Let suppose that $S_t$ is a process defined as:

$$ \begin{cases}dS_t = \mu S_tdt+m(v_t)dW^1_t\\ dv_t = \mu_v(v_t)dt + \sigma_v(v_t)dW^2_t\end{cases}$$

where the two Brownian motion have correlation $\rho$. In an article, a property is mentioned and says:

Let $\delta>0$ and $n\in\mathbb{N}^*$, then we have: $$\begin{cases}\mathbb{E}\bigg(\ln^n(S_{t+\delta}/S_t)\bigg|\sigma(S_s,v_s),0\leq s \leq t\bigg) \perp S_t\\ \frac{\partial}{\partial S_t}\bigg[\mathbb{E}\bigg(\ln^n(S_{t+\delta}/S_t)\bigg|\sigma(S_s,v_s),0\leq s \leq t\bigg)\bigg] = 0\end{cases}.$$

Does anyone have an idea about a proof of this statement ?

Thank you very much!

Let suppose that $S_t$ is a process defined as:

$$ \begin{cases}dS_t = \mu S_t\,dt+m(v_t)\,dW^1_t\\ dv_t = \mu_v(v_t)\,dt + \sigma_v(v_t)\,dW^2_t\end{cases}$$

where the two Brownian motions have correlation $\rho$. In an article, a property is mentioned and says:

Let $\delta>0$ and $n\in\mathbb{N}^*$, then we have: $$\begin{cases}\mathbb{E}\bigg(\ln^n(S_{t+\delta}/S_t)\bigg|\sigma(S_s,v_s),0\leq s \leq t\bigg) \perp S_t, \\ \frac{\partial}{\partial S_t} \bigg[\mathbb{E}\bigg( \ln^n(S_{t+\delta}/S_t) \bigg| \sigma(S_s,v_s),0\leq s \leq t\bigg)\bigg] = 0. \end{cases}$$

Does anyone have an idea about a proof of this statement ?

Thank you very much!

Let suppose that $S_t$ is a process defined as:

$$ \begin{cases}dS_t = \mu S_tdt+m(v_t)dW^1_t\\ dv_t = \mu_v(v_t)dt + \sigma_v(v_t)dW^2_t\end{cases}$$

where the two brownian motion have correlation $\rho$. In an article, a property is mentionned and says:

Let $\delta>0$ and $n\in\mathbb{N}^*$, then we have: $$\begin{cases}\mathbb{E}\bigg(\ln^n(S_{t+\delta}/S_t)\bigg|\sigma(S_s,v_s),0\leq s \leq t\bigg) \perp S_t\\ \frac{\partial}{\partial S_t}\bigg[\mathbb{E}\bigg(\ln^n(S_{t+\delta}/S_t)\bigg|\sigma(S_s,v_s),0\leq s \leq t\bigg)\bigg] = 0\end{cases}.$$

Does anyone have an idea about a proof of this statement ?

Thank you very much!

Let suppose that $S_t$ is a process defined as:

$$ \begin{cases}dS_t = \mu S_tdt+m(v_t)dW^1_t\\ dv_t = \mu_v(v_t)dt + \sigma_v(v_t)dW^2_t\end{cases}$$

where the two Brownian motion have correlation $\rho$. In an article, a property is mentioned and says:

Let $\delta>0$ and $n\in\mathbb{N}^*$, then we have: $$\begin{cases}\mathbb{E}\bigg(\ln^n(S_{t+\delta}/S_t)\bigg|\sigma(S_s,v_s),0\leq s \leq t\bigg) \perp S_t\\ \frac{\partial}{\partial S_t}\bigg[\mathbb{E}\bigg(\ln^n(S_{t+\delta}/S_t)\bigg|\sigma(S_s,v_s),0\leq s \leq t\bigg)\bigg] = 0\end{cases}.$$

Does anyone have an idea about a proof of this statement ?

Thank you very much!

Let suppose that $S_t$ is a process defined as:

$$ \begin{cases}dS_t = \mu S_tdt+m(v_t)dW^1_t\\ dv_t = \mu_v(v_t)dt + \sigma_v(v_t)dW^2_t\end{cases}$$

where the two brownian motion have correlation $\rho$. In an article, a property is mentionned and says:

Let $\delta>0$ and $n\in\mathbb{N}^*$, then we have: $$\frac{\partial}{\partial S_t}\bigg[\mathbb{E}\bigg(\ln^n(S_{t+\delta}/S_t)\bigg|\sigma(S_s,v_s),0\leq s \leq t\bigg)\bigg] = 0.$$

Does anyone have an idea about a proof of this statement ?

Thank you very much!

Let suppose that $S_t$ is a process defined as:

$$ \begin{cases}dS_t = \mu S_tdt+m(v_t)dW^1_t\\ dv_t = \mu_v(v_t)dt + \sigma_v(v_t)dW^2_t\end{cases}$$

where the two brownian motion have correlation $\rho$. In an article, a property is mentionned and says:

Let $\delta>0$ and $n\in\mathbb{N}^*$, then we have: $$\begin{cases}\mathbb{E}\bigg(\ln^n(S_{t+\delta}/S_t)\bigg|\sigma(S_s,v_s),0\leq s \leq t\bigg) \perp S_t\\ \frac{\partial}{\partial S_t}\bigg[\mathbb{E}\bigg(\ln^n(S_{t+\delta}/S_t)\bigg|\sigma(S_s,v_s),0\leq s \leq t\bigg)\bigg] = 0\end{cases}.$$

Does anyone have an idea about a proof of this statement ?

Thank you very much!