Closed-Form Pricing of an Inverse-Price Option under Geometric Brownian Motion

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


# 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:

$$K\mathbb{E}^{\mathbb{Q}}\left[\frac{1}{S_T}\mathbb{I}_{S_T>K}\right]=\frac{K}{S_0}\mathbb{E}^{\mathbb{Q}}\left[\frac{S_0}{S_T}\mathbb{I}_{S_T>K}\right]=\\=\frac{K}{S_0}\mathbb{E}^{\mathbb{Q}}\left[\exp\left\{-rT+0.5\sigma^2T-\sigma W_T\right\}\mathbb{I}_{S_T>K}\right]=\\=\frac{K}{S_0}e^{-rT+0.5\sigma^2T}\mathbb{E}^{\mathbb{Q}}\left[\exp\left\{-\sigma W_T\right\}\mathbb{I}_{S_T>K}\right]=\\=\frac{K}{S_0}e^{-rT+0.5\sigma^2T}\mathbb{E}^{\mathbb{Q}}\left[\exp\left\{-\sigma \sqrt{T}Z\right\}\mathbb{I}_{Z>-d_2}\right]=\\=\frac{K}{S_0}e^{-rT+0.5\sigma^2T}\int_{-d2}^{\infty}\left(\exp\left\{-\sigma \sqrt{T}h\right\}\frac{1}{\sqrt{2\pi}}\exp\left\{\frac{-h^2}{2}\right\}\right)dh=\\=\frac{K}{S_0}e^{-rT+0.5\sigma^2T}\int_{-d2}^{\infty}\left(\frac{1}{\sqrt{2\pi}}\exp\left\{\frac{-h^2-2\sqrt{T}\sigma+\sigma^2T - \sigma^2T }{2}\right\}\right)dh=\\=\frac{K}{S_0}e^{-rT+\sigma^2T}\int_{-d2}^{\infty}\left(\frac{1}{\sqrt{2\pi}}\exp\left\{\frac{-(h+\sigma \sqrt{T})^2}{2}\right\}\right)dh=\\=\frac{K}{S_0}e^{-rT+\sigma^2T}\mathbb{P}\left(Z-\sigma\sqrt{T}>-d2\right)=\\=\frac{K}{S_0}e^{-rT+\sigma^2T}\mathbb{P}\left(Z<d2-\sigma\sqrt{T}\right)=\\=\frac{K}{S_0}e^{-rT+\sigma^2T}N(d_3)$$

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.

## Answer by Kevin (score 7, accepted)

https://quant.stackexchange.com/a/59384

### $\frac{1}{S_t}$ 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 $r-\sigma^2$.