just saw this question, great answer. Was wondering: say I want to price a regular call option on $S_t^2$. We can derive the process for $S_t^2$ as Actstu did above: $d(S_T^2)^*=[(2\alpha + \sigma^2)-2r] (S_t^2)^*dt + 2\sigma (S_t^2)^* dW_t$. Then, you could set $2\alpha+\sigma^2-2r:=\tilde{r}$ and $2\sigma:=\tilde{\sigma}$ and by inspection apply the regular B-S formula whereby in $d_1$ and $d_2$, you replace $r$ with $\tilde{r}$ and $\sigma$ with $\tilde{\sigma}$. The problem is that $S_t^2$ is not a martingale under the risk-neutral measure, so intuitively B-S shouldn't work? $\endgroup$ – Jan Stuller Jul 20 at 11:46
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    $\begingroup$ @JanStuller Thank you very much! You're aksing about pricing a claim paying $\max\{S_T^2-K,0\}$, right? Have a look here. You simply define the new numeraire $S_t^2$ and obtain something like $N_{0,T}^2\mathbb Q^2[\{S_T\geq K\}]-K\Phi(d_2)$. The equations for the former terms are derived in the linked answer $\endgroup$ – KeSchn Jul 20 at 12:18
  • $\begingroup$ The second part of that option price should of course be discounted, that is $Ke^{-rT}\Phi(d_2)$ and not just $K\Phi(d_2)$. $\endgroup$ – KeSchn Jul 20 at 14:37

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