# Confusion related to Kalman filters density view

I was reading this book related to Kalman filters and I didn't understand a couple of things. I have also attached the screenshot of the pages from the book where I had confusion.

The book is Shumway, R.H. and Stoffer, D.S. Time Series Analysis and Its Applications: With R Examples

From the first page, I didn't understand how this was derived

I mean how come we have $p(x_t|x_{t-1}) = f_w(x_t-\phi_{x_{t-1}})$ and has mean 0 and covariance Q. I mean it should be for the $w_t$ which has mean 0 and covariance Q not $p(x_t|x_{t-1})$

From the second page, I didn't understand the following

I mean obviously $p(\mu_t|Y_{t-1} => N(\mu_t^{t-1},P_t^{t-1}))$. But I didn't get how we get that from the integral

From the third page, I didn't get how this was derived

I know I might be asking a lot, but I have spent several days to exactly grasp the derivation of the equations in Kalman filters. I would really appreciate if someone could help me. Thanks in advance

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Your post is on a border of quotation abuse; please give attribution to the book author. –  mbq Jan 4 '13 at 14:27

Regarding your first question, $x_t|x_{t-1}$ has the same variance as $w_t$; if in the transition equation $x_t = \Phi x_{t-1} + w_t$ you hold $x_{t-1}$ fixed, the only source of variance in $x_t$ is just $w_t$.
Yeah $w_t$ and $x_t$ have same variance. But the mean is different. So we can only say that the variance is equal not the distribution of $x_t$ and $w_t$ itself. So how can we write $p(x_t|x_{t-1})= f_w$ –  user34790 Jan 4 '13 at 23:32