In probability theory, conditional dependence is a relationship between two or more events that are dependent when a third event occurs.^[1][2] For example, if ${\displaystyle A}$ and ${\displaystyle B}$ are two events that individually increase the probability of a third event ${\displaystyle C,}$ and do not directly affect each other, then initially (when it has not been observed whether or not the event ${\displaystyle C}$ occurs)^[3][4]

$${\displaystyle \operatorname {P} (A\mid B)=\operatorname {P} (A)\quad {\text{ and ))\quad \operatorname {P} (B\mid A)=\operatorname {P} (B)}$$

${\displaystyle A{\text{ and ))B}$

But suppose that now ${\displaystyle C}$ is observed to occur. If event ${\displaystyle B}$ occurs then the probability of occurrence of the event ${\displaystyle A}$ will decrease because its positive relation to ${\displaystyle C}$ is less necessary as an explanation for the occurrence of ${\displaystyle C}$ (similarly, event ${\displaystyle A}$ occurring will decrease the probability of occurrence of ${\displaystyle B}$). Hence, now the two events ${\displaystyle A}$ and ${\displaystyle B}$ are conditionally negatively dependent on each other because the probability of occurrence of each is negatively dependent on whether the other occurs. We have^[5]

$${\displaystyle \operatorname {P} (A\mid C{\text{ and ))B)<\operatorname {P} (A\mid C).}$$

Conditional dependence of A and B given C is the logical negation of conditional independence ${\displaystyle ((A\perp \!\!\!\perp B)\mid C)}$.^[6] In conditional independence two events (which may be dependent or not) become independent given the occurrence of a third event.^[7]

Example

In essence probability is influenced by a person's information about the possible occurrence of an event. For example, let the event ${\displaystyle A}$ be 'I have a new phone'; event ${\displaystyle B}$ be 'I have a new watch'; and event ${\displaystyle C}$ be 'I am happy'; and suppose that having either a new phone or a new watch increases the probability of my being happy. Let us assume that the event ${\displaystyle C}$ has occurred – meaning 'I am happy'. Now if another person sees my new watch, he/she will reason that my likelihood of being happy was increased by my new watch, so there is less need to attribute my happiness to a new phone.

To make the example more numerically specific, suppose that there are four possible states ${\displaystyle \Omega =\left\{s_{1},s_{2},s_{3},s_{4}\right\},}$ given in the middle four columns of the following table, in which the occurrence of event ${\displaystyle A}$ is signified by a ${\displaystyle 1}$ in row ${\displaystyle A}$ and its non-occurrence is signified by a ${\displaystyle 0,}$ and likewise for ${\displaystyle B}$ and ${\displaystyle C.}$ That is, ${\displaystyle A=\left\{s_{2},s_{4}\right\},B=\left\{s_{3},s_{4}\right\},}$ and ${\displaystyle C=\left\{s_{2},s_{3},s_{4}\right\}.}$ The probability of ${\displaystyle s_{i))$ is ${\displaystyle 1/4}$ for every ${\displaystyle i.}$

Event	${\displaystyle \operatorname {P} (s_{1})=1/4}$	${\displaystyle \operatorname {P} (s_{2})=1/4}$	${\displaystyle \operatorname {P} (s_{3})=1/4}$	${\displaystyle \operatorname {P} (s_{4})=1/4}$	Probability of event
${\displaystyle A}$	0	1	0	1	${\displaystyle {\tfrac {1}{2))}$
${\displaystyle B}$	0	0	1	1	${\displaystyle {\tfrac {1}{2))}$
${\displaystyle C}$	0	1	1	1	${\displaystyle {\tfrac {3}{4))}$

and so

Event	${\displaystyle s_{1))$	${\displaystyle s_{2))$	${\displaystyle s_{3))$	${\displaystyle s_{4))$	Probability of event
${\displaystyle A\cap B}$	0	0	0	1	${\displaystyle {\tfrac {1}{4))}$
${\displaystyle A\cap C}$	0	1	0	1	${\displaystyle {\tfrac {1}{2))}$
${\displaystyle B\cap C}$	0	0	1	1	${\displaystyle {\tfrac {1}{2))}$
${\displaystyle A\cap B\cap C}$	0	0	0	1	${\displaystyle {\tfrac {1}{4))}$

In this example, ${\displaystyle C}$ occurs if and only if at least one of ${\displaystyle A,B}$ occurs. Unconditionally (that is, without reference to ${\displaystyle C}$), ${\displaystyle A}$ and ${\displaystyle B}$ are independent of each other because ${\displaystyle \operatorname {P} (A)}$—the sum of the probabilities associated with a ${\displaystyle 1}$ in row ${\displaystyle A}$—is ${\displaystyle {\tfrac {1}{2)),}$ while

$${\displaystyle \operatorname {P} (A\mid B)=\operatorname {P} (A{\text{ and ))B)/\operatorname {P} (B)={\tfrac {1/4}{1/2))={\tfrac {1}{2))=\operatorname {P} (A).}$$

${\displaystyle C}$

$${\displaystyle \operatorname {P} (A\mid C)=\operatorname {P} (A{\text{ and ))C)/\operatorname {P} (C)={\tfrac {1/2}{3/4))={\tfrac {2}{3))}$$

$${\displaystyle \operatorname {P} (A\mid C{\text{ and ))B)=\operatorname {P} (A{\text{ and ))C{\text{ and ))B)/\operatorname {P} (C{\text{ and ))B)={\tfrac {1/4}{1/2))={\tfrac {1}{2))<\operatorname {P} (A\mid C).}$$

${\displaystyle C}$

${\displaystyle A}$

${\displaystyle B,A}$

${\displaystyle B}$

${\displaystyle C.}$