Independence vs correlation in stochastic vol models

Question

I am struggling a bit with some basic stuff lately:

Consider a SV model \begin{align} dS_t &= \sigma_t S_t dW_t \\ d\sigma_t &= b(\sigma_t,t) dZ_t \end{align} with $dW_t dZ_t = 0$.

I know that zero correlation does not imply independence, and in fact $S_t$ is clearly not independent of $\sigma_t$.

However, I cannot see from the above SDEs how $\sigma_t$ can depend on $S_t$, in fact I think it doesn't.

But if $\sigma_t$ were independent of $S_t$, then the local volatility function $$ LV(K,T) := E_t [ \sigma^2_T | S_T = K] = E_t [ \sigma^2_T] $$ would not depend on $K$. But this would imply a flat local vol function which doesn't make sense.

What is wrong in my reasoning?

Answer 1

Consider the random variables $S_t$ and $\sigma_t$. Let their marginal cumulative distribution functions be $F_{S,t}$ and $F_{\sigma,t}$. The variables are said to be independent if their joint distribution function satisfies
$F_{S,\sigma,t}=F_{S,t}F_{\sigma,t}$. Independence is naturally symmetric. The best way to think about it intuitively is that knowing the value of one random variable gives no information about the other variable. Note that dependence between the random variables might not be in any sense causal.

Without specifying $b$ in your example it is difficult to formally prove independence/dependence. However, when $\sigma_t$ is persistent (autocorrelated), since $S_t=S_0+\int_0^{t}\sigma_sS_sdW_s$ you would expect high values of $\sigma_t$ to increase the dispersion of $S_t$, i.e. make observing extreme values of $S_t$ more likely. Alternatively, a high value of $\sigma_t$ is more likely when $S_t$ takes a value in the tails. Therefore you would not expect these two processes to generally be independent.