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?

  • $\begingroup$ There is dependence because the volatility process affects the dispersion of $S_t$. Note that $\sigma_t$ is a persistent process and that $S_t=S_0+\int_0^tS_s\sigma_sdW_s$. If $\sigma_t$ is high you would expect $Var(S_t)$ to be high. $\endgroup$
    – fes
    Jun 16, 2022 at 11:34
  • $\begingroup$ But I am asking about the (in)dependence of the vol process on the spot. $\endgroup$
    – user34971
    Jun 16, 2022 at 11:47
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    $\begingroup$ @markleeds I'm starting to feel very thick :) but $b$ is not a function of $S$, so how can the change in vol depend on (the change) in $S$? $\endgroup$
    – user34971
    Jun 16, 2022 at 13:02
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    $\begingroup$ Per the model, there is no formulistic dependence of $\sigma_t$ on $S_t$, but $S_t$ depends on $\sigma_t$. From a Bayesian standpoint, though, the distribution of $\Delta S_t$ yields information for the distribution of $\sigma_t$ $\endgroup$ Jun 16, 2022 at 13:08
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    $\begingroup$ @Frido: I think Fes's answer is great because he doesn't even involve the second equation. So, to simplify his explanation: When $\sigma_t$ is larger, the change in $S_t$ will generally be larger. Therefore, everything else remaining constant, $\sigma_t$ should be expected to be greater, the greater the value of the change in $S_t$. But the the value of the change in $S_t$ should be positively correlated with the value of $S_t$ as as long as $\sigma_t$ is not going back and forth in different directions in a volatile way. I think the latter is what fes means by persistence. $\endgroup$
    – mark leeds
    Jun 16, 2022 at 18:26

1 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.

  • $\begingroup$ How do you define “persistence”? $\endgroup$ Jun 16, 2022 at 13:04
  • $\begingroup$ @Daneel Olivaw I meant autocorrelated. $\endgroup$
    – fes
    Jun 16, 2022 at 13:06
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    $\begingroup$ +1 Thanks, haven't accepted it yet (hope you don't mind). Let me think abt it a bit more and see also what others answer if any. $\endgroup$
    – user34971
    Jun 16, 2022 at 13:59
  • $\begingroup$ @FridoRolloos Sure take your time:) $\endgroup$
    – fes
    Jun 16, 2022 at 14:07
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    $\begingroup$ I suspect that the fact that $$\log S_T/S_t = -\frac12 \int_t^T \sigma^2_u du + \int_t^T \sigma_u dW_u$$ at least shows that the integrated variance depends on among others $S_T$. Is there a way to show a similar relationship between $S_T$ and $\sigma_T$? Because that is I think what I am having trouble with. $\endgroup$
    – user34971
    Jun 16, 2022 at 19:13

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