# Joint distribution's chain rule gives proportional

The chain rule of the joint distribution is:

$$\mathrm P(A_n, \ldots , A_1) = P(A_n | A_{n-1}, \ldots , A_1) \cdot P( A_{n-1}, \ldots , A_1)\mbox{.}$$

I've seen in many places (for instance in Naive Bayes classifier article), that is not an equality but a proportion as:

$$\mathrm P(A_n, \ldots , A_1) \varpropto P(A_n | A_{n-1}, \ldots , A_1) \cdot P( A_{n-1}, \ldots , A_1)\mbox{.}$$

Why do they say that is proportional?

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These could/should be equality signs. –  Did Sep 19 '12 at 14:57
If they could be equality signs, what is the meaning that they want to express using a 'proportion to' sign? –  Kits89 Sep 19 '12 at 15:15
Typo, page written too fast, irrelevance, in this context, of equality with respect to proportionality, whatever. –  Did Sep 19 '12 at 15:34
I don't understand it. For me would be more comfortable to keep using equality, if it's not relevant. Unless I want to express some nuance.. –  Kits89 Sep 19 '12 at 15:42
The Wikipedia article you link to used to have equality; the change to proportionality is relatively recent; it's the sole edit ever made by that user. I suspect that it's based on a confusion between $p(C,F_1,\dotsc,F_n)$ and $p(C\mid F_1,\dotsc,F_n)$, since these equations are in the end used to establish a proportionality for the latter, including a constant of proportionality $Z$. I think the edit should be reversed. –  joriki Sep 19 '12 at 16:37