What does the \(\propto\) symbol mean?

When we use the \(\propto\) notation, what we mean is the following. If

\[f(x)\propto g(x)\]

This implies that:

\[k \times f(x) = g(x)\]

where k is an unknown constant (i.e., k does not depend on x). We can multiply our function \(f(x)\) by any constant and the proportionality relationship between \(f(x)\) and \(g(x)\) will remain true.

This concept is better illustrated directly with an example. Normal distributions are the basis of many statistical tests. Here is the pdf of a normal distribution with parameters \(\mu\) and \(\sigma^2\):

\[f(x)=\frac{1}{\sqrt{2\pi \sigma^2}} exp(-\frac{1}{2\pi \sigma^2} (x-\mu)^2)\]

If we are interested in the parameter \(\mu\) only, then we can write:

\[f(\mu|x,\sigma^2) = K \times exp(-\frac{1}{2\pi \sigma^2} (x-\mu)^2)) \propto exp(-\frac{1}{2\pi \sigma^2} (x-\mu)^2)\]

where \(K=\frac{1}{\sqrt{2\pi \sigma^2}}\) is a constant because it does not vary for different \(\mu\) values. Further notice that we are conditioning on our data x and the other parameter \(\sigma^2\). This implies that they are constants from the perspective of \(f(\mu|x,\sigma^2)\).

On the other hand, if we were interested in the parameter \(\sigma^2\), then what would \(f(\sigma^2|x,\mu)\) be proportional to?

\[f(x)=K\times \frac{1}{\sqrt{\sigma^2}} exp(-\frac{1}{2\pi \sigma^2} (x-\mu)^2)\propto \frac{1}{\sqrt{\sigma^2}} exp(-\frac{1}{2\pi \sigma^2} (x-\mu)^2)\]

where \(K=\frac{1}{\sqrt{2\pi}}\) is a constant that does not vary for different \(\sigma^2\) values.

Determining which parts of \(f(x)\) are important to keep track and which can be simply substituted by a K constant is important for Bayesian analysis, as you will discover when dealing with conjugate prior-likelihood pairs.

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