Price of European option with payoff $V(T)=\max (S^2(T) - K,0)$

Deriving the dynamics of $X(t) = S^2(t)$.

In the Black-Scholes world, the stock price $S(t)$ has $\mathbb{Q}$-dynamics:

$$ dS_t = rS(t)dt + \sigma S(t) dW^{\mathbb{Q}}(t) \tag{1} $$

Let $X(t) = S^2(t)$. The square of the stock price has the $\mathbb{Q}$-dyamics:

$$ \begin{align*} dX_t &= 2S(t)dS(t) + \frac{1}{2}\cdot 2 \cdot dS_t \cdot dS_t\\ &= 2S(t) ( rS(t)dt + \sigma S(t) dW^{\mathbb{Q}}(t)) + \sigma^2 S^2(t) dt \\ &= S^2(t)[(2r + \sigma^2)dt + 2\sigma dW^{\mathbb{Q}}(t)]\\ &= X(t)[(2r + \sigma^2)dt + 2\sigma dW^{\mathbb{Q}}(t)] \end{align*} $$

Let $f(x) = \log x$. By Ito’s formula:

$$ \begin{align*} d(\log X_t) &= \frac{1}{X(t)}dX(t) + \frac{1}{2}\cdot \left(-\frac{1}{X^2(t)}\right)dX(t)\cdot dX(t)\\ &= (2r + \sigma^2)dt + 2\sigma dW^{\mathbb{Q}}(t) - \frac{1}{2}\cdot 4\sigma^2 dt\\ &= (2r - \sigma^2)dt + 2\sigma dW^{\mathbb{Q}}(t) \\ X(t) &= X(0) \exp[(2r - \sigma^2)t + 2\sigma W^{\mathbb{Q}}(t)] \tag{2} \end{align*} $$

Computing the expectations.

By the risk neutral pricing formula, the price of the claim $V(T) = \max(S^2(T) - K,0)$ at time $t$ is:

$$ \begin{align*} V(t,T) &= \mathbb{E}^{\mathbb{Q}}[e^{-r(T-t)}(S^2(T) - K)\cdot 1_{S^2(T) > K}|\mathcal{F}_t] \\ &= e^{-r(T-t)}\mathbb{E}^{\mathbb{Q}}[S^2(T) 1_{S^2(T) > K}|\mathcal{F}_t] - Ke^{-r(T-t)}\mathbb{E}^{\mathbb{Q}}[1_{S^2(T) > K}|\mathcal{F}_t] \end{align*}\tag{3} $$

Computing $\mathbb{E}^{\mathbb{Q}}[1_{S(T) > K}|\mathcal{F}_t]$.

Define :

$$ \begin{align*} \tau &:= T - t\\ d_{1}(\tau,x) &= \frac{\log\frac{x}{\sqrt{K}} + \left(r + \frac{3}{2}\sigma^2\right)\tau}{\sigma \sqrt{\tau}}\\ d_{2}(\tau,x) &= \frac{\log\frac{x}{\sqrt{K}} + \left(r - \frac{\sigma^2}{2}\right)\tau}{\sigma \sqrt{\tau}} \end{align*} $$

The second expectation $\mathbb{E}^{\mathbb{Q}}[1_{S(T) > K}|\mathcal{F}_t]$ is computed in the standard way.

$$ \begin{align*} \mathbb{E}^{\mathbb{Q}}[1_{S^2(T) > K}|\mathcal{F}_t] &= \mathbb{Q}(S^2(T) > K|\mathcal{F_t})\\ &=\mathbb{Q}\left(\log S^2(T) > \log K|\mathcal{F_t}\right)\\ &=\mathbb{Q}\left(\log S^2(t) + (2r - \sigma^2)\tau + 2\sigma \sqrt{\tau} Z > \log K|\mathcal{F_t}\right)\\ &=\mathbb{Q}\left( 2\sigma \sqrt{\tau} Z > \log K - \log S^2(t) - 2(r - \sigma^2/2)\tau|\mathcal{F_t}\right)\\ &=\mathbb{Q}\left(Z > \frac{\log K - \log S^2(t) - 2(r - \sigma^2/2)\tau}{2\sigma \sqrt{\tau}}|\mathcal{F_t}\right)\\ &=\mathbb{Q}\left(Z < \frac{\log S^2(t) - \log K + 2(r - \sigma^2/2)\tau}{2\sigma \sqrt{\tau}}|\mathcal{F_t}\right)\\ &=\mathbb{Q}\left(Z < \frac{\log \frac{S^2(t)}{K} + 2(r - \sigma^2/2)\tau}{2\sigma \sqrt{\tau}}|\mathcal{F_t}\right)\\ &=\mathbb{Q}\left(Z < \frac{\log \frac{S(t)}{\sqrt{K}} + (r - \sigma^2/2)\tau}{\sigma \sqrt{\tau}}|\mathcal{F_t}\right)\\ &=\Phi(d_{2}(\tau,S(t))) \end{align*} \tag{4} $$

Computing $\mathbb{E}^{\mathbb{Q}}[S^2(T) 1_{S^2(T) > K}|\mathcal{F}_t]$.

