# Simplying Bayes Theorem expression: SIS particle filter posteriori

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.

• $p(z_k|z_{k-1})$ 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 Commented Jan 9, 2020 at 14:53
• 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 $N(0,1)$ ... Commented Jan 9, 2020 at 15:43
• ... 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 $1$. Commented Jan 9, 2020 at 15:45
• 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? Commented Jan 10, 2020 at 8:27