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3.5 Independent events

In the previous section we saw that ${{\rm P}(A|B)}$ , the conditional probability of ${A}$ given ${B}$ , was in general not equal to ${{\rm P}(A)}$ , the unconditional probability of ${A}$ . In the special case when

\displaystyle{\rm P}(A|B)={\rm P}(A),

we say that ${A}$ is independent of ${B}$ . Independence means that knowing that the event ${B}$ has occurred does not change the chance that ${A}$ will occur.

Note that, using the definitions of conditional probability, ${{\rm P}(A\cap B)={\rm P}(A|B){\rm P}(B)}$ so if ${A}$ is independent of ${B}$ , then ${{\rm P}(A\cap B)={\rm P}(A){\rm P}(B)}$ . We therefore get the following equivalent definition of independence.

${A}$ and ${B}$ are independent events if and only if ${\rm P}(A\cap B)={\rm P}(A){\rm P}(B).$

Exercise 3.23.

If two coins are thrown and the four possible outcomes are equally likely, show that the events “Head on first coin” and “Head on second coin” are independent.

Solution.

$\displaystyle\Omega$	$\displaystyle=$	$\displaystyle\{H_{1}H_{2},H_{1}T_{2},T_{1}H_{2},T_{1}T_{2}\}$
$\displaystyle{\rm P}(\mbox{H on first})$	$\displaystyle=$	$\displaystyle{\rm P}(\{H_{1}H_{2},H_{1}T_{2}\})=\frac{2}{4}=\frac{1}{2}$
$\displaystyle{\rm P}(\mbox{H on second})$	$\displaystyle=$	$\displaystyle{\rm P}(\{H_{1}H_{2},T_{1}H_{2}\})=\frac{2}{4}=\frac{1}{2}$
$\displaystyle{\rm P}(\mbox{H on both})$	$\displaystyle=$	$\displaystyle{\rm P}(\{H_{1}H_{2}\})=\frac{1}{4}.$

We observe

\displaystyle{\rm P}(\mbox{H on both})

\displaystyle=

\displaystyle{\rm P}(\mbox{H on first}){\rm P}(\mbox{H on second})

so independent.

Similar calculations show the independence of any pair of events with one referring to the first coin and the other to the second coin.

Example 3.24.

The unfair coin. Suppose that the coin is biased and that the probability of a head occurring on any throw is ${\theta}$ , which can be any number between ${0}$ and ${1}$ . Use independence to determine the probabilities of the ${4}$ outcomes when throwing the coin twice.

Solution.

$\displaystyle\Omega$	$\displaystyle=$	$\displaystyle\{H_{1}H_{2},H_{1}T_{2},T_{1}H_{2},T_{1}T_{2}\}$
$\displaystyle{\rm P}(\{H_{1}H_{2}\})$	$\displaystyle=$	$\displaystyle{\rm P}(\{H_{1}\}){\rm P}(\{H_{2}\})=\theta\theta=\theta^{2}$
$\displaystyle{\rm P}(\{H_{1}T_{2}\})$	$\displaystyle=$	$\displaystyle{\rm P}(\{H_{1}\}){\rm P}(\{T_{2}\})=\theta(1-\theta)$
$\displaystyle{\rm P}(\{T_{1}H_{2}\})$	$\displaystyle=$	$\displaystyle{\rm P}(\{T_{1}\}){\rm P}(\{H_{2}\})=(1-\theta)\theta$
$\displaystyle{\rm P}(\{T_{1}T_{2}\})$	$\displaystyle=$	$\displaystyle{\rm P}(\{T_{1}\}){\rm P}(\{T_{2}\})=(1-\theta)(1-\theta)=(1-% \theta)^{2}.$

Note: when ${\theta=1/2}$ each of the above is ${1/4}$ .

Exercise 3.25.

Suppose that the probability of mothers being hypertensive (high blood pressure) is ${0.1}$ and fathers is ${0.2}$ . Find the probability of a child’s parents both being hypertensive, assuming both events are independent.

Solution.

	$\displaystyle{\rm P}(\mbox{both }H)$	$\displaystyle=$	$\displaystyle{\rm P}(\mbox{mother }H){\rm P}(\mbox{father }H)$
		$\displaystyle=$	$\displaystyle 0.1\times 0.2=0.02$

Note: we would expect these two events to be independent if the primary determinants of hypertensivity were genetic, however if the primary determinants were environmental then we might expect the two events not to be independent.

