Conditioning on an event

Conditioning on an event of probability zero

If ${\displaystyle P(B)=0}$, then according to the definition, ${\displaystyle P(A\mid B)}$ is undefined.

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

Instead of conditioning on X being exactly x, we could condition on it being closer than distance ${\displaystyle \epsilon }$ away from x. The event ${\displaystyle B=\{x-\epsilon <X<x+\epsilon \}}$ will generally have nonzero probability and hence, can be conditioned on. We can then take the limit

${\displaystyle \lim _{\epsilon \to 0}P(A\mid x-\epsilon <X<x+\epsilon ).}$

For example, if two continuous random variables X and Y have a joint density ${\displaystyle f_{X,Y}(x,y)}$, then by L'Hôpital's rule and Leibniz integral rule, upon differentiation with respect to ${\displaystyle \epsilon }$:

${\displaystyle {\begin{aligned}\lim _{\epsilon \to 0}P(Y\in U\mid x_{0}-\epsilon <X<x_{0}+\epsilon )&=\lim _{\epsilon \to 0}{\frac {\int _{x_{0}-\epsilon }^{x_{0}+\epsilon }\int _{U}f_{X,Y}(x,y)\mathrm {d} y\mathrm {d} x}{\int _{x_{0}-\epsilon }^{x_{0}+\epsilon }\int _{\mathbb {R} }f_{X,Y}(x,y)\mathrm {d} y\mathrm {d} x}}\\&={\frac {\int _{U}f_{X,Y}(x_{0},y)\mathrm {d} y}{\int _{\mathbb {R} }f_{X,Y}(x_{0},y)\mathrm {d} y}}.\end{aligned}}}$

The resulting limit is the conditional probability distribution of Y given X and exists when the denominator, the probability density ${\displaystyle f_{X}(x_{0})}$, is strictly positive.

It is tempting to define the undefined probability ${\displaystyle P(A\mid X=x)}$ using this limit, but this cannot be done in a consistent manner. In particular, it is possible to find random variables X and W and values x, w such that the events ${\displaystyle \{X=x\}}$ and ${\displaystyle \{W=w\}}$ are identical but the resulting limits are not:^[10]

${\displaystyle \lim _{\epsilon \to 0}P(A\mid x-\epsilon \leq X\leq x+\epsilon )\neq \lim _{\epsilon \to 0}P(A\mid w-\epsilon \leq W\leq w+\epsilon ).}$

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

Conditioning on a discrete random variable

Let X be a discrete random variable and its possible outcomes denoted V. For example, if X represents the value of a rolled die then V is the set ${\displaystyle \{1,2,3,4,5,6\}}$. Let us assume for the sake of presentation that X is a discrete random variable, so that each value in V has a nonzero probability.

For a value x in V and an event A, the conditional probability is given by ${\displaystyle P(A\mid X=x)}$. Writing

${\displaystyle c(x,A)=P(A\mid X=x)}$

for short, we see that it is a function of two variables, x and A.

For a fixed A, we can form the random variable ${\displaystyle Y=c(X,A)}$. It represents an outcome of ${\displaystyle P(A\mid X=x)}$ whenever a value x of X is observed.

The conditional probability of A given X can thus be treated as a random variable Y with outcomes in the interval ${\displaystyle [0,1]}$. From the law of total probability, its expected value is equal to the unconditional probability of A.

Partial conditional probability

The partial conditional probability ${\displaystyle P(A\mid B_{1}\equiv b_{1},\ldots ,B_{m}\equiv b_{m})}$ is about the probability of event ${\displaystyle A}$ given that each of the condition events ${\displaystyle B_{i}}$ has occurred to a degree ${\displaystyle b_{i}}$ (degree of belief, degree of experience) that might be different from 100%. Frequentistically, partial conditional probability makes sense, if the conditions are tested in experiment repetitions of appropriate length ${\displaystyle n}$.^[11] Such ${\displaystyle n}$-bounded partial conditional probability can be defined as the conditionally expected average occurrence of event ${\displaystyle A}$ in testbeds of length ${\displaystyle n}$ that adhere to all of the probability specifications ${\displaystyle B_{i}\equiv b_{i}}$, i.e.:

${\displaystyle P^{n}(A\mid B_{1}\equiv b_{1},\ldots ,B_{m}\equiv b_{m})=\operatorname {E} ({\overline {A}}^{n}\mid {\overline {B}}_{1}^{n}=b_{1},\ldots ,{\overline {B}}_{m}^{n}=b_{m})}$^[11]

