Looking for a rigorous derivation of Kalman Filter, i eventually come up with this two derivations:
The first one derives Kalman filter best state estimate, at step $t$, $\mathbf{\widehat{x}}_t$ as
$$\mathbf{\widehat{x}}_t=\mathbf{x}_t^t =E(\,\mathbf{X}_t\mid \{\mathbf{y}\}_1^t)$$
From equivalence (9) and definition of $\mathbf{x}_t^t$ in pdf number 1
But looking at the second one, best estimate appears to be calculated from
$$\mathbf{\widehat{x}}_t=arg\,\underset{\mathbf{x}_t}{max} \; P(\mathbf{X}_t=\mathbf{x}_t\mid \{\mathbf{y}\}_1^t)$$
From equivalences (3) and (6) in pdf number 2 (Here i used subscript $t$ in place of $n$)
Given the initial condition $\mathbf{x}_0$, output measurement vectors $\mathbf{y}_i$ till step $t$,
The starting point of the second pdf appears more intuitive to me (best estimate $\hat{\mathbf{x}}_t$, which better explains outcomes $\mathbf{y}_i$ till time $t$, is the most probable value of probability distribution of $\mathbf{x}_t$ given the knowledge of {$\mathbf{y}_1,\,...,\mathbf{y}_t$}), but i don't get if and how should be the same of the starting point of the first pdf
Can someone please show me the equivalence of the two starting points? Clearly they lead to the same conclusion (i.e. Kalman filter)
Important to say is that $\mathbf{x}_i$ and $\mathbf{y}_i$ have gaussian distributions, as made clear by both references
Any help is appreciated, thanks
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Thinking about it, my explanation would be very simple: since $(\mathbf{X}_t\mid \{\mathbf{y}\}_1^t)$ has a gassian distribution, then picking the most probable value is equivalent to take the expected value, this reasoning may be correct for a non-conditioned random variable but i don't know if it could also hold for the 'conditional case'