How to derive the price of a square-or-nothing call option?

Question

At maturity $T$, the holder of a "square-or-nothing" call option written on an underlying $S_t$ receives a payoff of the form $$ \phi(S_T) = \frac{S_T^2}{K} \pmb{1}_{\{S_T \geq K\}} = \begin{cases}\frac{S_T^2}{K}, &\text{ if }\ \ S_T \geq K, \\ 0, & \text{otherwise}.\end{cases} $$

Assume a Black-Scholes diffusion framework where the underlying's risk-neutral drift $\mu$ and volatility $\sigma$ are given.

Can one derive a closed-form pricing formula for such an option?

Answer 1

I provided an answer, based on an elementary approach, to an exactly same question yesterday. However, that question has disappeared, even though I like to keep a record for what I wrote. I would suggest that people do not delete their questions as they may be helpful for others. Here, I re-post that answer.

We assume that, under the risk-neutral measure, \begin{align*} S_T= S_0e^{(\mu-\frac{1}{2}\sigma^2)T + \sigma \sqrt{T}Z}, \end{align*} where $Z$ is a standard normal random variable. Let \begin{align*} d_1 = \frac{\ln \frac{S_0}{K} + (\mu+\frac{1}{2}\sigma^2)T }{\sigma\sqrt{T}}, \end{align*} and \begin{align*} d_2 = \frac{\ln \frac{S_0}{K} + (\mu-\frac{1}{2}\sigma^2)T }{\sigma\sqrt{T}}. \end{align*} Then, \begin{align*} e^{-rT}E\left(\frac{S_T^2}{K}\pmb{1}_{S_T >K} \right) &= \frac{e^{-rT} S_0^2}{K}E\left(e^{(2\mu-\sigma^2)T + 2\sigma \sqrt{T}Z}\pmb{1}_{S_0e^{(\mu-\frac{1}{2}\sigma^2)T + \sigma \sqrt{T}Z} >K} \right)\\ &= \frac{e^{(-r +2\mu-\sigma^2)T} S_0^2}{K}E\left(e^{2\sigma \sqrt{T}Z}\pmb{1}_{Z >-d_2} \right)\\ &=\frac{e^{(-r +2\mu-\sigma^2)T} S_0^2}{K}\int_{-d_2}^{\infty} e^{2\sigma \sqrt{T}z} \frac{1}{\sqrt{2\pi}} e^{-\frac{z^2}{2}}dz\\ &=\frac{e^{(-r +2\mu-\sigma^2)T} S_0^2}{K} \frac{1}{\sqrt{2\pi}}\int_{-d_2}^{\infty} e^{-\frac{1}{2}(z- 2\sigma \sqrt{T})^2 + 2 \sigma^2 T} dz\\ &= \frac{e^{(-r +2\mu+\sigma^2)T} S_0^2}{K}\frac{1}{\sqrt{2\pi}}\int_{-d_2- 2\sigma \sqrt{T}}^{\infty} e^{-\frac{1}{2}x^2} dx\\ &= \frac{e^{(-r +2\mu+\sigma^2)T} S_0^2}{K}\Phi(d_2+ 2\sigma \sqrt{T})\\ &= \frac{e^{(-r +2\mu+\sigma^2)T} S_0^2}{K}\Phi(d_1+ \sigma \sqrt{T}), \end{align*} where $\Phi$ is the cumulative distribution function of a standard normal random variable.

Answer 2

See this excellent paper by @MarkJoshi which defines/discusses the use of power numeraires.

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Starting from a dynamics specified under the risk-neutral measure $\mathbb{Q}$

\begin{align} &\frac{dS_t}{S_t} = (r-q) dt + \sigma dW_t^{\mathbb{Q}}\\ \iff& S_T\ \vert\ \mathcal{F}_t = S_t e^{(r-q-\frac{\sigma^2}{2})(T-t) + \sigma(W_T-W_t)} \tag{EQ.0} \end{align}

Let us consider the asset (power numéraire)

\begin{align} N_{t,T} :=&\ \ e^{-r(T-t)} \mathbb{E}^{\mathbb{Q}} \left[S_T^2\ \vert\ \mathcal{F}_t \right] \\ =&\ \ S_t^2 e^{\left( r - 2q + \sigma^2 \right)(T-t)} > 0, \forall t \tag{EQ.1} \in [0,T] \end{align}

and define the unique equivalent martingale measure $\mathbb{Q}^N$ which uses this asset as numéraire.

From the Girsanov theorem (or using the rationale described in the aforementioned paper), it is straightforward to infer the dynamics of the risky asset $S_t$ under the measure $\mathbb{Q}^N$

$$ \frac{dS_t}{S_t} = (r - q + 2\sigma^2) dt + \sigma dW_t^{\mathbb{Q}^N} \tag{EQ.2}$$

since the Radon-Nikodym happens to compute as

$$ \left. \frac{d\mathbb{Q}^N}{d\mathbb{Q}} \right\vert_{\mathcal{F}_T} = \frac{N_{T,T}\ B_0}{N_{0,T}\ B_T} = \frac{S_T^2}{N_{0,T}\ e^{rT}} = \mathcal{E} \left( 2\sigma W_T^{\mathbb{Q}} \right) \tag{EQ.3} $$

where the notation $\mathcal{E}(X_t)=\exp(X_t-\frac{1}{2}\langle X,X \rangle_t)$ denotes the stochastic exponential.

Now, the price of square-or-nothing option can be evaluated as \begin{align} V_0 &= e^{-rT} \mathbb{E}^\mathbb{Q} \left[ \frac{S_T^2}{K} \pmb{1}_{\{ S_T \geq K \}} \ \vert\ \mathcal{F}_0 \right] \\ &= e^{-rT} \mathbb{E}^\mathbb{Q^N} \left[ \frac{S_T^2}{K} \pmb{1}_{\{ S_T \geq K \}} \left(\frac{d\mathbb{Q}^N}{d\mathbb{Q}}\big|_{\mathcal{F}_T}\right)^{-1}\ \vert\ \mathcal{F}_0 \right]\ \ \ \text{(change of numéraire)} \\ &= N_{0,T} \mathbb{E}^\mathbb{Q^N} \left[ \frac{1}{K} \pmb{1}_{\{ S_T \geq K \}}\ \vert\ \mathcal{F}_0 \right]\ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{(use (EQ.3))} \\ &= \frac{N_{0,T}}{K} \mathbb{Q^N}(S_T \geq K) \end{align}

At this stage, because we have shown in $\text{(EQ.2)}$ that $S_T$ was lognormally distributed under $\mathbb{Q}^N$, plugging the definition $\text{(EQ.1)}$ of $N_{0,T}$ finally allows us to re-write the above equation as

\begin{align} V_0 &= \frac{S_0^2}{K} e^{\left( r - 2q + \sigma^2 \right)T} \Phi( d ) \\ d &= \frac{\ln \left(\frac{S_0}{K}\right) + \left(r-q+\frac{3}{2}\sigma^2\right)T }{\sigma\sqrt{T}}\\ \Phi(x) &= \mathbb{P}(X \leq x),\ X \sim N(0,1) \end{align} which is exactly the result given in @Gordon's answer and in Mark Joshi's paper, see middle of page 3.