analog of principle of inclusion-exclusion - MathOverflow most recent 30 from http://mathoverflow.net2013-06-18T22:13:21Zhttp://mathoverflow.net/feeds/question/1334http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/1334/analog-of-principle-of-inclusion-exclusionanalog of principle of inclusion-exclusionWill Orrick2009-10-20T01:27:13Z2009-10-29T00:49:59Z
<p>When I teach elementary probability to my finite math students, a common error is to mix up the concepts of disjointness and independence. At some point I thought that it might be helpful to some students to draw the analogy between the two concepts implied by the following pair of statements:</p>
<ul>
<li><p>To compute the probability of the union of disjoint events, you add the probabilities of the events.</p></li>
<li><p>To compute the probability of the intersection of independent events, you multiply the probabilities of the events.</p></li>
</ul>
<p>I also teach them is that when events are not disjoint, you can still compute the probability of their union by applying the principle of inclusion-exclusion. Hence the question: Is there a useful analog of the principle of inclusion-exclusion for computing the probability of the intersection of non-independent events?</p>
<p><em>Edit:</em>
I am incorporating the following clarification that I made in a comment responding to the answer of Anna Varvak:</p>
<blockquote>
<p>In inclusion-exclusion, one alternately adds and subtracts intersections. Intersections measure the degree to which disjointness fails. Can we write the right-hand side of Bayes Theorem as alternate multiplications and divisions of something, where "something" measures the degree to which independence fails?</p>
</blockquote>
http://mathoverflow.net/questions/1334/analog-of-principle-of-inclusion-exclusion/1337#1337Answer by Michael Lugo for analog of principle of inclusion-exclusionMichael Lugo2009-10-20T01:39:53Z2009-10-20T01:39:53Z<p>Inclusion-exclusion doesn't depend on independence. You want</p>
<p>P(A union B) = P(A) + P(B) - P(A intersect B).</p>
<p>Now, P(A intersect B) is P(A) P(B) if A and B are independent. But that holds even if A and B aren't independent. You might rewrite it as</p>
<p>P(A union B) = P(A) + P(B) - P(A) P(B|A)</p>
<p>if your students are okay with conditional probability.</p>
http://mathoverflow.net/questions/1334/analog-of-principle-of-inclusion-exclusion/1338#1338Answer by Anna Varvak for analog of principle of inclusion-exclusionAnna Varvak2009-10-20T01:42:36Z2009-10-20T01:42:36Z<p>Sounds like Bayes Theorem, which in its intuitive form is </p>
<p>P(A and B) = P(A) * P(B, given A)
= P(B) * P(A, given B)</p>
<p>This really is quite intuitive, and I find that students understand it quite easily: for both A and B to happen, we can describe it in two ways: either A happens and then B happens (given that A happened), or B happens and then A happens (given that B happened).</p>
<p>For more than two events, there's the extended Bayes theorem:</p>
<p>P(A and B and C) = P(A) * P(B, given A) * P(C, given A and B)</p>
http://mathoverflow.net/questions/1334/analog-of-principle-of-inclusion-exclusion/1344#1344Answer by Reid Barton for analog of principle of inclusion-exclusionReid Barton2009-10-20T02:39:43Z2009-10-20T02:39:43Z<p>Writing B \ A for the event "B occurs but A does not" (as in the difference of sets) we have...</p>
<p>P(A ∪ B) = P(A) + P(B \ A)</p>
<p>P(A ∩ B) = P(A) × P(B | A)</p>
<p>Just fun with symbols I think...</p>
http://mathoverflow.net/questions/1334/analog-of-principle-of-inclusion-exclusion/1394#1394Answer by Kenny Easwaran for analog of principle of inclusion-exclusionKenny Easwaran2009-10-20T07:42:26Z2009-10-20T07:42:26Z<p>I suppose one attempt would be to just reverse the inclusion-exclusion equation.</p>
<p>Since P(A union B) = P(A) + P(B) - P(A intersect B) we also have</p>
<p>P(A intersect B) = P(A) + P(B) - P(A union B)</p>
<p>Of course, that totally ignores the analogy you want to make between disjointness and independence.</p>
<p>Maybe I can try to reverse-engineer something. Let P(A intersect B) = P(A)P(B)/f(A,B), where f is the function giving the unknown factor. We see that f(A,B)=P(A)P(B)/P(A intersect B).</p>
<p>For the three-variable case we have P(A intersect B intersect C) = P(A)P(B)P(C) divided by (f(A,B)f(A,C)f(B,C) times g(A,B,C). Plugging in the formula for f(A,B), we see that</p>
<p>P(A intersect B intersect C) = P(A intersect B)P(A intersect C)P(B intersect C) divided by (P(A)P(B)P(C)) times g(A,B,C). So g is the three-way intersection times the individual probabilities, divided by the two-way intersections.</p>
<p>I must admit, I don't see any interesting pattern developing here with f and g, the way I did in the inclusion-exclusion case, but maybe someone else does?</p>
http://mathoverflow.net/questions/1334/analog-of-principle-of-inclusion-exclusion/2294#2294Answer by Will Orrick for analog of principle of inclusion-exclusionWill Orrick2009-10-24T15:28:11Z2009-10-29T00:49:59Z<p>I'll take a stab at answering my own question.</p>
<p>The missing “something” in the edited version of my question appears to be the mutual
information of one or more events, denoted I(A,B,C,...). More precisely,
it appears to be the quantity e<sup>−I(A,B,C,...)</sup>. It would be good if
someone knowledgeable about information theory could weigh in here—I haven't
found a definition in the literature of the mutual information of more than
two events, but the one I give below seems to be the “right” one.</p>
<p>In the information theory textbooks I looked at, the information of a single
event E is defined as I(E) = −log P(E). The mutual information of two
events, E and F, is defined as
I(E,F) = −log [P(E) P(F) / P(E ∩ F)],
and is a measure of the degree to which E and F fail to be independent. That is, I(E,F)
is zero if E and F are independent, positive if they are positively correlated, and negative if
they are negatively correlated.</p>
<p>The appropriate generalization to three events seems (following the
suggestion of Kenny Easwaran) to be
I(E,F,G) = −log [P(E) P(F) P(G) P(E ∩ F ∩ G)
/ P(E ∩ F) P(E ∩ G) P(F ∩ G)]
which makes some sense as a measure of the failure of independence since,
in order for E, F, and G to be independent, it is required, not only that
P(A ∩ B) = Pr(A) P(B) for all pairs of events, but also that
P(E ∩ F ∩ G) = P(E) P(F) P(G).
