Correlating Lognormal and Normal Processes in Simulation

# Correlation of a lognormal asset and a normal asset


# 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?

## Answer by Quantuple (score 7, accepted)

https://quant.stackexchange.com/a/27768

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