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In the book Beyond the Kalman Filter: Particle Filters for Tracking Applications on page 39 the weight update equation for the particle filter is derived.

The derivations begins by introducing the posteriori $$ p({x}_k|{z}_k) = \frac{p({z}_k|{x}_k,{z}_{k-1})p({x}_k|{z}_{k-1})}{p({z}_k|{z}_{k-1})} $$ which can be rearranged as $$ \frac{p(z_k|x_k)p(x_k|x_{k-1})}{p(z_k|z_{k-1})}p(x_{k-1}|z_{k-1}) $$ The book then writes that $$ p({x}_k|{z}_k) \propto p(z_k|x_k)p(x_k|x_{k-1})p(x_{k-1}|z_{k-1}) $$

I am not entirely sure I follow the last step. My understanding is that $p(z_k|z_{k-1})$ is assumed equal to some constant value therefore the simplification holds. But I am not entirely understanding how that assumption can be made.

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  • $\begingroup$ $p(z_k|z_{k-1})$ is a constant of proportionality. It is used in Bayes Theorem to give a valid probability i.e a value between 0 and 1. However it can often be difficult or impossible to calculate hence it we can write the last step as being proportional to the numerator $\endgroup$
    – vigos
    Commented Jan 9, 2020 at 14:53
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    $\begingroup$ This symbol "$\propto$" just means "equal up to multiplication with a constant". However, once you fix all the inputs $x_t, x_{t-1}, z_t, z_{t-1}$, everything is a constant, right? People usually use that "$\propto$" in order to express that the 'missing constant' is not in the center of interest right now (because maybe they want to form the derivative in $x_t$ or so, then $z_t$ and $z_{t-1}$ are just constants being dragged along the whole computation). Another reason for using this is that if you can show that $p(x|z) = c/\sqrt{2\pi}*e^{-x^2/2}$ then $p(x|z)$ is actually $N(0,1)$ ... $\endgroup$ Commented Jan 9, 2020 at 15:43
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    $\begingroup$ ... because both sides of the equation are probability distributions in $x$ and integrating over $x$ gives that the constant $c$ (that often depends on $z$ and is therefore not really a constant (but for the integral over $x$ it is!)) is actually $1$. $\endgroup$ Commented Jan 9, 2020 at 15:45
  • $\begingroup$ Does $p(z_k|z_{k-1})$ need so be constant for this to work? Or does this simplification work just on the principle that $p(z_k|z_{k-1})$ is a scalar value? $\endgroup$ Commented Jan 10, 2020 at 8:27

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