Black–Scholes Pricing for an Inverse-Asset Call Payoff

# How to derive a valuation formula for an "Asymptotic" Call option?


# How to derive a valuation formula for an "Asymptotic" Call option?












Suppose you have an "Asymptotic" Call option with a payoff equal to $ K * max[1 - K/S_{T}, 0] $, where $K$ is the strike price and $S_{T}$ is the underlying asset's price at maturity.

How would one derive a closed-form valuation formula for this option using the Black-Scholes-Merton framework?

## Answer by Andrea (score 3)

https://quant.stackexchange.com/a/82104

One can apply Ito to $S^{-1}$ and obtain

$d S^{-1} = (\sigma^2 - r) S^{-1} dt - \sigma S^{-1} dW$

The payoff is a put on $S^{-1}$

$K^2 \max (K^{-1} - S^{-1}, 0)$

So the price comes from the Black-Scholes put formula

$K^2 \, \text{Put}\left (\frac{1}{S_0}, \frac{1}{K}, r, q=2r-\sigma^2, \sigma, T \right)$

## Answer by user35980 (score 2)

https://quant.stackexchange.com/a/82061

Interesting. Looks like some kind of deformed digital. By BSM framework I assume you mean $S_T$ follows GBM. Here goes: $$V=e^{-rT}E_Q\left[K\max\left(1-\frac{K}{S_T},0\right)\right] =e^{-rT}E_Q\left[K\left(1-\frac{K}{S_T}\right)1_{S_T>K}\right]$$ which can be simplified to $$ V=Ke^{-rT}E_Q\left[1_{S_T>K}\right] - K^2e^{-rT}E_Q\left[\left(\frac{1}{S_T}\right)1_{S_T>K}\right].$$ So the first term is a $K$-strike digital call with payout $K$. The second term is the "exotic" part which can also be solved for in closed form. So I think your option looks like a regular $K$-strike digital call with a funky payout.

See Kurt G's answer for the explicit solution.

## Answer by Kurt G. (score 2)

https://quant.stackexchange.com/a/82072

The correct pricing formula is $$\tag1 e^{-rT}KN(d_2)-e^{-2rT\,+\,\sigma^2T}\frac{K^2}{S_0}N(d_3) $$ where \begin{align*}\tag2 d_2&=\frac{\ln(S_0/K)+(r-\sigma^2/2)T}{\sigma\sqrt{T}}\,,\\[2mm] d_3&=\frac{\ln(S_0/K)+(r-3\,\sigma^2/2)T}{\sigma\sqrt{T}}=d_2-\sigma\sqrt{T}\,.\tag3 \end{align*} Note that the first term in (1) appears in the BSM formula with a minus sign.

To understand the second term observe that it comes from

$$\tag4 \frac{e^{-2rT}K^2}{S_0}\mathbb E\left[e^{\color{red}+\sigma W_T\color{red}+\sigma^2T/2}1_{\{S_T>K\}}\right]\,. $$ This can be evaluated as usual by completing the square: \begin{align*} &\int_{-\infty}^{d_2} e^{+\sigma\sqrt{T}x+\sigma^2T/2}\frac1{\sqrt{2\pi}}e^{-x^2/2}\,dx= e^{\sigma^2T}\int_{-\infty}^{d_2} e^{+\sigma\sqrt{T}x-\sigma^2T/2}\frac1{\sqrt{2\pi}}e^{-x^2/2}\,dx\\[2mm]&= e^{\sigma^2T}\int_{-\infty}^{d_2}\frac1{\sqrt{2\pi}}e^{-\frac{(x-\sigma\sqrt{T})^2}2}\,dx=e^{\sigma^2T}N(d_2-\sigma\sqrt{T})\,.\tag5 \end{align*} The following code helps to verify this numerically

```
from numpy import log, exp, sqrt
from scipy.integrate import quad
from scipy.stats import norm

def option( S, K, T, r, sig ):
    d2 = (log(S/K) + r*T - sig**2*T/2)/sig/sqrt(T)
    d3 = (log(S/K) + r*T - 3*sig**2*T/2)/sig/sqrt(T)
    Nd2 = norm.cdf(d2)
    Nd3 = norm.cdf(d3)
    return( exp(-r*T)*K*Nd2 - exp(-2*r*T + sig**2*T)*K**2/S*Nd3 )

def payoff( x, S, K, T, r, sig ):
    Sx = S*exp( x*sig*sqrt(T) - sig**2*T/2 + r*T )
    return( exp(-r*T)*K*max( 1 - K/Sx, 0 )*norm.pdf(x) )

K = 1.4
sig = 0.2
T = 3
S = 1.3
r = 0.02

PVnum = quad( payoff, -50, 50, args = (S,K,T,r,sig) )
print( PVnum[1] )
print( PVnum[0] )
print( option(S,K,T,r,sig) )
```