let $M_t$ be the following stochastic integral
$$ M_t = \int_0^t \sigma_s dW_s $$
where $\sigma_t$ is a sufficiently regular deterministic function and $W_t$ is a standard Wiener process (that is $W_t \sim \mathcal{N}(0,t)$ with independent increments).
It can be shown that $M_t$ is martingale with distribution $M_t \sim \mathcal{N}(0, \Sigma_t)$ where (Ito's isometry) I have defined the variance
$$ \Sigma_t = \int_0^t \sigma^2_s ds $$
Could you kindly check if the following two preliminary assertions are true (everywhere $0 \leq s<t<T$):
$M_t$ has independent increments, that is $M_s$ is independent from $M_t - M_s$.
Covariance: ${\mathbb E}[M_t M_s] = \Sigma_s$.
Proof of 2.: If 1. holds, reasoning as in the Wiener case: \begin{align} {\mathbb E}[M_t M_s] &= {\mathbb E}[(M_t - M_s + M_s )M_s] \\ &= {\mathbb E}[(M_t - M_s)M_s] + {\mathbb E}[M^2_s] \\ &= {\mathbb E}[M_t - M_s] \cdot {\mathbb E}[M_s] + {\mathbb E}[M^2_s] \\ &= {\mathbb E}[M^2_s] \\ &= \Sigma_s \end{align}
Finally, my question: Conditional expectation: $${\mathbb E}[M_t|W_T] = ?$$
Edit I’m aware of the result ${\mathbb E}[W_t|W_T]=\frac{t}{T}W_T$ using brownian bridge.
Thanks for your kind attention.
Edit2 This question has a follow-up which might be of interest as well: Regression of stochastic integral on Wiener process