Let $\sigma_t$ be a stochastic volatility. Then is it correct $$ \mathbb{E}\left[\exp\left(\int_0^t \sigma_s\,dB_s \right) \right] = \mathbb{E}\left[\exp\left(\frac{1}{2}\int_0^t \sigma_s^2\,ds\right) \right]? $$ If this is correct, then how can we prove?
What I have tried.
By using the property of the exponential martingale, we have $$ \mathbb{E}\left[e^{\int_0^t \sigma_s\,dB_s } \right] = \mathbb{E}\left[e^{\frac{1}{2}\int_0^t \sigma_s^2\,ds}\left( 1+\int_0^t Z_s \sigma_s\,dB_s \right) \right], $$ where $Z_s = \exp\left(\int_0^t \sigma_s\,dB_s - \frac{1}{2} \int_0^t \sigma_s^2\,ds \right)$. In my intuition, conditional expectation $$ \begin{split} \mathbb{E}\left[ e^{\frac{1}{2}\int_0^t \sigma_s^2\,ds} \int_0^t Z_s \sigma_s\,dB_s \middle| \cal{F}_{t-} \right] &= e^{\frac{1}{2}\int_0^{t-} \sigma_s^2\,ds} \int_0^{t-} Z_s \sigma_s dB_s \\ &+ e^{\frac{1}{2}\int_0^{t} \sigma_s^2\,ds} \int_{t-}^{t} Z_s \sigma_s \mathbb{E}[dB_s\mid \cal{F}_{t-}] \end{split} $$ is zero by induction, where the equality is from $\sigma_t$ is $\cal{F}_{t-}$-measurable. However, this seems to be not rigorous enough to consider as a proof.