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

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?

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.

• Hey, hope I can revive this post a little bit. Could you elaborate a little on how you went from $$\frac{e^{(-r +2\mu-\sigma^2)T} S_0^2}{K} \frac{1}{\sqrt{2\pi}}\int_{-d_2}^{\infty} e^{-\frac{1}{2}(z- 2\sigma \sqrt{T})^2 + 2 \sigma^2 T} dz$$ to $$\frac{e^{(-r +2\mu+\sigma^2)T} S_0^2}{K}\frac{1}{\sqrt{2\pi}}\int_{-d_2- 2\sigma \sqrt{T}}^{\infty} e^{-\frac{1}{2}x^2} dx$$ I am either not familiar with this type of transformation or I'm missing something simple. Thanks in advance! Commented Mar 7, 2019 at 15:19
• Let $x=z-2\sigma\sqrt{T}$, and change the integral limits. Commented Mar 7, 2019 at 15:22
• Yeah, it's the changing of the integral limits that is confusing me. Could you elaborate what exactly happens there? Like is there maybe one extra step in between you could write out to make it more obvious what happens? Commented Mar 7, 2019 at 15:26
• I think this is from the first year calculus. If $z=-d_2$, what is $x$?, and, if $z=\infty$, what will be $x$? Commented Mar 7, 2019 at 15:33
• Ah, I imagined it was something rather simple. Thanks alot for the help. Commented Mar 7, 2019 at 15:35

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

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.

• @MarkJoshi and Quantuple, thanks. That is an excellent paper. Commented May 25, 2016 at 13:50
• It is :) you are welcome Commented May 25, 2016 at 15:43
• I made some changes to (EQ.3) and the step for numeraire change. Commented May 25, 2016 at 16:15
• Yes and you are perfectly right, thank you for that :) Commented May 25, 2016 at 16:20
• The link to Joshi's paper seems to be dead. Can anyone find it? Commented Dec 18, 2017 at 0:37