In Sergii Kuchuk and Yuliya Mishura paper, Pricing the European Call Option in the Model with Stochastic Volatility Driven by Ornstein-Uhlenbeck Process, Exact Formulas, the model can be represented by $$ dS_t = \mu S_t dt + \sigma(Y_t) S_t dB_t $$ $$ dY_t = -\alpha Y_t dt + k dW_t $$ One of the assumptions in the paper is the volatility function $ \sigma : \mathbb{R} \to \mathbb{R}_+ $ is measureable and bounded away from zero by a constant $ c $ and satisfies the conditions $ \int_{0}^{T} \sigma^2(Y_t) dt < \infty $. Later it is said that given the square integrability of $ \sigma(Y_s) $ the solution of the first equation of the model is $$ S_t = S_0 exp \bigg{(} \mu t - \frac{1}{2} \int_{0}^{t} \sigma^2(Y_S) ds + \int_{0}^{t} \sigma(Y_s) dB_s \bigg{)} $$
Question
- Does the statement "given the square integrability of $ \sigma(Y_s) $" imply $ \int_{0}^{T} \sigma^2(Y_t) dt < \infty $? I know that $ \sigma(Y_s) $ is square integrable if $ \mathbb{E} \big{(} \int_{0}^{t} \sigma(Y_s)^2 ds \big{)} < \infty $. Can we imply if $ \mathbb{E} \big{(} \int_{0}^{t} \sigma(Y_s)^2 ds \big{)} < \infty $ then $ \int_{0}^{T} \sigma^2(Y_t) dt < \infty $?
- Why $ \sigma $ should be bounded away from zero?