# Pricing an Option with payoff $\left(1-\frac{K}{S_t}\right)^{+}$

Let $$S_t=S_0 \exp\left\{rt+0.5\sigma^2t+\sigma W_t\right\}$$ be the usual GBM model for a Stock price under the money-market numeraire.

Suppose we want to price an option with payoff at maturity: $$C_T=(1-\frac{K}{S_T})^{+}$$

Using the fundamental theorem, we have:

$$C_0=e^{-rT}\mathbb{E}^{\mathbb{Q}}\left[\left(1-\frac{K}{S_T}\right)\mathbb{I}_{S_T>K}\right]=e^{-rT}\mathbb{E}^{\mathbb{Q}}\left[\mathbb{I}_{S_T>K}-\frac{K}{S_T}\mathbb{I}_{S_T>K}\right]=\\=e^{-rt}N(d_2)-e^{-rT}K\mathbb{E}^{\mathbb{Q}}\left[\frac{\mathbb{I}_{S_T>K}}{S_T}\right]=\\=e^{-rT}N(d_2)-e^{-rT}K\int_{K}^{\infty}\left(\frac{1}{h}f_{S_T}(h)\right)dh=\\=e^{-rT}N(d_2)-e^{-rT}K\int_{K}^{\infty}\left(\frac{1} {h^2 \sqrt{t}\sigma \sqrt{2\pi}} \exp\left\{{-\frac{(\ln(h/S_0)-(r-0.5\sigma^2)t)^2}{2\sigma^2t}}\right\}\right)dh$$

Question 1: Now is there an easy way to solve the above integral analytically?

Question 2: Is there a smarter way to price this type of option, i.e. via a different Numeraire or something along similar lines?

Thank you so much for any hints,

Edit: For completeness, I found a hint in this question here, which leads to an alternative way of solving the pricing problem. Using that hint, the integral term can be simplified as follows:

$$K\mathbb{E}^{\mathbb{Q}}\left[\frac{1}{S_T}\mathbb{I}_{S_T>K}\right]=\frac{K}{S_0}\mathbb{E}^{\mathbb{Q}}\left[\frac{S_0}{S_T}\mathbb{I}_{S_T>K}\right]=\\=\frac{K}{S_0}\mathbb{E}^{\mathbb{Q}}\left[\exp\left\{-rT+0.5\sigma^2T-\sigma W_T\right\}\mathbb{I}_{S_T>K}\right]=\\=\frac{K}{S_0}e^{-rT+0.5\sigma^2T}\mathbb{E}^{\mathbb{Q}}\left[\exp\left\{-\sigma W_T\right\}\mathbb{I}_{S_T>K}\right]=\\=\frac{K}{S_0}e^{-rT+0.5\sigma^2T}\mathbb{E}^{\mathbb{Q}}\left[\exp\left\{-\sigma \sqrt{T}Z\right\}\mathbb{I}_{Z>-d_2}\right]=\\=\frac{K}{S_0}e^{-rT+0.5\sigma^2T}\int_{-d2}^{\infty}\left(\exp\left\{-\sigma \sqrt{T}h\right\}\frac{1}{\sqrt{2\pi}}\exp\left\{\frac{-h^2}{2}\right\}\right)dh=\\=\frac{K}{S_0}e^{-rT+0.5\sigma^2T}\int_{-d2}^{\infty}\left(\frac{1}{\sqrt{2\pi}}\exp\left\{\frac{-h^2-2\sqrt{T}\sigma+\sigma^2T - \sigma^2T }{2}\right\}\right)dh=\\=\frac{K}{S_0}e^{-rT+\sigma^2T}\int_{-d2}^{\infty}\left(\frac{1}{\sqrt{2\pi}}\exp\left\{\frac{-(h+\sigma \sqrt{T})^2}{2}\right\}\right)dh=\\=\frac{K}{S_0}e^{-rT+\sigma^2T}\mathbb{P}\left(Z-\sigma\sqrt{T}>-d2\right)=\\=\frac{K}{S_0}e^{-rT+\sigma^2T}\mathbb{P}\left(Z

So the final result would be:

$$C_0=e^{-rT}N(d_2)-\frac{K}{S_0}e^{-2rT+\sigma^2T}N(d_3)$$

Which is the same result as provided in the answer below.

• You can use the stock price as the numeraire and then it becomes a standard vanilla option. Commented Nov 15, 2020 at 16:09
• @Gordon: Than you. If I use the stock as Numeraire, then I get: $$C_0=S_0\mathbb{E}^\mathbb{Q}\left[ \frac{\left(1-\frac{K}{S_T}\right)^{+}}{S_T} \right]=\\=S_0\mathbb{E}^\mathbb{Q}\left[ \frac{1}{S_T}\mathbb{I}_{S_T>K}-\frac{K}{S_T^2}\mathbb{I}_{S_T>K} \right]$$ I am not sure that will make it easier to price than pricing under the money-market numeraire? Commented Nov 15, 2020 at 16:16

