# Expected value and variance of the stock log-returns under Local Volatility framework

I want to calculate the expected value and the variance of the stock process log-returns in the Local Volatility setting (and the realized/terminal correlation but let us begin in the one-dimentional setting). First some intro and my progress so far.

Consider the standard local volatility SDE: $$d S_t = \mu(t) S_t\, dt + \sigma(t,S_t) S_t dW_t$$ with deterministic drift $$\mu(t)$$ and some deterministic local volatility function $$\sigma(t,x)$$. Using Ito's lemma we can derive the following SDE for the $$\ln S_t$$: $$d \ln S_t = \ln S_0 + \left( \mu(t) - \frac{1}{2}\sigma^2(t,S_t)\right) dt + \sigma(t, S_t)dW_t$$ which can be written in the integral form as $$\ln S_t = \ln S_0 + \int_0^t \left( \mu(u) - \frac{1}{2}\sigma^2(u,S_u)\right) du + \int_0^t \sigma(u,S_u) dW_u$$

I am interested in the log-returns over some (small) time interval $$\Delta t$$, thus, from the above we get $$\ln\left(\frac{S_{t+\Delta t}}{S_t}\right) = \ln(S_{t+\Delta t}) - \ln(S_{t}) = \\ = \int_t^{t+\Delta t} \left( \mu(u) - \frac{1}{2}\sigma^2(u,S_u)\right) du + \int_t^{t+\Delta t} \!\! \sigma(u,S_u) dW_u$$

Proceeding to the quantities that I want to calculate, namely, the expected value and variance of the log-returns, we have $$\mathbb{E}\left[ \ln\left(\frac{S_{t+\Delta t}}{S_t}\right) \right] =\\ = \mathbb{E}\left[ \int_t^{t+\Delta t} \left( \mu(u) - \frac{1}{2}\sigma^2(u,S_u)\right) du + \int_t^{t+\Delta t} \!\! \sigma(u,S_u) dW_u \right] = \\ = \int_t^{t+\Delta t} \mu(u) du - \frac{1}{2} \, \mathbb{E} \left[ \int_t^{t+\Delta t} \sigma^2(u,S_u) du\right] + \mathbb{E} \left[ \int_t^{t+\Delta t} \!\! \sigma(u,S_u) dW_u \right]$$ Since the last inegral is zero, we arrive at the following $$\begin{eqnarray}\label{ev} \mathbb{E}\left[ \ln\left(\frac{S_{t+\Delta t}}{S_t}\right) \right] = \int_t^{t+\Delta t} \mu(u) du - \frac{1}{2} \, \int_t^{t+\Delta t} \mathbb{E} \left[ \sigma^2(u,S_u) \right] du = \quad ... \,\, ??? \end{eqnarray}$$ Here is the place where I get stuck. Question 1: Is there a way to further simplify the second integral? It comes back in the variance derivation below, where I continue with this form of the expected value.

Further, the variance calculation goes as follows: $$\textrm{var} \left[ \ln\left(\frac{S_{t+\Delta t}}{S_t}\right) \right] = \mathbb{E}\left[ \ln\left(\frac{S_{t+\Delta t}}{S_t}\right) - \mathbb{E}\left[ \ln\left(\frac{S_{t+\Delta t}}{S_t}\right) \right] \right]^2 = \\ = \mathbb{E}\left[ \int_t^{t+\Delta t} \left( \mu(u) - \frac{1}{2}\sigma^2(u,S_u)\right) du + \int_t^{t+\Delta t} \!\! \sigma(u,S_u) dW_u \\ - \int_t^{t+\Delta t} \mu(u) du + \frac{1}{2} \, \int_t^{t+\Delta t} \mathbb{E} \left[ \sigma^2(u,S_u) \right] du \right]^2 = \\ = \mathbb{E}\left[ \int_t^{t+\Delta t} \!\! \sigma(u,S_u) dW_u -\frac{1}{2} \int_t^{t+\Delta t} \left( \sigma^2(u,S_u) - \mathbb{E} \left[ \sigma^2(u,S_u) \right] \right) du \right]^2 = ...$$ Now, we take the square and get the three components $$... = \mathbb{E} \left[ \int_t^{t+\Delta t} \!\! \sigma(u,S_u) dW_u \right]^2 \\ - \mathbb{E} \left[ \int_t^{t+\Delta t} \!\! \sigma(u,S_u) dW_u \times \int_t^{t+\Delta t} \left( \sigma^2(u,S_u) - \mathbb{E} \left[ \sigma^2(u,S_u)\right]\right) du \right] \\ + \frac{1}{4} \, \mathbb{E} \left[ \int_t^{t+\Delta t} \left( \sigma^2(u,S_u) - \mathbb{E} \left[ \sigma^2(u,S_u)\right]\right) du \right]^2$$

By virtue of the Ito's isometry the first component equals $$\mathbb{E} \left[ \int_t^{t+\Delta t} \!\! \sigma(u,S_u) dW_u \right]^2 = \int_t^{t+\Delta t} \mathbb{E} \left[ \sigma^2(u,S_u) \right] du$$ which is the integral we have already seen in the expected value formula. Thus, the whole variance of the log-return is (so far) given by $$\textrm{var} \left[ \ln\left(\frac{S_{t+\Delta t}}{S_t}\right) \right] = \int_t^{t+\Delta t} \mathbb{E} \left[ \sigma^2(u,S_u) \right] du \\ - \mathbb{E} \left[ \int_t^{t+\Delta t} \!\! \sigma(u,S_u) dW_u \times \int_t^{t+\Delta t} \left( \sigma^2(u,S_u) - \mathbb{E} \left[ \sigma^2(u,S_u)\right]\right) du \right] \\ + \frac{1}{4} \, \mathbb{E} \left[ \int_t^{t+\Delta t} \left( \sigma^2(u,S_u) - \mathbb{E} \left[ \sigma^2(u,S_u)\right]\right) du \right]^2$$

Question 2: In general, how to proceed further? Does the expected value of the product of the integrals simplify to something? Third component resembles the variance of the squared local volatility function - can we use it somehow?

Note that in case of constant volatility the variance of the log-return simplifies to $$\sigma^2 \Delta t$$, which is exactly what we define as volatility in the Black-Scholes setting. But I want to know what is the (empirical/realized) volatility in case of the Local Volatility model setting (as defined by the time scaled variance of the log-returns of the stock process).

Any advice/hint (also partial) would be appreciated!