For a product space $\Omega = \Omega_1 \times ... \times \Omega_n$ and its distribution (a probability function) $P$,

the $i$th marginal distribution is:

$$P_i(\omega_i)=\sum_{t\in \Omega: t_i=\omega_i} P(t)$$ where $\omega_i \in \Omega_i$.

What does the condition ${t\in \Omega: t_i=\omega_i}$ mean?

Here it's claimed that the marginal distributions appear in the "margins" of the following table:

enter image description here

I don't see the connection to the equation. What is the $i$th margin distribution in that table?


1 Answer 1


$t$ is an $n$-tuple $(t_1,t_2,\ldots,t_n)$ whose $i$-th component is a member of $\Omega_i$. The sum is over all $t$ such that the $i$-th component is a fixed element $\omega$ of $\Omega_i$, not $\omega_i$ as it is written. The notation leaves much to be desired.

  • $\begingroup$ I still don't get it. Would need to see a small practical example of how to do the "drawing" of the elements. Or to know how to know how many terms the sum has. $\endgroup$
    – mavavilj
    Nov 8, 2015 at 14:23
  • 1
    $\begingroup$ Dilip, "$\omega_i$" appears correctly in the notation, which has the desirable property of being explicit about which margin is involved. Unfortunately it does overburden the poor subscript "$i$", which it uses both for naming various objects ($\Omega_i$, $P_i$, and $\omega_i$) and to represent the image of the canonical projection $\pi_i:\Omega\to\Omega_i$ via the shorthand $t_i = \pi_i(t)$. Presumably this is because the author thinks of each $t\in\Omega$ implicitly as a tuple $(t_1,\ldots,t_i,\ldots,t_n)$. $\endgroup$
    – whuber
    Nov 8, 2015 at 14:59

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