# Expectation of the realized volatility

I was reading Zhang and Wang 2023 and I have some doubts regarding it. The realized Stochastic Volatility Model is expressed as follows:

$$\begin{matrix} y_t = \exp \big( \frac{h_t}{2} \big) \epsilon_t \quad \quad \quad & & & \epsilon_t\sim \text{N}(0,1), \\[12pt] z_t = \varsigma+h_t+u_t \quad \quad \ \ \ & & & \ \ u_t\sim N(0,\sigma^2_u), \\[12pt] h_{t+1} = \mu+\phi(h_t-\mu)+\eta_t \ & & & \ \ \ \eta_t\sim N(0,\sigma^2_\eta), \\[12pt] h_1 = \mu+\epsilon_0 \quad \quad \quad \quad \quad & & & \ \ \ \ \ \ \epsilon_0 \sim N(0,\frac{\sigma^2_\eta}{1-\phi^2}). \\[12pt] \end{matrix}$$

To ensure the strict stationarity and iterative nature of the stochastic process, the persistence parameter $$|\phi| < 1$$ is assumed in the logarithmic volatility equation $$h_t$$ and set $$h_1$$

My question: Why do we have the expectation: $$\mathbb{E}(z_t) = \varsigma+ \frac{\mu}{1-\phi}?$$

I cannot see why the expectation of $$h_t$$ should be $$\frac{\mu}{1-\phi}$$. $$h_t$$ should follow an AR(1) process and the expected value of an AR(1) process with drift is $$\frac{\mu}{1-\phi}$$ but in this case we have an additional term given by $$-\phi \mu$$. Am I missing something? Can someone explain why the $$\mathbb{E}(z_t)$$ is given by the above result?

my attempt was: since the AR(1) process is stationary given that $$|\phi|<1$$ the mean is constant over time and [\begin{align*} \mathbb{E}(h) - \phi\mathbb{E}(h) &= \mu - \phi\mu \\ (1-\phi)\mathbb{E}(h) &= \mu(1-\phi). \end{align*}] which gives $$\mathbb{E}(h) = \mu$$ why they got $$\frac{\mu}{1-\phi}$$?

• Cross-posted at Cross Validated. Sep 22 at 5:42