# Why is posterior density proportional to prior density times likelihood function?

According to Bayes' theorem, $P(y|\theta)P(\theta) = P(\theta|y)P(y)$. But according to my econometric text, it says that $P(\theta|y) \propto P(y|\theta)P(\theta)$. Why is it like this? I don't get why $P(y)$ is ignored.

• Notice that it does not say that the two are equal, but proportional (up to a factor, that is, $1/P(y)$) – jpmuc Jul 15 '13 at 13:16
• $P(y)$ is not being ignored but treated as a constant because it is a function of the data $y$ which are fixed for the problem at hand. If $A(x) = cB(x)$ where $c$ is a constant (meaning not dependent on $x$), then we can write $A(x) \propto B(x)$ which simply means that $\frac{A(x)}{B(x)}$ is a (unspecified) constant. Note that the extrema of $A(x)$ and $B(x)$ occur at the same locations so that things like maximum a posteriori probability (MAP or MAPP) estimates can be found from $P(y\mid\theta)P(\theta)$ without the need to know (or compute) $P(y)$. – Dilip Sarwate Jul 15 '13 at 13:24

$Pr(y)$, the marginal probability of $y$, is not "ignored." It is simply constant. Dividing by $Pr(y)$ has the effect of "rescaling" the $Pr(y|\theta)P(\theta)$ computations to be measured as proper probabilities, i.e. on a $[0,1]$ interval. Without this scaling, they are still perfectly valid relative measures, but are not restricted to the $[0,1]$ interval.

$Pr(y)$ is often "left out" because $Pr(y)=\int Pr(y|\theta)Pr(\theta)d\theta$ is often difficult to evaluate, and it is usually convenient enough to indirectly perform the integration via simulation.

Notice that

$$P(\theta | y) = \frac{P(\theta, y)}{P(y)} = \frac{P(y | \theta) P(\theta)}{P(y)}.$$

Since you're interested in calculating the density of $\theta$, any function that does not depend on this parameter ― such as $P(y)$ ― can be discarded. This gives you

$$P(\theta | y) \propto P(y | \theta) P(\theta).$$

The consequence of discarding $P(y)$ is that now the density $P(\theta | y)$ has lost some properties like integration to 1 over the domain of $\theta$. This is not a big deal since one is usually not interested in integrating likelihood functions, but in maximizing them. And when you're maximizing a function, multiplying this function by some constant (remember that, in the Bayesian approach, the data $y$ is fixed), doesn't change the $\theta$ that corresponds to the maximum point. It does change the value of the maximum likelihood, but then again, one is usually interested in the relative positioning of each $\theta$.