Pricing a Call with Multiple Brownian Volatility Drivers

# How to price a call option which depends on two Wiener processes?


# How to price a call option which depends on two Wiener processes?












Could someone explain to me why the regular call pricing formula works, just with $\sigma$ replaced by $\|\sigma\|$ in the case where the underlying asset depends on two Wiener processes?

For example, we might have had a model with two risky assets, and therefore two Wiener processes, in order to ensure no-arbitrage and completeness. And then we may be interested in a call option on only one of these stocks.

I am reading my textbook, and in it, it merely uses the regular formula for call option pricing, except it takes a norm of the two-dimensional volatility vector. But, it does not prove why this is the correct approach.

EDIT: Here's the picture of the text:

As you can see, he merely plugs in $\sigma_{S^f}$ (two-dimensional) in place of $\sigma$, in the regular call price formula, which is given as follows:

## Answer by LocalVolatility (score 3)

https://quant.stackexchange.com/a/32557

Let $\sigma \in \mathbb{R}_n^n$ and let $W$ be an n-dimensional standard Brownian motion. Define a new one-dimensional process $\tilde{W}$ by

\begin{equation} \tilde{W}_t = \frac{1}{\vert \vert \sigma \vert \vert} \sum_{i = 1}^n \sigma_i W_t^{(i)}. \end{equation}

Then it is easy to show that $\tilde{W}$ is a continuous martingale, starting at $\tilde{W}_0 = 0$ and with quadratic variation $\mathrm{d} \langle \tilde{W} \rangle_t = \mathrm{d}t$. It follows by Levy's characterization theorem that $\tilde{W}$ is a standard one-dimensional Brownian motion. Consequently, $S$ follows a geometric Brownian motion with respect to $\tilde{W}$ and diffusion coefficient $\vert \vert \sigma \vert \vert$, i.e

\begin{eqnarray} \mathrm{d}S_t & = & r S_t \mathrm{d}t + S_t \sum_{i = 1}^n \sigma_i \mathrm{d}W_t^i\\ & = & r S_t \mathrm{d}t + \vert \vert \sigma \vert \vert S_t \mathrm{d}\tilde{W}_t, \end{eqnarray}

and the usual Black-Scholes result applies.

Alternatively, you could note that

\begin{equation} \sum_{i = 1}^n \sigma_i W_t^{(i)} \end{equation}

is a sum of $n$ independent normal random variables. Thus, it is normally distributed with mean zero and variance

\begin{equation} \sum_{i = 1}^n \sigma_i^2 t = \vert \vert \sigma \vert \vert^2 t. \end{equation}

Consequently,

\begin{equation} \ln S_t \sim \mathcal{N} \left( S_0 + \left( r - \frac{1}{2} \vert \vert \sigma \vert \vert^2 \right) t, \vert \vert \sigma \vert \vert^2 t\right) \end{equation}

and again the usual Black-Scholes analysis applies.