conditional expectation and variance of integral stochastic of a Geometric Brownian motion

Question

Let $\sigma_s$ be a Geometric Brownian Motion (GBM), ie, $$ \sigma_s =\sigma_0 \exp(\frac{-\alpha^2}{2}s+\alpha W_s) $$ where $W_s$ is a Standard Brownian Motion. Calculate the conditional Expectation and variance of: $$ I_T = \int^T_0 \sigma_s\, ds $$ That is, $$ E[I_T\mid W_T] $$ and $$ \operatorname{Var}[I_T\mid W_T] $$ Could someone give me a help with that? Any advice is appreciated.

Answer 1

Here, I showed that $$\mathbb E[f((W_s)_{s\leq t})\mid W_t] = \mathbb E[f((Y_s^t+\frac stx)_{s\leq t})]_{x=W_t},$$ where $(Y_s^t)_{s\leq t}$ is a Brownian bridge.

Therefore $$\mathbb E\Big[\int_0^t\sigma_sds\mid W_t\Big] = \sigma_0\mathbb E\Big[\int_0^te^{\alpha W_s-\frac{\alpha^2}2s}ds\mid W_t\Big] = \sigma_0\mathbb E\Big[\int_0^te^{\alpha Y_s^t + \alpha x\frac st-\frac{\alpha^2}2s}ds\Big]_{x=W_t}.$$

Since $Y_s^t \sim N(0,s-\frac{s^2}t)$, we have that $$\mathbb E[e^{\alpha Y_s^t}] = e^{\frac12\alpha^2(s-\frac{s^2}t)}$$ and thus $$\mathbb E\Big[\int_0^t\sigma_sds\mid W_t\Big] = \sigma_0\int_0^te^{\alpha W_t\frac st -\frac{s^2}{2t}}ds. $$

Analogously, $$\mathbb E\Big[\Big(\int_0^t\sigma_sds\Big)^2\mid W_t\Big] = \sigma_0^2\mathbb E\Big[\Big(\int_0^te^{\alpha W_s-\frac{\alpha^2}2s}ds\Big)^2\mid W_t\Big]=\sigma_0^2\mathbb E\Big[\Big(\int_0^te^{\alpha Y_s^t + \alpha x\frac st-\frac{\alpha^2}2s}ds\Big)^2\Big]_{x=W_t} = 2\sigma_0^2\mathbb E\Big[\int_0^t\int_0^se^{\alpha Y_s^t + \alpha x\frac st-\frac{\alpha^2}2s+\alpha Y_r^t + \alpha x\frac rt-\frac{\alpha^2}2r}drds\Big]_{x=W_t}.$$

Since $\Bigg(\begin{matrix} Y_s^t \\ Y_r^t \end{matrix}\Bigg) \sim N(0,Q)$, where $Q = \Bigg(\begin{matrix} s-\frac{s^2}t & r - \frac{rs}t \\ r-\frac{rs}t & r - \frac{r^2}t \end{matrix}\Bigg)$, you can obtain the right formula for the variance.