Conditional probability: Difference between revisions

Revision as of 01:03, 5 July 2025

Revision as of 01:12, 5 July 2025

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[[File:Probability tree diagram.svg|thumb|On a [[Tree diagram (probability theory)|tree diagram]], branch probabilities are conditional on the event associated with the parent node. (Here, the overbars indicate that the event does not occur.)]]

[[File:Venn Pie Chart describing Bayes' law.png|thumb|Venn Piepie Chartchart describing conditional probabilities]]

=== Conditioning on an event ===

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The case of greatest interest is that of a random variable {{mvar|Y}}, conditioned on a continuous random variable {{mvar|X}} resulting in a particular outcome {{mvar|x}}. The event <math>B = \{ X = x \}</math> has probability zero and, as such, cannot be conditioned on.

Instead of conditioning on {{mvar|X}} being ''exactly'' {{mvar|x}}, we could condition on it being closer than distance <math>\epsilonvarepsilon</math> away from {{mvar|x}}. The event <math>B = \{ x-\epsilonvarepsilon < X < x+\epsilonvarepsilon \}</math> will generally have nonzero probability and hence, can be conditioned on.

We can then take the [[limit (mathematics)|limit]]

{{NumBlk|::|<math>\lim_{\epsilonvarepsilon \to 0} P(A \mid x-\epsilonvarepsilon < X < x+\epsilonvarepsilon).</math>|{{EquationRef|1}}}}

For example, if two continuous random variables {{mvar|X}} and {{mvar|Y}} have a joint density <math>f_{X,Y}(x,y)</math>, then by [[L'Hôpital's rule]] and [[Leibniz integral rule]], upon differentiation with respect to <math>\epsilonvarepsilon</math>:

:<math>

\begin{aligned}

\lim_{\epsilonvarepsilon \to 0} P(Y \in U \mid x_0-\epsilonvarepsilon < X < x_0+\epsilonvarepsilon) &=

\lim_{\epsilonvarepsilon \to 0} \frac{\int_{x_0-\epsilonvarepsilon}^{x_0+\epsilonvarepsilon} \int_U f_{X, Y}(x, y) \, \mathrm{d}y \, \mathrm{d}x}{\int_{x_0-\epsilonvarepsilon}^{x_0+\epsilonvarepsilon} \int_\mathbb{R} f_{X, Y}(x, y) \, \mathrm{d}y \, \mathrm{d}x} \\[6pt]

&= \frac{\int_U f_{X, Y}(x_0, y) \, \mathrm{d}y}{\int_\mathbb{R} f_{X, Y}(x_0, y) \, \mathrm{d}y}.

\end{aligned}

</math>

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It is tempting to ''define'' the undefined probability <math>P(A \mid X=x)</math> using limit ({{EquationNote|1}}), but this cannot be done in a consistent manner. In particular, it is possible to find random variables {{mvar|X}} and {{mvar|W}} and values {{mvar|x}}, {{mvar|w}} such that the events <math>\{X = x\}</math> and <math>\{W = w\}</math> are identical but the resulting limits are not:

:<math>\lim_{\epsilonvarepsilon \to 0} P(A \mid x-\epsilonvarepsilon \le X \le x+\epsilonvarepsilon) \neq \lim_{\epsilonvarepsilon \to 0} P(A \mid w-\epsilonvarepsilon \le W \le w+\epsilonvarepsilon).</math>

The [[Borel–Kolmogorov paradox]] demonstrates this with a geometrical argument.

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=== Example ===

When [[Morse code]] is transmitted, there is a certain probability that the "dot" or "dash" that was received is erroneous. This is often taken as interference in the transmission of a message. Therefore, it is important to consider when sending a "dot", for example, the probability that a "dot" was received. This is represented by: <math>P(\text{dot sent } | \text{ dot received}) = P(\text{dot received } | \text{ dot sent}) \frac{P(\text{dot sent})}{P(\text{dot received})}.</math> In Morse code, the ratio of dots to dashes is 3:4 at the point of sending, so the probabilityprobabilities of a "dot" and "dash" are <math>P(\text{dot sent}) = \frac {3}{7} \text{ and \ } P(\text{dash sent}) = \frac {4}{7}</math>. If it is assumed that the probability that a dot is transmitted as a dash is 1/10, and that the probability that a dash is transmitted as a dot is likewise 1/10, then Bayes's rule can be used to calculate <math>P(\text{dot received})</math>.

