Talk:Law of total probability

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Notations

I read somewhere that $P(A|B):={\frac {P(A\cap B)}{P(B)}}$

in which case the two "different" notions of total probability would be aequivalent, since obvisously

$P(A\cap B_{n})=P(A\mid B_{n})P(B_{n})$ for $n=1,2,3,\ldots$

Note however that i have very little knowledge of the whole domain, so maybe i just missed the point...

--lu 10:42, 26 Apr 2005 (UTC)

conditional probability

The conditional probablity of A given B is only defined if the probability of B is non-zero. This is really unsatisfactory in this context, since you end up having to partition the space, avoiding zero-measure sets.

There must be a way around this tecnicallity. Or is it necessary? --67.103.110.175 02:55, 27 May 2006 (UTC)[reply]

OK, self, the statement should run more like, given Bi, mutually exclusive, whose probabilities sum to one ... Mathworld[1] So you can ignore that part of the space with zero probability, or include it, according to the situation. All that is important is that the part of the space with positive probability be covered. --67.103.110.175 02:55, 27 May 2006 (UTC)[reply]

It is absolutely necessary to be able to condition on continuous random variables. That entails conditioning on events of probability zero and getting nonzero conditional probabilities. Michael Hardy 21:52, 27 May 2006 (UTC)[reply]

Thanks for reverting and keeping the page accurate. However, in this case, it is a vanity and disservice not to mention that P(A|B) will not be defined for P(B)=0, in the discrete case. As for the continuous case, these are going to be a countable collection of events covering the space, so zero probability events remain insubstantial. Ozga 00:02, 28 May 2006 (UTC)[reply]

They are not at all insubstantial in the continuous case. When one finds E(Pr(A | X)). where X is a continuous random variable, the event on which one conditions always has probability 0. Michael Hardy 01:04, 30 May 2006 (UTC)[reply]

As in E(Pr(A | X))=Pr(A), as one reintegrates up the slices of A? --Ozga 17:28, 31 May 2006 (UTC)[reply]

I find this sentence confusing: "In the discrete case, the statements above are equivalent to the following statement, which also holds in the continuous case and is not the same as the law of alternatives." I think it should be reworded. Also, it is rather tricky to define the conditional probability of an event given the value of a continuous random variable. I think it is more or less

\mathrm {Pr} (A|X=x)=\lim _{\epsilon \rightarrow 0}{\frac {\mathrm {Pr} (A\cap \{x-\epsilon <X<x+\epsilon \})}{\mathrm {Pr} (\{x-\epsilon <X<x+\epsilon \})}},

provided that the density function of X is continuous at x, but take that with a grain of salt. --130.94.162.64 23:48, 15 June 2006 (UTC)[reply]

Indeed, it can be considered "tricky", since probably most people who grasp the idea intuitively do not know the Radon-Nikodym theorem, on which one of the usual definitions relies. But nonetheless perfectly doable. Michael Hardy 00:56, 16 June 2006 (UTC)[reply]

I still find that sentence confusing. From reading the entire article, I get the impression that the "law of total probability" has a perfectly unambiguous meaning that can be applied in a consistent way to both the discrete and continuous cases. The disclaimer that "nomenclature is not wholly standard" seems both a little unnecessary and rather foreboding to me. Perhaps instead, one could mention (in the beginning of that section) that in the discrete case it is also called the "law of alternatives". Also, the article should have references. --130.94.162.64 18:46, 17 June 2006 (UTC)[reply]

I think that's exactly what we should do. MisterSheik 22:34, 16 February 2007 (UTC)[reply]

Definition

From my user talk page 83.67.217.254 14:39, 26 August 2007 (UTC)[reply]

Your edit was horribly wrong. We DO NOT want to condition on the event's actually occurring; we DO want to allow general random variables and not just indicator variables of events; and the fact that we sometimes condition on events of probability 0 and therefore need to talk about densiety functions in order to apply the result to those cases in no way means the identity is wrong. Michael Hardy 14:27, 26 August 2007 (UTC)[reply]

Hey, thanks for the flowers Michael! :-) You were referring to this edit, which you reverted with a charming and welcoming comment. I'll be back later. 83.67.217.254 14:39, 26 August 2007 (UTC)[reply]

In the meantime you may want to reflect on the fact that the definition of conditional probability does not involve random variables, but events. In other words, in the definition of P(A|B), both A and B are events, not random variables. 83.67.217.254 14:44, 26 August 2007 (UTC)[reply]

You are writing nonsense. The definition there is the definition of conditional probability given an event. There is also such a thing as conditional probability given a random variable. That is the concept needed here. I have a Ph.D. in statistics and I know this material. Michael Hardy 17:02, 28 September 2007 (UTC)][reply]

And I repeat: the edit I reverted was horribly horribly wrong. Michael Hardy 17:04, 28 September 2007 (UTC)[reply]

I agree the definition is inconsistent —Preceding unsigned comment added by 221.47.185.2 (talk) 01:54, 28 September 2007 (UTC)[reply]