Let $X_t$ be an Ito's process where
$dX_t=b_tdt+\sigma_tdW_t=\tilde{b}_tdt+\tilde{\sigma}_tdW_t$. Prove
$b_t=\tilde{b}_t$ and $\sigma_t=\tilde{\sigma}_t$ a.s
Here my solution for $b_t=\tilde{b}_t$ a.s.
- $X_t=X_0 + \int^t_0b_sds + \int^t_0\sigma_sdW_s.$ Therefore setting $X_0=E(X_t)$ and taking the Expectation of the previous equation, $EX_t=EX_t + E\int^t_0b_sds + E\int^t_0\sigma_sdW_s \implies E\int^t_0b_sds =0$, since $E\int^t_0\sigma_sdW_s=0$
Hence doing the same for $\tilde{b}_t$, I would get $E\int^t_0b_sds =E\int^t_0\tilde{b}_sds$ therefore $E\int^t_0(\tilde{b}_s-b_s)ds=0 \implies b_t=\tilde{b}_t$ a.s
What do you think?
But then how would I prove $\sigma_t=\tilde{\sigma}_t$ a.s ?
Without further assumptions on $(b_t)_{t \geq 0}$ and $(\tilde{b}_t)_{t \geq 0}$, the claim is not true. Consider for example
$$X_t := t = \int_0^t 1 \, ds = \int_0^t 1_{\{s \neq 1\}} \, ds$$
i.e. $\sigma_t = \tilde{\sigma}_t =0$ and $b_t = 1$, $\tilde{b}_t = 1_{\{t \neq 1\}}.$ Obviously, $$b_1 = 1 \neq 0 = \tilde{b}_1.$$
Some remarks concerning your proof: First of all, you cannot simply set $X_0 = \mathbb{E}(X_t)$. $X_0$ is a given random variable! Moreover, $\mathbb{E}X=0$ does not imply $X=0$ a.s.; this means that $$\mathbb{E}\left( \int_0^t (\tilde{b}_s-b_s) \, ds \right)=0$$ does not imply $\int_0^t \tilde{b}_s \, ds = \int_0^t b_s \, ds$ (or even $b_t = \tilde{b}_t$ a.s.).