: <math>P(\text{dot received}) = P(\text{dot received } \cap \text{ dot sent}) + P(\text{dot received } \cap \text{ dash sent})</math>

: <math>P(\text{dot received}) = P(\text{dot received } \mid \text{ dot sent})P(\text{dot sent}) + P(\text{dot received } \mid \text{ dash sent})P(\text{dash sent})</math>

: <math>P(\text{dot received}) = \frac{9}{10}\times\frac{3}{7} + \frac{1}{10}\times\frac{4}{7} = \frac{31}{70}</math>

Now, <math>P(\text{dot sent } \mid \text{ dot received})</math> can be calculated:

: <math>P(\text{dot sent } \mid \text{ dot received}) = P(\text{dot received } \mid \text{ dot sent}) \frac{P(\text{dot sent})}{P(\text{dot received})} = \frac{9}{10}\times \frac{\frac{3}{7}}{\frac{31}{70}} = \frac{27}{31}</math><ref>{{Cite web|title=Conditional Probability and Independence|url=http://www.math.ntu.edu.tw/~hchen/teaching/StatInference/notes/lecture4.pdf|access-date=2021-12-22}}</ref>

== Statistical independence ==

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is also equivalent. Although the derived forms may seem more intuitive, they are not the preferred definition as the conditional probabilities may be undefined, and the preferred definition is symmetrical in ''A'' and ''B''. Independence does not refer to a disjoint event.<ref>{{Cite book|last=Tijms|first=Henk|url=https://www.cambridge.org/core/books/understanding-probability/B82E701FAAD2C0C2CF36E05CFC0FF3F2|title=Understanding Probability|date=2012|publisher=Cambridge University Press|isbn=978-1-107-65856-1|edition=3|___location=Cambridge|doi=10.1017/cbo9781139206990}}</ref>

It should also be noted that given the independent event pair [''A '',''B''] and an event ''C'', the pair is defined to be [[Conditional independence|conditionally independent]] if the product holds true:<ref>{{Cite book|last=Pfeiffer|first=Paul E.|url=https://www.worldcat.org/oclc/858880328|title=Conditional Independence in Applied Probability|date=1978|publisher=Birkhäuser Boston|isbn=978-1-4612-6335-7|___location=Boston, MA|oclc=858880328}}</ref>

: <math>P(AB \mid C) = P(A \mid C)P(B \mid C).</math>

This theorem could beis useful in applications where multiple independent events are being observed.

'''Independent events vs. mutually exclusive events'''

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=== Assuming conditional probability is of similar size to its inverse ===

{{Main|Confusion of the inverse}}

[[File:Bayes theorem visualisation.svg|thumb|450x450px|A geometric visualization of Bayes' theorem. In the table, the values 2, 3, 6 and 9 give the relative weights of each corresponding condition and case. The figures denote the cells of the table involved in each metric, the probability being the fraction of each figure that is shaded. This shows that <math>P(A|B) P(B) = P(B|\mid A) P(A)</math> i.e. <math>P(A|\mid B) = \frac{P(B|\mid A)} {P(A)|P(\cap B)}</math> . Similar reasoning can be used to show that <math>P(\bar A|\mid B) = \frac{P(B|\mid\bar A) P(\bar A)}{P(B)}</math> etc.]]

In general, it cannot be assumed that ''P''(''A''|''B'')&nbsp;≈&nbsp;''P''(''B''|''A''). This can be an insidious error, even for those who are highly conversant with statistics.<ref>Paulos, J.A. (1988) ''Innumeracy: Mathematical Illiteracy and its Consequences'', Hill and Wang. {{ISBN|0-8090-7447-8}} (p. 63 ''et seq.'')</ref> The relationship between ''P''(''A''|''B'') and ''P''(''B''|''A'') is given by [[Bayes' theorem]]:

:<math>\begin{align}