# Delta Hedging with a Different Underlying

In Bouzoubaa and Osseiran page 68 equation 5.3, the authors discuss delta hedging a call written for asset $$S_1$$ using a different but correlated underlying asset $$S_2$$. The authors provide the following formula.

$$\frac{\partial S_2}{\partial S_1}=\rho_{1,2}\frac{\sigma_2S_2}{\sigma_1S_1},$$

where $$\rho_{1,2}$$ is the correlation between $$S_1$$ and $$S_2$$ and $$\sigma$$ refers to volatility (standard deviation).

My Question

Should this formula be

$$\frac{\partial S_2}{\partial S_1}=\rho_{1,2}\frac{\sigma_2}{\sigma_1},$$

instead? Because this would be the $$\hat{\beta}$$ in an OLS regression (see here). Intuitively, if we fit

$$S_2 = \hat{\alpha} + \hat{\beta} S_1+\varepsilon,$$

then

$$\frac{\partial S_2}{\partial S_1}=\hat{\beta}=\rho_{1,2}\frac{\sigma_2}{\sigma_1}.$$

Am I missing something? What are the terms $$S_1$$ and $$S_2$$ doing in the printed formula?

I'd tackle this using some handwavery application of the bivariate conditional normal distribution to the geometric brownian motions:

For a multivariate normally distributed random variable $$X$$, partitioned into two blocks $$X_1,X_2$$, i.e. $$X\equiv\left(X_1,X_2\right)^T$$ with zero mean and covariance matrix

$$\Sigma\equiv\begin{pmatrix} \Sigma_{11} & \Sigma_{12} \\ \Sigma_{21} & \Sigma_{22} \end{pmatrix}$$

the conditional distribution of $$X_1$$ given $$X_2$$ is also Normal, with mean end variance

\begin{align} \mathrm{E}(X_1|X_2=x_2)&=\Sigma_{12}\Sigma_{22}^{-1}x_2\\ \mathrm{V}(X_1|X_2=x_2)&=\Sigma_{11}-\Sigma_{12}\Sigma_{22}^{-1}\Sigma_{21} \end{align}

In the bivariate case $$\mathrm{E}(x_1|x_2)=\frac{\sigma_1\sigma_2\rho}{\sigma_2^2}x_2=\rho\frac{\sigma_1}{\sigma_2}x_2$$ and $$\mathrm{V}(x_1|x_2)=\sigma_1^2(1-\rho^2)$$

### How it applies to your question

Let $$S_1,S_2$$ follow a bivariate geometric Brownian motion:

\begin{align} \frac{dS_i}{S_i}&=\mu_i dt+\sigma_idW_i \end{align} with $$dW_1dW_2=\rho dt$$. Then

$$\begin{pmatrix}dS_1&dS_2\end{pmatrix}^T\sim\mathbf{N}\left(\mathbf{0}\times dt,\begin{pmatrix}S_1^2\sigma_1^2&S_1S_2\sigma_1 \sigma_2\rho\\ S_1S_2 \sigma_1 \sigma_2 \rho&S_2^2\sigma_2^2\end{pmatrix}dt\right)$$

Applying the above:

$$\mathrm{E}\left(x_1|x_2\right)=\frac{S_1S_2\sigma_1\sigma_2\rho}{S_2^2\sigma_2^2}x_2=\frac{S_1}{S_2}\frac{\sigma_1}{\sigma_2}\rho x_2$$

and

$$\mathrm{E}\left(x_2|x_1\right)=\frac{S_1S_2\sigma_1\sigma_2\rho}{S_1^2\sigma_1^2}x_1=\frac{S_2}{S_1}\frac{\sigma_2}{\sigma_1}\rho x_1$$

i.e.

$$\mathrm{E}(dS_2)/dS_1=\frac{S_1S_2\sigma_1\sigma_2\rho}{S_1^2\sigma_1^2}=\frac{S_2}{S_1}\frac{\sigma_2}{\sigma_1}\rho$$

which yields the statement from the book.

HTH?