Usually the following formulation of the Law of Total Probability is used:

If events $\{H_i\}$ with positive probabilities form the partition of a sample space then for any event $A$ on the same probability space we have $P(A) = \sum_{i} P(A|H_i)P(H_i)$

Do we rigorously need the partition here? Can we use the following alternative formulation instead?

If pairwise disjoint events $\{H_i\}$ with positive probabilities form the cover of the event $A$ (i.e. $A \subseteq \bigcup_i H_i$) then we have $P(A) = \sum_{i} P(A|H_i)P(H_i)$.

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    $\begingroup$ Your alternative formulation is correct! $\endgroup$ – Mike Earnest Apr 3 '18 at 23:09

Hint: What is $\Pr ( A \vert H_i )$ when $A \cap H_i = \emptyset$?

This should tell you the connection between the two statements.

  • $\begingroup$ Yes, when $A \cap H_i = \varnothing$ then $P(A|H_i) = 0$. $\endgroup$ – Rodvi Apr 4 '18 at 9:35
  • $\begingroup$ @Rodvi Good. Now you should hopefully be able to see why the sum in your first statement is the same as the sum in the second. $\endgroup$ – Theoretical Economist Apr 4 '18 at 9:37
  • $\begingroup$ I see. Just want to say that the second statement may be more convenient in situation when we don't know the partition of $\Omega$. In that case it will be enough to find the cover of $A$ to apply LOTP using the second statement. $\endgroup$ – Rodvi Apr 4 '18 at 9:53
  • $\begingroup$ @Rodvi To be honest I can’t imagine a situation where you might not know the partition, though that might be due to a lack of imagination on my part. However, the two statements are equivalent, so you are free to use whichever you find more convenient. $\endgroup$ – Theoretical Economist Apr 4 '18 at 9:55

Once you have pairwise disjoint non-empty sets $\{ H_i \}$ that cover $A$, that gives you a partition --i.e., the class $\{ H_i \cap A \}$ partitions $A$. So your second statement is not really an alternative result at all - it is just a way of stating the same result.

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    $\begingroup$ I disagree. Pairwise disjoint non-empty sets $\{H_i \}$ that cover $A$ give us the partition of $A$. While in the first formulation above $\{H_i\}$ form the partition of sample space $\Omega$. And $A \ne \Omega$. $\endgroup$ – Rodvi Apr 4 '18 at 9:32

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