How Rescaling Changes Volatility Comparisons in Coupled SDEs

# Stochastic Calculus Rescale Exercise


# Stochastic Calculus Rescale Exercise












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?

## Answer by quallenjäger (score 2)

https://quant.stackexchange.com/a/31654

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.

## Answer by loxol (score 1)

https://quant.stackexchange.com/a/25128

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 $A_t$. $\sigma_A^2\sigma_B > \sigma_A \leftrightarrow \sigma_A\sigma_B > 1$

## Answer by Nico (score 0)

https://quant.stackexchange.com/a/18615

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,

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

hence

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

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