Exercise 3.26.

If ${{\rm P}(A)=0.2}$ and ${{\rm P}(B)=0.3}$ find ${{\rm P}(A\cap B)}$ if

i.

${A}$ and ${B}$ are independent,
ii.

${A}$ and ${B}$ are exclusive.

Solution.

i.

${{\rm P}(A\cap B)={\rm P}(A){\rm P}(B)=(0.2)(0.3)=0.06}$
ii.

${{\rm P}(A\cap B)={\rm P}(\emptyset)=0}$ .

Exercise 3.27.

Consider all mother and their first child’s blood pressure measurements. Let ${A=\{\mbox{mother's DBP}\geq 95\}}$ and ${B=\{\mbox{child's DBP}\geq 80\}}$ . Suppose we know that ${{\rm P}(A)=0.1}$ , ${{\rm P}(B)=0.2}$ and ${{\rm P}(A\cap B)=0.05}$ . Are mother and child’s blood pressures independent?

Solution.

\displaystyle A\mbox{ indep }B

\displaystyle\iff{\rm P}(A\cap B)={\rm P}(A){\rm P}(B).

But ${0.05\neq 0.1\times 0.2}$ so not independent.

Example 3.28.

Suppose that eye colours are only blue or green. Suppose also that there is a simple genetic coding for this: if both your eye colour alleles are ${G}$ then you have green eyes, otherwise you have blue eyes. Suppose each allele is ${G}$ with probability 0.1 (and you may assume independence). What is the probability that a random population member has green eyes? What is the probability that two randomly selected (unrelated) people both have green eyes?

Solution.

For an individual to have green eyes, both alleles must be ${G}$ . By independence,

{{\rm P}(GG)=\frac{1}{10}\frac{1}{10}=\frac{1}{100}.}

For two independent individuals, again by independence, we calculate

{{\rm P}(\mbox{both green})={\rm P}(GG){\rm P}(GG)=\frac{1}{100}\frac{1}{100}=% \frac{1}{10000}.}

Genes are passed from parent to child — each parent passes a randomly selected one of their two alleles to the child. If both parents have green eyes, what is the eye colour of the child?

Solution.

Both parents are $G G$ , both pass a ${G}$ to the child, the child is ${GG}$ and has green eyes.

Suppose we don’t know the parents’ eye colour, but we know a brother had green eyes. What can we now say about the probability of the subject having green eyes? What about the probability of a pair of brothers having green eyes?

Solution.

If we know the brother has green eyes, then each parent had at least one ${G}$ allele.
Since each parent passes this $G$ allele to the subject with probability ${1/2}$ , the probability that each parent passes a ${G}$ to the child is at least ${1/2}$ .
Hence

{{\rm P}(GG\,|\,\mbox{brother }GG)\geq\frac{1}{2}\frac{1}{2}=\frac{1}{4}.}

Now

$\displaystyle{\rm P}(\mbox{two brothers }GG)$	$\displaystyle=$	$\displaystyle{\rm P}(\mbox{brother one }GG){\rm P}(\mbox{brother two }GG\,\|\,% \mbox{brother one }GG)$
	$\displaystyle\geq$	$\displaystyle\frac{1}{100}\frac{1}{4}$
	$\displaystyle=$	$\displaystyle\frac{1}{400}.$

Note that if the population prevalence of ${G}$ is much smaller than ${1/10}$ , then ${{\rm P}(GG)}$ decreases significantly, but the calculation for ${{\rm P}(GG\,|\,{\rm brother}GG)}$ is unchanged. This is relevant to the sad case of Sally Clark http://en.wikipedia.org/wiki/Sally_Clark, in which a statistically illiterate doctor used independence of siblings’ cot death to convict her incorrectly of murdering her babies. If we swap “green eyes” for “suffers cot death”, then essentially the doctor took ${{\rm P}(G)=1/92}$ . Calculate the probabilities of two independent people, and two siblings, having green eyes with these numbers. The extreme low probability under independence was used as evidence to convict (which itself is known as the Prosecutor’s Fallacy).