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Cross-Asset Delta Hedging from Correlated Price Changes

Article Quant Q&A · Author: Sekots Reivan

Summary

The document explains why hedging an option on one asset with a correlated second asset can require a sensitivity that differs from the ordinary OLS slope between their price levels. Under a bivariate geometric Brownian motion, the instantaneous changes in the two prices have covariance that includes both assets’ price levels, their volatilities, and their correlation. Conditioning one price change on the other gives a hedge ratio proportional to correlation, the volatility ratio, and the ratio of current prices.

The key distinction is the modeled quantity: an OLS regression of price levels yields a slope based on level covariance, while the displayed hedge relation concerns conditional instantaneous price changes. The answer derives the relation using conditional normal distributions and illustrates the reverse conditional expectation as well. This is a model-based local hedge relationship; its practical accuracy depends on assumptions such as correlated diffusion dynamics and stable inputs, and it should not be mistaken for an unrestricted empirical regression result.

Key ideas

  • A cross-asset hedge ratio can be derived from the conditional expectation of one asset’s instantaneous price change given another’s.
  • Under correlated geometric Brownian motion, the ratio includes the current price levels as well as correlation and relative volatility.
  • An OLS slope for price levels is not automatically the relevant sensitivity for hedging instantaneous price movements.
  • The derivation relies on a bivariate normal diffusion model and is therefore conditional on those modeling assumptions.

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Full text
# Delta Hedging with a Different Underlying


# 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?

## Answer by Kermittfrog (score 4, accepted)

https://quant.stackexchange.com/a/69477

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?

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.