# Why $p(y_n|y_{0:n-1})$ is a constant in Kalman filter derivation?

In a derivation of Kalman filter, It says that in Equation:

$p(x_n|y_{0:n})=\frac{p(y_n|x_n)p(x_b|y_{0:n-1})}{p(y_n|y_{0:n-1})}$

the denominator $p(y_n|y_{0:n-1})$ is a constant. (See the article,Page 11 equation (28), and it said the denominator is constant in Page 12).

Can anybody explain it to me ?

## 1 Answer

$p(y_n|y_{0:n-1})$ is a constant with respect to $x_n$. In general, for a posterior density

$$p(x | y) = \frac{p(x, y)}{p(y)} = \frac{p(x, y)}{\int_\mathbb{R} p(x, y) dx}$$

If you integrate over all possible $x$s, you'll get 1. Thus, $p(y)$ is just a normalizing constant (so that density function integrates to 1) with respect to $x$. If you omit this constant, you'll get unnormalized density function. Sometimes it's enough to know the posterior up to a multiplicative constant (for example, one can find a mode of a posterior by maximizing the numerator).