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Expected Cross-Products of Brownian Motions with Random Correlation

Article Quant Q&A · Author: user1559897
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Summary

The document explains how the expected product of two Brownian motions relates to their instantaneous correlation. Their quadratic covariation accumulates correlation over time, so when correlation is deterministic, the product expectation is the same under the original and risk-neutral measures.

When correlation is random, the expectations instead depend on the expected correlation under each measure. A change of measure can change that expectation, so equality is not guaranteed. This distinction resolves the apparent failure: the shared pathwise covariation relation remains, but expectations of a random quantity need not match across probability measures. The discussion is conceptual and does not specify conditions under which the two measures would yield equal expected correlation.

Key ideas

  • The product of two Brownian motions has quadratic covariation equal to the time integral of their instantaneous correlation.
  • With deterministic correlation, the product expectation agrees under the original and risk-neutral measures.
  • With random correlation, expectations are taken under different measures and may differ.
  • The pathwise covariation identity does not imply equality of expectations across measures.

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# Can anyone explain one result regarding the correlation of Brownian motions?


# Can anyone explain one result regarding the correlation of Brownian motions?












Let $B_i(t), B_j(t)$ be two Brownian motion on a filtered probability space. Let {$\theta_1, \theta_2$} be a solution to the market risk of price equations and using {$\theta_1, \theta_2$}, the risk neutral Brownian motions are written as $\tilde{B_i}(t), \tilde{B_j}(t)$. It is easily shown that $$d\tilde{B_i}(t)d \tilde{B_j}(t) = \rho_{ij}(t)$$

Now suppose $\rho_{ij}(t)$ is not random, then $$dB_i(t){B_j}(t) = B_i(t)dB_i+ B_j(t)dB_i+dB_idB_j = B_i(t)dB_i+ B_j(t)dB_i+ \rho_{ij}(t)dt$$ Thus, $E[B_iB_j]=\int_0^{t}\rho_{ij}(u)du$

Similarly, $$\tilde{E}[\tilde{B_i}\tilde{B_j}]=\int_0^{t}\rho_{ij}(u)du.$$ Thus $$\tilde{E}[\tilde{B_i}\tilde{B_j}]=E[B_iB_j].$$

However, if $\rho_{ij}(t)$ is random, $$\tilde{E}[\tilde{B_i}\tilde{B_j}]\neq E[B_iB_j].$$ In this case which of the previous steps fail to hold?

## Answer by Gordon (score 1)

https://quant.stackexchange.com/a/30028

If $\rho_{i, j}(t)$ is deterministic, they are indeed equal. However, if $\rho_{i, j}(t)$ is random, then \begin{align*} E\left(B_i(t) B_j(t) \right) &= E\left(\int_0^t \rho_{i,j}(u)du \right)\\ &= \int_0^t E(\rho_{i, j}(u)) du, \end{align*} and, similarly, \begin{align*} \tilde{E}\left(\tilde{B}_i(t) \tilde{B}_j(t) \right) &= \int_0^t \tilde{E}(\rho_{i, j}(u)) du. \end{align*} They may be different.

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