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Pricing of vanilla options

I will present the simplest way to obtain the prices for vanilla put and call options in the Black-Scholes modes. The trick is to use expectation pricing with optimal choice of numeraire.

Choice of numeraire: recall that no-arbitrage is equivalent to demanding all assets denominated with respect to the riskless bond are martingales. Instead of requiring for all assets that \( A/B \) is a martingale, one can equivalently measure values in units of stock instead of the bond, and demand that all \(A/S\) are martingales. The process that one denominates with respect to is called the numeraire, and one is free to choose it.

Call option

A vanilla call option on stock \(S\) with strike \(K\) and expiry \(T\) is defined by the pay-off at \(T\):

$$ C(t=T) = (S_T-K) 1_{S_T\geq K}, $$

where \(1_{S_T\geq K}\) is the indicator function. Let us denote the numeraire as \(N\). Then, the martingale property implies:

$$ \frac{C_0}{N_0} = \mathbb{E}\left[\frac{C_T}{N_T}\right] = \mathbb{E}\left[\frac{(S_T-K) }{N_T}1_{S_T\geq K}\right] = \mathbb{E}\left[\frac{S_T }{N_T}1_{S_T\geq K}\right] - \mathbb{E}\left[\frac{K}{N_T}1_{S_T\geq K}\right], $$

where all expectations are taken in the appropriate risk-neutral measure. The key point here is that we can treat the two terms separately and choose different numeraires for each. First, consider the second term. As \(K\) is a constant, it makes sense to choose the riskless bond as numeraire. In the bond numeraire risk-neutral measure, the stock follows the process:

$$ \frac{dS}{S} = rdt + \sigma dW \quad\Rightarrow\quad S_T = S_0 \exp\left( (r-\frac{1}{2}\sigma^2)T + \sigma\sqrt{T}\mathcal{N}(0,1) \right), $$

where we took an Ito integral to obtain \(S_T\). The contribution of the second term then takes the form:

$$ C(t=0) \supset \frac{B_0}{B_T} K \mathbb{E}[1_{S_T\geq K}] = Ke^{-rT} \int_{-\infty}^{\infty} \frac{dx}{\sqrt{2\pi}} e^{-x^2/2} 1_{S_T\geq K} = Ke^{-rT} \int_{x^*}^{\infty} \frac{dx}{\sqrt{2\pi}} e^{-x^2/2} $$

where

$$ S_0 \exp\left( (r-\frac{1}{2}\sigma^2)T + \sigma\sqrt{T}x^* \right) = K \quad\Leftrightarrow\quad x^* = \frac{\log(K/S_0) - (r-\sigma^2/2) T}{\sigma\sqrt{T}} $$

In terms of the cumulative normal distribution function \(\Phi(x)\), we have

$$ C(t=0) \supset Ke^{-rT} \Phi(-x^*), \quad \text{where} \quad \Phi(x) = \int_{-\infty}^{x}\frac{dy}{\sqrt{2\pi}} e^{-y^2/2}. $$

For the first term, we choose \(S_t\) as numeraire, and to determine the risk-neutral measure we demand \(B/S\) to be a driftless process. In the real-world measure:

$$ d\left(\frac{B}{S}\right) = \frac{dB}{S} + B d\left(\frac{1}{S}\right) + dB d\left(\frac{1}{S}\right) = \frac{B}{S} [rdt -\mu dt - \sigma dW + \sigma^2 dt], $$

and therefore in the risk-neutral measure:

$$ \frac{dS}{S} = (r+\sigma^2)dt + \sigma dW \quad\Rightarrow\quad S_T = S_0 \exp( (r + \frac{1}{2}\sigma^2)T + \sigma\sqrt{T}\mathcal{N}(0,1) ). $$

Evaluating the first expectation yields the following contribution to the call price:

$$ C(t=0) \supset S_0 \mathbb{E}[1_{S_T\geq K}] = S_0 \int_{-\infty}^{\infty} \frac{dx}{\sqrt{2\pi}} e^{-x^2/2} 1_{S_T\geq K} = S_0 \int_{y^*}^{\infty} \frac{dx}{\sqrt{2\pi}} e^{-x^2/2} = S_0 \Phi(-y^*) $$

where

$$ y^* = \frac{\log(K/S_0) - (r+\sigma^2/2) T}{\sigma\sqrt{T}} $$

Putting everything together, one obtains:

$$ C(t=0) = S_0 \Phi(a_+) - e^{-rT} K \Phi(a_-), \quad \text{where} \quad a_{\pm} = \frac{\log(S_0/K) + (r \pm \sigma^2 /2) T}{\sigma \sqrt{T}}. $$

Put option

This is straightforward to deduce from the call option, the pay-off at expiry is:

$$ P(t=T) = (K-S_T) 1_{K\geq S_T}, $$

and so

$$ P(t=0) = e^{-rT} K \Phi(-a_-) - S_0 \Phi(-a_+). $$