My textbook says the following:
In order to make probability statements about $\theta$ given $y$, we must begin with a model providing a joint probability distribution for $\theta$ and $y$. The joint probability mass or density function can be written as a product of two densities that are often referred to as the prior distribution $p(\theta)$ and the sampling distribution (or data distribution) $p(y|\theta)$, respectively:
$$p(\theta, y) = p(\theta)p(y|\theta)$$
Simply conditioning on the known value of the data $y$, using the basic property of conditional probability known as Bayes' rule, yields the posterior density:
$$p(\theta|y) = \dfrac{p(\theta, y)}{p(y)} = \dfrac{p(\theta)p(y|\theta)}{p(y)}, \tag{1.1}$$
where $p(y) = \sum_\theta p(\theta)p(y|\theta)$, and the sum is over all possible values of $\theta$ (or $p(y) = \int p(\theta) p(y | \theta) \ d\theta$ in the case of continuous $\theta$). An equivalent form of (1.1) omits the factor $p(y)$, which does not depend on $\theta$ and, with fixed $y$, can thus be considered a constant, yielding the unnormalised posterior density, which is the right side of (1.2):
$$p(\theta|y) \propto p(\theta)p(y|\theta)$$
Page 7, Bayesian Data Analysis, Third Edition, by Gelman et al.
If we have
$$p(\theta|y) = \dfrac{p(\theta, y)}{p(y)} = \dfrac{p(\theta)p(y|\theta)}{p(y)} \tag{1.1},$$
then we can multiply through by $p(y)$ to get
$$p(\theta, y) = p(\theta)p(y|\theta).$$
So I'm wondering why we change the equals sign to a proportional ($\propto$) sign? Mathematically, why are we doing this? As I demonstrated above, there doesn't seem to be anything that algebraically indicates that that we must do this?
I would greatly appreciate it if people could please take the time to clarify this.
