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So if i want to calcualte the correlation between a pair of assets, my intuition is that i should calculate whatever correlation i plan on using;

When we look at correlation, it's normally the correlation of the log returns - which makes sense from a MC standpoint, since it's the correlated random numbers that create the returns.

If i want to simulate a set of paths of the pair of assets, and one will be simulated using a lognormal returns process and the other a random normal walk (absolute), then i should convert each time series into just the random number sequences which, when put through the process i will use, will recreate them - and then take the correlation of these numbers (assuming i'm using only historical correlation)?

i.e. work out the correlation of the underlying random sequences.

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    $\begingroup$ Seems to me like you are mixing instantaneous correlation (i.e. the linear correlation between the Brownian motions driving 2 stochastic processes) and terminal correlation (i.e. the linear correlation between two random variables e.g. two log-returns). The first corresponds to the $\rho$ in $ d\langle W_1, W_2 \rangle_t = \rho dt $ and the second corresponds to $\rho = \text{corr}(X_1, X_2)$. It so happens that for 2 assets $X_t$ and $Y_t$ modelled using GBMs with $d\langle W^X, W^Y \rangle_t = \rho dt$ then $\text{corr}(d\ln(X_t),d\ln(Y_t)) = \rho$. $\endgroup$
    – Quantuple
    Jun 23, 2016 at 12:27
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    $\begingroup$ When you look at the (linear) dependence between 2 time series people usually measure/refer to the terminal correlation (Pearson correlation coefficient). Yet, when you have to model the dependence between the same 2 time series using an SDE, you'll need to define/set up the instantaneous correlation coefficient between the 2 driving Brownian motions. As hinted above, under the BS model both are correlation values are the same. However, as soon as stochastic volatility kicks in, this is not true anymore $\endgroup$
    – Quantuple
    Jun 23, 2016 at 12:29
  • $\begingroup$ @Quantuple - i understand this, what i'm after really is just something to use as either an initial guess, or (depending on what exactly i'm modelling) just use it as it. If i want to obtain $\mathrm{d}\left< W_1,W_2 \right> = \rho \mathrm{d}t$ where the two processes are not both GBMs, but one is absolute instead, then can i do something like $corr(\mathrm{d}\ln(X_t), \mathrm{d} Y_t) = \rho$? $\endgroup$
    – will
    Jun 23, 2016 at 12:47

1 Answer 1

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Let $(X_t)_{t\geq 0}$ denote a Geometric Brownian Motion $$ \frac{dX_t}{X_t} = \mu_X dt + \sigma_X dW^X_t,\ \ \ X(0) = X_0$$ such that $X_t$ is lognormally distributed $\forall t > 0$ $$ X_t = X_0 e^{(\mu_X - \frac{1}{2}\sigma_X ^2)t + \sigma_X W_t^X}$$

Let $(Y_t)_{t\geq 0}$ denote an Arithmetic Brownian Motion $$ dY_t = \mu_Y dt + \sigma_Y dW_t^Y,\ \ \ Y(0)=Y_0 $$ such that $Y_t$ is normally distributed $\forall t > 0$ $$Y_t = Y_0 + \mu_Y t + \sigma_Y W_t^Y $$

Consider an instantaneous correlation between the driving Brownian motions $W^X$ and $W^Y$ $$ \rho := \frac{d\langle W^X, W^Y\rangle_t}{dt} $$

[Proposition] Specifying an instantaneous correlation $\rho$ between the two driving Brownian motions means that the two normal variables $\ln X_t$ and $Y_t$ exhibit a terminal correlation $\rho$.

[Proof] To see this, notice that \begin{align} \ln X_t &= \ln X_0 + (\mu_X - \frac{1}{2}\sigma^2_X)t + \sigma_X W^X_t \\ &= \ln X_0 + (\mu_X - \frac{1}{2}\sigma^2_X)t + \sigma_X (\rho W^Y_t + \sqrt{1-\rho^2}W^{Y,\perp}_t) \end{align} hence the covariance writes \begin{align} \text{cov}(\ln X_t,Y_t) &= \text{cov}(\ln X_0 + (\mu_X - \frac{1}{2}\sigma^2_X)t + \sigma_X (\rho W^Y_t + \sqrt{1-\rho^2}W^{Y,\perp}_t), Y_0 + \mu_Y t + \sigma_Y W_t^Y) \\ &= \text{cov}(\sigma_X (\rho W^Y_t + \sqrt{1-\rho^2}W^{Y,\perp}_t), \sigma_Y W_t^Y) \\ &= \text{cov}(\sigma_X \rho W^Y_t, \sigma_Y W_t^Y) + \text{cov}(\sigma_X \sqrt{1-\rho^2}W^{Y,\perp}_t, \sigma_Y W_t^Y) \\ &= \rho \sigma_X \sigma_Y \underbrace{\text{cov}(W^Y_t, W^Y_t)}_{=t} + \sqrt{1-\rho^2}\sigma_X \sigma_Y \underbrace{\text{cov}(W^{Y,\perp}_t,W_t^Y)}_{=0} \\ &= \rho \sigma_X \sigma_Y t \end{align} by bilinearity of the covariance operator and using the fact that $W^{Y,\perp}_t \perp W^Y_t$. In parallel we have: $$ \text{var}(\ln X_t) = \sigma_X^2 t $$ $$ \text{var}(Y_t) = \sigma_Y^2 t $$ so that $$ \text{corr} = \frac{\text{cov}(\ln X_t,Y_t)}{\sqrt{\text{var}(\ln X_t)\text{var}(Y_t)}} = \frac{\rho \sigma_X \sigma_Y t}{\sigma_X \sqrt{t} \sigma_Y \sqrt{t}} = \rho $$ which concludes the demonstration

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  • $\begingroup$ Note that the difference between instantaneous and terminal correlations I mentioned in my comments only matters when volatility is allowed to be stochastic. $\endgroup$
    – Quantuple
    Jun 23, 2016 at 20:59
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    $\begingroup$ decorrelation happens whenever one is not a scalar multiplier of the other. $\endgroup$
    – Gordon
    Jun 23, 2016 at 22:33
  • $\begingroup$ @Gordon Just for the sake of clarification, what do you mean by "one" and "the other". $\endgroup$
    – Quantuple
    Jun 24, 2016 at 1:09
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    $\begingroup$ The volatility functions. $\endgroup$
    – Gordon
    Jun 24, 2016 at 1:10
  • $\begingroup$ Exactly what i was after, thanks. I think what gordon's saying is that this won't (exactly) in a local vol model, except for very specific surfaces. $\endgroup$
    – will
    Jun 24, 2016 at 12:43

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