Talk:Law of total probability

message: "because \Pr(A\mid B_n)\, is finite."
Shouldn't that be 0/0, infinite, or undefined?
128.163.8.202 (talk) 10:40, 16 September 2015 (UTC)
thread_title: mistake?

message: I like the statement of the definition of total probability. It is readable and specialist terms are well referenced which enable the lay reader to gain some understanding of what is meant. Is there a good example of this proposition working in practice?
Blueawr (talk) 19:04, 31 August 2012 (UTC)
thread_title: Statement

message: 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)
replies: text: This is good. The article really should related to this. Jackzhp (talk) 15:55, 12 March 2011 (UTC)
thread_title: Notations

message: 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)
replies: text: 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)
replies: text: 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)

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

text: \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)

text: 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)
replies: text: 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)
replies: text:  I think that's exactly what we should do. MisterSheik 22:34, 16 February 2007 (UTC)
thread_title:  conditional probability 

message: The text states:
"The law of total probability can also be stated for conditional probabilities. Taking the B_n as above, and assuming X is not mutually exclusive with A or any of the B_n". Note that is assumes that  X is not mutually exclusive with any of the B_n. Doesn't this imply that X is the sample space if each B_i has a single event?   — Preceding unsigned comment added by 134.58.253.57 (talk) 08:06, 11 May 2012 (UTC)
thread_title:  conditional probability no2 

message: Substituted at 20:03, 1 May 2016 (UTC)
thread_title: Assessment comment

message: Why does this article use "Pr(A)" rather than "P(A)" as most articles (and references) do?  — Preceding unsigned comment added by 99.46.205.49 (talk) 14:14, 7 November 2017 (UTC)
thread_title:  Pr(A) vs. P(A)? 

message: I suppose there is a tradition in mathematics teaching that a proposition is called a "law" if establishing it as a theorem or a definition is less important that drilling into students heads.  However, since this is a topic in mathematics, it would be appropriate to have a section that indicates a proof of this theorem.  It seems to me that any coherent discussion of mathematical probability should mention probability spaces and probability measures, although it need not do so in the introduction.
The content of mathematical articles in Wikipedia is schizophrenic.  Some are detailed and formal.  Some, like the current article are informal; they treat the topic in a Platonic sense, as if the subject is a physically real phenomena and the mathematics is merely an observer's description instead of a definition.  To a non-specialist the Platonic view is more useful than the formal view.  However, mathematical articles should acknowledge the basic organization of mathematics - definitions, assumptions, theorems, even if they choose not to present a lot of formal detail.  
I see no reason for the current article to describe the "law" of total probability with the more general term "proposition" if the specific term "theorem" is appropriate.
Tashiro~enwiki (talk) 08:42, 31 August 2018 (UTC)
thread_title: "Law" or "Theorem" 

message: ok so events of zero probability can partition like for a continuous random variable Y, the events (Y=k), for k in the range of Y actually partition Y's probability space's sample space
but how would you use the 'alternative' to compute P(A), for some A (an event i.e. an element of Y's probability space's sigma-algebra)? Thewriter006 (talk) 22:54, 28 March 2021 (UTC)
thread_title:  the alternative 'these terms are simply omitted from the summation' 

message: In the original article the derivation assumes independence, but you don't need to assume any independence to arrive at the original equation. Having the independence comment makes it seems like the derivation is only valid with the assumption.
\begin{aligned}
P(A|C) & = \sum_{n}P(A|C \cap B_n)P(B_n) \\
& = \sum_{n} \frac{P(A \cap C \cap B_n)P(B_n \cap C)}{P(C \cap B_n)P(C|B_n)} \\
& = \sum_{n} \frac{P(A \cap C \cap B_n)\cancel{P(B_n \cap C)}}{\cancel{P(C \cap B_n)}P(C|B_n)} \\
& = \sum_{n} \frac{P(A \cap C \cap B_n)}{P(C|B_n)} \\
& = \sum_{n} \frac{P(B_n)P(C|B_n)P(A|C \cap B_n)}{P(C|B_n)} \\
& = \sum_{n} \frac{P(B_n)\cancel{P(C|B_n)}P(A|C \cap B_n)}{\cancel{P(C|B_n)}} \\
& = \sum_{n} P(B_n)P(A|C \cap B_n)
\end{aligned}
 71.178.191.95 (talk) 00:24, 17 August 2023 (UTC)
thread_title:  the law of total probability for conditional probabilities doesn't need to assume independence