Let $\tilde{\mathbb{Q}}$ be another probability measure related to $\mathbb{Q}$ defined by the Radon-Nikodym derivative:

$$ \begin{align*} \frac{d\tilde{\mathbb{Q}}}{d\mathbb{Q}} &= \frac{N(T)/B(T)}{N(0)/B(0)} \\ &= \frac{S^2(T)/e^{rT}}{S^2(0)/1}\\ &= \exp\left[(r-\sigma^2)T + 2\sigma W^{\mathbb{Q}(T)}\right]\\ &= \exp\left[2\sigma W^{\mathbb{Q}(T)}-\frac{1}{2}4\sigma^2 T\right]\exp\left[rT + \sigma^2 T\right] \end{align*} $$

Using Girsanov’s theorem, we can express $W^{\tilde{\mathbb{Q}}}(t) = W^{\mathbb{Q}}(t) - 2\sigma t$ and the constant expression $\exp\left[rT + \sigma^2 T\right]$ is taken care of at the end, after we have found a suitable CDF.

By the change of measure theorem, the first expectation can be expressed as follows:

$$ \begin{align*} \mathbb{E}^{\tilde{\mathbb{Q}}}[1_{S^2(T)>K}|\mathcal{F}_t] &= \mathbb{E}^{\mathbb{Q}}\left[\frac{(d\tilde{\mathbb{Q}}/d\mathbb{Q})_T} {(d\tilde{\mathbb{Q}}/d\mathbb{Q})_t}1_{S^2(T)>K}|\mathcal{F}_t\right]\\ \mathbb{E}^{\tilde{\mathbb{Q}}}[1_{S^2(T)>K}|\mathcal{F}_t] &= \mathbb{E}^{\mathbb{Q}}\left[\frac{S^2(T)e^{-rT}} {S^2(t)e^{-rt}}1_{S^2(T)>K}|\mathcal{F}_t\right]\\ S^2(t)\mathbb{E}^{\tilde{\mathbb{Q}}}[1_{S^2(T)>K}|\mathcal{F}_t] &= \mathbb{E}^{\mathbb{Q}}\left[S^2(T)e^{-r(T-t)} 1_{S^2(T)>K}|\mathcal{F}_t\right] \tag{5} \end{align*} $$

Computing $\mathbb{E}^{\tilde{\mathbb{Q}}}[1_{S^2(T)>K}|\mathcal{F}_t]$.

We can now solve for $\tilde{\mathbb{Q}}(S^2(T) > K)$ as:

$$ \begin{align*} \mathbb{E}^{\tilde{\mathbb{Q}}}[1_{S^2(T)>K}|\mathcal{F}_t] &= \tilde{\mathbb{Q}}\left(S^2(T) > K\right)\\ &=\tilde{\mathbb{Q}}\left(\log S^2(T) > \log K\right)\\ &=\tilde{\mathbb{Q}}\left(\log S^2(t) + (2r - \sigma^2)\tau + 2\sigma W^{\mathbb{Q}}(\tau) > \log K\right)\\ &=\tilde{\mathbb{Q}}\left(\log S^2(t) + (2r - \sigma^2)\tau + 2\sigma (W^{\tilde{\mathbb{Q}}}(\tau) + 2\sigma \tau) > \log K\right)\\ &=\tilde{\mathbb{Q}}\left(\log S^2(t) + (2r + 3\sigma^2)\tau + 2\sigma W^{\tilde{\mathbb{Q}}}(\tau) > \log K\right)\\ &=\tilde{\mathbb{Q}}\left(2\sigma W^{\tilde{\mathbb{Q}}}(\tau) > \log K - \log S^2(t) - (2r + 3\sigma^2)\tau \right)\\ &=\tilde{\mathbb{Q}}\left(W^{\tilde{\mathbb{Q}}}(1) > \frac{\log \sqrt{K} - \log S(t) - (r + \frac{3}{2}\sigma^2)\tau}{\sigma\sqrt{\tau}}\right)\\ &=\tilde{\mathbb{Q}}\left(W^{\tilde{\mathbb{Q}}}(1) < \frac{\log \frac{S(t)}{\sqrt{K}} + (r + \frac{3}{2}\sigma^2)\tau}{\sigma\sqrt{\tau}}\right)\\ &=\Phi(d_1(\tau,S(t)) \end{align*} $$

Conclusion.

The price of the European power call option having payoff $V(T)=\max (S^\alpha(T) - K,0)$, where $\alpha = 2$ is:

$$ V(t,T) = S^2(t)\Phi(d_1(\tau,S(t))) - Ke^{-r(T-t)}\Phi(d_2(\tau,S(t))) $$