Let
$dX_t = \mu dt + \sigma dW_t$. I need to calculate the solution of this SDE. So I know that the solution of this sde is the brownian motion with drift: $X_t = x_0 + \mu t +\sigma B(t) $
Now I need to calculate the conditional distribution $X_{t+\Delta t} | X_t $
I know that since $B(t) \sim N (0,t) $ and if $Y(t)=\mu t + \sigma B(t)$ then $Y \sim N(\mu t, {\sigma}^2 t)$ however here we would be having an additional $+x_0$. so would the distribution of $X(t) \sim N(\mu t+x_0, {\sigma}^2 t)$ ?
However I still don’t know how to yield the conditional distribution $X_{t+\Delta t} | X_t $
And I further would need to calculate $\mathbb{E} (X_{t+\Delta t} | F_t)$ and
Edit with the information from the comments:
$X_{t+\Delta t} = X_t + \mu \Delta t + \sigma (W(t+\Delta t) - W(t))$
So I set $V_t = \mu \Delta t + \sigma (W(t+\Delta t)-W(t))$ And I came to the conclusion that $V_t \sim N(\mu \Delta t, {\sigma}^2 \Delta t)$ (right?) and $X_t \sim N(\mu t+x_0, {\sigma}^2 t)$, and $X_{t+\Delta t} = X_t + V_t$. But would $X_t$ and $V_t$ be independent?