# Conditional dependence

In probability theory, conditional dependence is a relationship between two or more events that are dependent when a third event occurs. For example, if $A$ and $B$ are two events that individually increase the probability of a third event $C,$ and do not directly affect each other, then initially (when it has not been observed whether or not the event $C$ occurs) $\operatorname {P} (A\mid B)=\operatorname {P} (A)\quad {\text{ and }}\quad \operatorname {P} (B\mid A)=\operatorname {P} (B)$ ($A{\text{ and }}B$ are independent).
But suppose that now $C$ is observed to occur. If event $B$ occurs then the probability of occurrence of the event $A$ will decrease because its positive relation to $C$ is less necessary as an explanation for the occurrence of $C$ (similarly, event $A$ occurring will decrease the probability of occurrence of $B$). Hence, now the two events $A$ and $B$ are conditionally negatively dependent on each other because the probability of occurrence of each is negatively dependent on whether the other occurs. We have $\operatorname {P} (A\mid C{\text{ and }}B)<\operatorname {P} (A\mid C).$
Conditional dependence of A and B given C is the logical negation of conditional independence $((A\perp \!\!\!\perp B)\mid C)$. In conditional independence two events (which may be dependent or not) become independent given the occurrence of a third event.

## Example

In essence probability is influenced by a person's information about the possible occurrence of an event. For example, let the event $A$ be 'I have a new phone'; event $B$ be 'I have a new watch'; and event $C$ 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 $C$ 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 $\Omega =\left\{s_{1},s_{2},s_{3},s_{4}\right\},$ given in the middle four columns of the following table, in which the occurrence of event $A$ is signified by a $1$ in row $A$ and its non-occurrence is signified by a $0,$ and likewise for $B$ and $C.$ That is, $A=\left\{s_{2},s_{4}\right\},B=\left\{s_{3},s_{4}\right\},$ and $C=\left\{s_{2},s_{3},s_{4}\right\}.$ The probability of $s_{i}$ is $1/4$ for every $i.$
| Event | $\operatorname {P} (s_{1})=1/4$ | $\operatorname {P} (s_{2})=1/4$ | $\operatorname {P} (s_{3})=1/4$ | $\operatorname {P} (s_{4})=1/4$ | Probability of event |
| $A$ | 0 | 1 | 0 | 1 | ${\tfrac {1}{2}}$ |
| $B$ | 0 | 0 | 1 | 1 | ${\tfrac {1}{2}}$ |
| $C$ | 0 | 1 | 1 | 1 | ${\tfrac {3}{4}}$ |
and so
| Event | $s_{1}$ | $s_{2}$ | $s_{3}$ | $s_{4}$ | Probability of event |
| $A\cap B$ | 0 | 0 | 0 | 1 | ${\tfrac {1}{4}}$ |
| $A\cap C$ | 0 | 1 | 0 | 1 | ${\tfrac {1}{2}}$ |
| $B\cap C$ | 0 | 0 | 1 | 1 | ${\tfrac {1}{2}}$ |
| $A\cap B\cap C$ | 0 | 0 | 0 | 1 | ${\tfrac {1}{4}}$ |
In this example, $C$ occurs if and only if at least one of $A,B$ occurs. Unconditionally (that is, without reference to $C$), $A$ and $B$ are independent of each other because $\operatorname {P} (A)$—the sum of the probabilities associated with a $1$ in row $A$—is ${\tfrac {1}{2}},$ while $\operatorname {P} (A\mid B)=\operatorname {P} (A{\text{ and }}B)/\operatorname {P} (B)={\tfrac {1/4}{1/2}}={\tfrac {1}{2}}=\operatorname {P} (A).$ But conditional on $C$ having occurred (the last three columns in the table), we have $\operatorname {P} (A\mid C)=\operatorname {P} (A{\text{ and }}C)/\operatorname {P} (C)={\tfrac {1/2}{3/4}}={\tfrac {2}{3}}$ while $\operatorname {P} (A\mid C{\text{ and }}B)=\operatorname {P} (A{\text{ and }}C{\text{ and }}B)/\operatorname {P} (C{\text{ and }}B)={\tfrac {1/4}{1/2}}={\tfrac {1}{2}}<\operatorname {P} (A\mid C).$ Since in the presence of $C$ the probability of $A$ is affected by the presence or absence of $B,A$ and $B$ are mutually dependent conditional on $C.$