Conditional dependence

A Bayesian network illustrating 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) $$\mathrm{P}(A\mid B)=\mathrm{P}(A)\text{ and }\mathrm{P}(B\mid A)=\mathrm{P}(B)$$ ($A\text{ and }B$ 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 $$\mathrm{P}(A\mid C\text{ and }B)<\mathrm{P}(A\mid C).$$

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 $\mathrm{\Omega }=\{s_1,s_2,s_3,s_4\},$ 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, $A=\{s_2,s_4\},B=\{s_3,s_4\},$ and $C=\{s_2,s_3,s_4\}.$ The probability of $s_i$ is $1/4$ for every $i.$

Event	$\mathrm{P}(s_1)=1/4$	$\mathrm{P}(s_2)=1/4$	$\mathrm{P}(s_3)=1/4$	$\mathrm{P}(s_4)=1/4$	Probability of event
$A$	0	1	0	1	$\frac{1}{2}$
$B$	0	0	1	1	$\frac{1}{2}$
$C$	0	1	1	1	$\frac{3}{4}$

and so

Event	$s_1$	$s_2$	$s_3$	$s_4$	Probability of event
$A\cap B$	0	0	0	1	$\frac{1}{4}$
$A\cap C$	0	1	0	1	$\frac{1}{2}$
$B\cap C$	0	0	1	1	$\frac{1}{2}$
$A\cap B\cap C$	0	0	0	1	$\frac{1}{4}$

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 $\mathrm{P}(A)$—the sum of the probabilities associated with a $1$ in row $A$—is $\frac{1}{2},$ while $$\mathrm{P}(A\mid B)=\mathrm{P}(A\text{ and }B)/\mathrm{P}(B)=\frac{1/4}{1/2}=\frac{1}{2}=\mathrm{P}(A).$$ But conditional on $C$ having occurred (the last three columns in the table), we have $$\mathrm{P}(A\mid C)=\mathrm{P}(A\text{ and }C)/\mathrm{P}(C)=\frac{1/2}{3/4}=\frac{2}{3}$$ while $$\mathrm{P}(A\mid C\text{ and }B)=\mathrm{P}(A\text{ and }C\text{ and }B)/\mathrm{P}(C\text{ and }B)=\frac{1/4}{1/2}=\frac{1}{2}<\mathrm{P}(A\mid C).$$ 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 $C.$

See also

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