# Correlation of a lognormal asset and a normal asset

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.

Is this correct?

• 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 $\text{corr}(d\ln(X_t),d\ln(Y_t)) = \rho$. Jun 23, 2016 at 12:27
• 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 Jun 23, 2016 at 12:29
• @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 $corr(\mathrm{d}\ln(X_t), \mathrm{d} Y_t) = \rho$?
– will
Jun 23, 2016 at 12:47

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

• Note that the difference between instantaneous and terminal correlations I mentioned in my comments only matters when volatility is allowed to be stochastic. Jun 23, 2016 at 20:59
• decorrelation happens whenever one is not a scalar multiplier of the other. Jun 23, 2016 at 22:33
• @Gordon Just for the sake of clarification, what do you mean by "one" and "the other". Jun 24, 2016 at 1:09
• The volatility functions. Jun 24, 2016 at 1:10
• 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.
– will
Jun 24, 2016 at 12:43