## $$\frac{1}{S_t}$$ is log-normal

If $$S_t$$ is a geometric Brownian motion, so is $$\frac{1}{S_t}$$ and indeed any power $$S_t^\alpha$$. Simply use Itô's Lemma and set $$f(t,x)=\frac{1}{x}$$, \begin{align*} \mathrm{d}f(t,S_t) &= \left(0-\mu S_t\frac{1}{S_t^2}+\frac{1}{2}\sigma^2S_t^2\frac{2}{S_t^3}\right)\mathrm{d}t-\sigma S_t \frac{1}{S_t^2}\mathrm{d}W_t \\ &=- \frac{1}{S_t}\left(\left(\mu -\frac{1}{2}\sigma^2\right)\mathrm{d}t+\sigma \mathrm{d}W_t\right). \end{align*}

Even simpler, you can see \begin{align*} S_t&=S_0\exp\left(\left(\mu-\frac{1}{2}\sigma^2\right)t+\sigma W_t\right) \\ \implies \frac{1}{S_t}&=S_0^{-1}\exp\left(-\left(\mu-\frac{1}{2}\sigma^2\right)t-\sigma W_t\right). \end{align*} The most trivial way is probably $$\ln\left(\frac{1}{S_t}\right)=-\ln(S_t)\sim N\left(-\ln(S_0)-\left(\mu-\frac{1}{2}\sigma^2\right)t,\sigma^2t \right).$$

## The rest is standard

Let $$X=e^{m+s Z}$$, where $$m=-\ln(S_0)-\left(r-\frac{1}{2}\sigma^2\right)T$$, $$s=\sigma\sqrt{T}$$ and $$Z\sim N(0,1)$$. Then, \begin{align*} \mathbb{E}\left[\max\left\{1-\frac{K}{S_T},0\right\}\right] &= K\mathbb{E}\left[\max\left\{\frac{1}{K}-X,0\right\}\right] \\ &= \Phi\left(-\frac{m+\ln(K)}{s}\right)-Ke^{m+0.5s^2}\Phi\left(-\frac{m+\ln(K)+s^2}{s}\right). \end{align*}

Then, $$e^{m+0.5s^2}= \frac{1}{S_0} e^{-\left(r-\sigma^2\right)T}$$ and of course, \begin{align*} \Phi\left(-\frac{m+\ln(K)}{s}\right) &=\Phi\left(\frac{\ln(S_0/K)+\left(r-\frac{1}{2}\sigma^2\right)T}{\sigma \sqrt{T}}\right)=:\Phi(d_0), \\ \Phi\left(-\frac{m+\ln(K)+s^2}{s}\right) &=\Phi\left(\frac{\ln(S_0/K)+\left(r-\frac{3}{2}\sigma^2\right)T}{\sigma \sqrt{T}}\right)=:\Phi(d_{-1}). \end{align*}

The final option price is then \begin{align*} V_0 = e^{-rT}\Phi\left(d_0\right)-\frac{K}{S_0}e^{-\left(2r-\sigma^2\right)T}\Phi\left(d_{-1}\right). \end{align*}

## Relationship to numéraires

You can see the drift $$r-\sigma^2$$ appearing in the terms $$e^{m+0.5s^2}$$ and $$\Phi\left(-\frac{m+\ln(K)+s^2}{s}\right)$$. This drift corresponds to a numéraire change as @Gordon suggests. Recall that the drift of $$S_t$$ under the stock measure $$\mathbb{S}$$ is $$r+\sigma^2$$, see here and your own question. This answer outlines power numéraires in great detail. The very end of this answer confirms that the drift of $$S_t$$ under a measure which uses the value process of $$S_t^{-1}$$ (that is $$V_t=e^{-r(T-t)}\mathbb{E}^\mathbb{Q}[S_T^{-1}|\mathcal{F}_t]$$) as numéraire is $$r-\sigma^2$$.

• Thanks so much, Kevin. Could we use $S_t^{-1}$ as Numeraire though? We'd have to suppose that $S_t^{-1}$ is a traded asset. Also, it's a not a Martingale under the measure $\mathbb{Q}$, so I am thinking it cannot be a traded asset? Commented Nov 15, 2020 at 17:03
• @JanStuller You can define any ''power numéraire'' (if that moments exists of course), even if the corresponding ''power asset'' isn't traded. A numéraire change is more of a mathematical trick than an economic argument. You're right with the martingale problem. You don't use $S_t^\alpha$ as numeraire, but the price process $V_t=e^{-r(T-t)}\mathbb{E}^\mathbb{Q}[S_T^\alpha|\mathcal{F}_t]$, whose discounted value is a $\mathbb{Q}$-martingale by construction (remember Joshi's 2017 paper). Commented Nov 15, 2020 at 17:06
• Ok, I get it now. Just for completeness, could you then pls edit the last sentence in your answer to say that "given that we'd use $V_t=e^{-r(T-t)}\mathbb{E}^\mathbb{Q}[S_T^{-1}|\mathcal{F}_t]$ as numeraire..."? Commented Nov 15, 2020 at 17:11
• @JanStuller one of my maths professor always said 'most errors occur in line two'! :D Commented Nov 16, 2020 at 19:01
• Haha, that is a legendary statement, I will make sure to remember that one! Thanks again :) Commented Nov 16, 2020 at 19:03