Based on that, partial conditional probability can be defined as

${\displaystyle P(A\mid B_{1}\equiv b_{1},\ldots ,B_{m}\equiv b_{m})=\lim _{n\to \infty }P^{n}(A\mid B_{1}\equiv b_{1},\ldots ,B_{m}\equiv b_{m}),}$

where ${\displaystyle b_{i}n\in \mathbb {N} }$^[11]

Jeffrey conditionalization^[12][13] is a special case of partial conditional probability, in which the condition events must form a partition:

${\displaystyle P(A\mid B_{1}\equiv b_{1},\ldots ,B_{m}\equiv b_{m})=\sum _{i=1}^{m}b_{i}P(A\mid B_{i})}$

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: ${\displaystyle P({\text{dot sent }}|{\text{ dot received}})=P({\text{dot received }}|{\text{ dot sent}}){\frac {P({\text{dot sent}})}{P({\text{dot received}})}}.}$ In Morse code, the ratio of dots to dashes is 3:4 at the point of sending, so the probability of a "dot" and "dash" are ${\displaystyle P({\text{dot sent}})={\frac {3}{7}}\ and\ P({\text{dash sent}})={\frac {4}{7}}}$. 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 ${\displaystyle P({\text{dot received}})}$.

${\displaystyle P({\text{dot received}})=P({\text{dot received }}\cap {\text{ dot sent}})+P({\text{dot received }}\cap {\text{ dash sent}})}$

${\displaystyle P({\text{dot received}})=P({\text{dot received }}\mid {\text{ dot sent}})P({\text{dot sent}})+P({\text{dot received }}\mid {\text{ dash sent}})P({\text{dash sent}})}$

${\displaystyle P({\text{dot received}})={\frac {9}{10}}\times {\frac {3}{7}}+{\frac {1}{10}}\times {\frac {4}{7}}={\frac {31}{70}}}$

Now, ${\displaystyle P({\text{dot sent }}\mid {\text{ dot received}})}$ can be calculated:

${\displaystyle P({\text{dot sent }}\mid {\text{ dot received}})=P({\text{dot received }}\mid {\text{ dot sent}}){\frac {P({\text{dot sent}})}{P({\text{dot received}})}}={\frac {9}{10}}\times {\frac {\frac {3}{7}}{\frac {31}{70}}}={\frac {27}{31}}}$^[15]

Assuming conditional probability is of similar size to its inverse

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 P(A|B) P(B) = P(B|A) P(A) i.e. P(A|B) = P(B|A) P(A)/P(B) . Similar reasoning can be used to show that P(Ā|B) = P(B|Ā) P(Ā)/P(B) etc.

In general, it cannot be assumed that P(A|B) ≈ P(B|A). This can be an insidious error, even for those who are highly conversant with statistics.^[18] The relationship between P(A|B) and P(B|A) is given by Bayes' theorem:

${\displaystyle {\begin{aligned}P(B\mid A)&={\frac {P(A\mid B)P(B)}{P(A)}}\\\Leftrightarrow {\frac {P(B\mid A)}{P(A\mid B)}}&={\frac {P(B)}{P(A)}}\end{aligned}}}$

That is, P(A|B) ≈ P(B|A) only if P(B)/P(A) ≈ 1, or equivalently, P(A) ≈ P(B).

Assuming marginal and conditional probabilities are of similar size

In general, it cannot be assumed that P(A) ≈ P(A|B). These probabilities are linked through the law of total probability:

${\displaystyle P(A)=\sum _{n}P(A\cap B_{n})=\sum _{n}P(A\mid B_{n})P(B_{n}).}$

where the events ${\displaystyle (B_{n})}$ form a countable partition of ${\displaystyle \Omega }$.

This fallacy may arise through selection bias.^[19] For example, in the context of a medical claim, let S_C be the event that a sequela (chronic disease) S occurs as a consequence of circumstance (acute condition) C. Let H be the event that an individual seeks medical help. Suppose that in most cases, C does not cause S (so that P(S_C) is low). Suppose also that medical attention is only sought if S has occurred due to C. From experience of patients, a doctor may therefore erroneously conclude that P(S_C) is high. The actual probability observed by the doctor is P(S_C|H).

Over- or under-weighting priors

Not taking prior probability into account partially or completely is called base rate neglect. The reverse, insufficient adjustment from the prior probability is conservatism.