I have the following system of SDE's

$ dA_t = \kappa_A(\bar{A}-A_t)dt + \sigma_A \sqrt{B_t}dW^A_t \\ dB_t = \kappa_B(\bar{B} - B_t)dt + \sigma_B \sqrt{B_t}dW^B_t $

If $\sigma_B > \sigma_A$ I would consider the volatility $B_t$ to be more volatile than $A_t$ because

$ d\langle A_\bullet\rangle_t = \sigma_A^2 B_t dt$ and $ d\langle B_\bullet\rangle_t = \sigma_B^2 B_t dt$

Now, if I rescale the process $B$ by $\sigma_A^2$ and define $\sigma_A^2B =\tilde{B}$, I get the an equivalent system of SDE's

$ dA_t = \kappa_A(\bar{A}-A_t)dt + \sqrt{\tilde{B}_t}dW^A_t \\ d\tilde{B}_t = \kappa_B(\sigma_A^2\bar{B} - \tilde{B}_t)dt + \sigma_A\sigma_B \sqrt{\tilde{B}_t}dW^B_t $

But now the claim "If $\sigma_B > \sigma_A$ I would consider the volatility $\tilde{B}_t$ to be more volatile than $A_t$" does not hold anymore. Consider $1>\sigma_B>\sigma_A$ and

$ d\langle A_\bullet\rangle_t = \tilde{B}_t dt$ and $ d\langle \tilde{B}_\bullet\rangle_t = \sigma_A^2\sigma_B^2 \tilde{B}_t dt$.

In this case the volatility $\tilde{B}$ of $A$ is more volatile than $A$ only if $\sigma_A^2\sigma_B^2>1$, which is completely different from the condition above (i.e., $\sigma_B > \sigma_A$).

What went wrong? Is there some error in the rescalling?

  • $\begingroup$ Changed $ d\langle A_\bullet\rangle_t = d\sigma_A^2 B_t dt$ to $ d\langle A_\bullet\rangle_t = \sigma_A^2 B_t dt$, and changed $ d\langle A_\bullet\rangle_t = d\tilde{B}_t dt$ to $ d\langle A_\bullet\rangle_t = \tilde{B}_t dt$. $\endgroup$ – Gordon Jul 2 '15 at 15:41

Loxol's answer is right. As long as you change your scale, your condition also changes, you can not more assume $\sigma_b>\sigma_a$ as you need also amend your condition in order to match your new rescaled process. Loxol already give the way of constructing new condition.

As you mentioned $\sigma_b>\sigma_a$ hold only for $B_t$ and $A_t$, this cannot be transferred to the new process. Consider $\sigma_b^2=1/2$ and $\sigma_a^2=1/4$, $B_t$ is more volatile than $A_t$. However, $\tilde B_t$ has $1/8$, which is not necassary volatile than $A_t$.

This is caused by changing the volatility structure with multiplication of $\sigma_A^2$. If $\sigma_A^2>1$, you emphasize the volatility of $B_t$ and for $\sigma_A^2<1$, you reduce the volatiliy. Only $\sigma_A^2=1$ the volatility structure remain unchanged, where your condition $\sigma_b>\sigma_a$ can be transferred to $\tilde B_t$. And the problem is just, if you choose $\sigma_A^2$<1, you are reducing the volatility of $\sigma_b$ and you cannot use $\sigma_b>\sigma_a$ for $\tilde B_t$. Therefore your assumption that $\sigma_b>\sigma_a$ implies $\tilde B_t$ is more volatile than $A_t$ is generally wrong.


If $\sigma_B > \sigma_A$, $B_t$ is more volatile than $A_t$. Now you define $\tilde{B_t}:=\sigma_A^2B_t $. The volatility of $\tilde{B_t}$ is equals to volatility of ${B_t}$ multiplied by $\sigma_A^2$. Therefore you can consider that if $\sigma_A^2\sigma_B > \sigma_A$, $\tilde{B_t}$ is more volatile than $A_t$. $\sigma_A^2\sigma_B > \sigma_A \leftrightarrow \sigma_A\sigma_B > 1$


When you scale your process you set the following: $\tilde{B}_t=f(B_t)=\sigma_A^2 B_t$ so then by means of Ito-Lemma, you get,

$df(B_t)=\partial_x f(B_t)dB_t+\frac{1}{2}\partial_{x^2}f(B_t)dt=\sigma_A^2 dB_t$


$d\tilde{B}_t=\sigma_A^2dB_t=\sigma_A^2 \kappa(\hat{B}_t-B_T)dt+\sigma_A^2\sigma_B dW_t^B$

but you can still symplify it and nothing has change such that your condition still holds.

  • $\begingroup$ My problem is the interpretation of both systems of SDE's. In the first, the volatility $B$ is more volatile than the process $A$ if and only if $\sigma_B>\sigma_A$ and in the second, the volatility $\tilde{B}$ is more volatile than the process $A$ if and only if $\sigma_B^2\sigma_A^2>1$. This throws me off since the conditions are completely changing. $\endgroup$ – Phun Jul 2 '15 at 7:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.