(Does anyone know if the definition of I(E,F,G) given above is standard?)</p>
<p>Now define C({A,B,C,...}) = P(A ∩ B ∩ C ∩ ...). The appropriate
generalization of the mutual information to an arbitrary number of events
seems to be
I(E,F,G,...) = −log [Π<sub>S</sub> C(S) / Π<sub>T</sub> C(T)]
where the product over S runs over all subsets of {E,F,G,...} of odd
cardinality and the product over T runs over all subsets of
{E,F,G,...} of even cardinality. (Again, does anyone know if this definition
is standard?)</p>
<p>With these definitions, we get the inclusion-exclusion-like rule,</p>
<p>−log P(E ∩ F ∩ G ∩ ...) = I(E) + I(F) + I(G) + ...
− I(E,F) − I(E,G) − ... + I(E,F,G) + ... − ...</p>
<p>This can be proved by a counting argument identical to the one used to
prove the principle of inclusion-exclusion. Negating and exponentiating
both sides produces an identity of the desired form. It would be nice
to also find a Möbius-inversion style proof.</p>
<p>As to whether this is “useful” in the same sense that the principle
of inclusion-exclusion is useful, I can't say. The definition of I(E,F,G,...)
is itself an inclusion-exclusion-like rule, so it's tautological that when you
invert it to find −log P(E ∩ F ∩ G ∩ ...) you will get an
inclusion-exclusion-like rule. I suppose the significance of all this depends
on how fundamental the mutual information is.</p>
<p><em>Addendum:</em>
The book <em>Elements of Information Theory</em> by Cover and Thomas mentions the problem of defining the mutual information
of three random variables in Problem 2.25. The mutual information of random variables is related to but somewhat different from the mutual information of events since, for example, I(X,Y,Z) involves taking the expectation over all events (X=x, Y=y, Z=z). The problem notes that, in contrast with the two random variable case, the mutual information of three random variables is not a non-negative quantity in general. This perhaps explains why not much theory has been developed around it. Interestingly, however, the mutual information of three random variables can be expressed in terms of the entropy of one, two, or three random variables via the inclusion-exclusion principle.</p>
http://mathoverflow.net/questions/1334/analog-of-principle-of-inclusion-exclusion/2647#2647Answer by Fritz for analog of principle of inclusion-exclusionFritz2009-10-26T17:05:30Z2009-10-26T17:05:30Z<p>In <b>belief propagation</b> there is a notion of inclusion-exclusion for computing the join probability distributions of a set of variables, from a set of factors or marginals over subsets of those variables. For example, suppose {X,Y,Z} is your set of variables, and you know the marginal probabilities for p<sub>X,Y</sub>(x,y) and p<sub>Y,Z</sub>(y,z). If these two are compatible, then the marginal p<sub>Y</sub> can be computed in either of the ways</p>
<p>p<sub>Y</sub>(y) = integral p<sub>X,Y</sub>(x,y) dx = integral p<sub>Y,Z</sub>(y,z) dz</p>
<p>Then a maximum entropy guess at the full joint distrubution is given by inclusion-exclusion over subsets of variables</p>
<p>p<sub>X,Y,Z</sub>(x,y,z) = p<sub>X,Y</sub>(x,y) p<sub>Y,Z</sub>(x,y) / p<sub>Y</sub>(y)</p>
<p>You might take a look at "Belief propagation" on wikipedia, or the more techinical article <i>"Constructing Free Energy Approximations and Generalized Belief Propagation Algorithms" by Yedidia, Freeman and Weiss</i>, which uses inclusion-exclusion in the form of 'counting numbers'.</p>