STATS8: Introduction to Biostatistics 24pt Probability

1 STATS8: Introduction to BiostatisticsProbabilityBabak ShahbabaDepartment of Statistics, UCII ntroduction We have used plots and summary statistics to learn about thedistribution of variables and to investigate their relationships. We now want to generalize our findings to the population. However, we almost always remain uncertain about the truedistributions and relationships in the population. Therefore, when we generalize our findings from a sample tothe whole population, we should explicitly specify the extentof our uncertainty. We now discuss Probability as a measure of uncertainty.

2 We use some examples from Commonly Used Genetic Terms Gene Single Nucleotide Polymorphisms (SNPs) Alleles Genotype Homozygous vs. heterozygous Phenotype Recessive vs. dominantRandom phenomena and their sample space A phenomenon is calledrandomif its outcome (value) cannotbe determined with certainty before it occurs. For example, coin tossing and genotypes are randomphenomena. The collection of all possible outcomesSis called tossing:S={H,T},Die rolling:S={1,2,3,4,5,6},Bi-allelic gene:S={A,a},Genotype:S={AA,Aa,aa}.Proba bility To each possible outcome in the sample space, we assign aprobabilityP, which represents how certain we are about theoccurrence of the corresponding outcome.

3 For an outcomeo, we denote the Probability asP(o), where0 P(o) 1. The total Probability of all outcomes in the sample space isalways tossing:P(H) +P(T) = 1,Die rolling:P(1) +P(2) +P(3) +P(4) +P(5) +P(6) = 1. Therefore, if the outcomes are equally probable, theprobability of each outcome is 1/nS, wherenSis the numberof possible events Aneventis a subset of the sample spaceS. A possible event for die rolling isE={1,3,5}. This is theevent of rolling an odd number. For the genotype example,E={AA,aa}is the event that aperson is homozygous. An event occurs when any outcome within that event occurs.

4 We denote the Probability of eventEasP(E). The Probability of an event is the sum of the probabilities forall individual outcomes included in that As a running example, we consider a bi-allelic geneAwithtwo allelesAanda. We assume that alleleais recessive and causes a specificdisease. Then only people with the genotypeaahave the We can define four events as follows:The homozygous event:HM={AA,aa},The heterozygous event:HT={Aa},The no-disease event:ND={AA,Aa},The disease event:D={aa}. Assume that the probabilities for different genotypes areP(AA) = ,P(Aa) = , andP(aa) = Then,P(HM) = + = ,P(HT) = ,P(ND) = + = ,P(D) = For any eventE, we define itscomplement,Ec, as the set ofall outcomes that are in the sample spaceSbut not inE.

5 For the gene-disease example, the complement of thehomozygous eventHM={AA,aa}is the heterozygous event{Aa}; we show this asHMc=HT. Likewise, the complement of the disease event,D={aa}, isthe no-disease event,ND={AA,Aa}; we show this asDc=ND. The Probability of the complement event is 1 minus theprobability of the event:P(Ec)= 1 P(E).Union For two eventsE1andE2in a sample spaceS, we define theirunionE1 E2as the set of all outcomes that are at least inone of the events. The unionE1 E2is an event by itself, and it occurs wheneither E1or E2(or both) occurs.

6 For example, the union of the heterozygous event,HT, andthe disease event,D, is{Aa} {aa}={Aa,aa}. When possible, we can identify the outcomes in the union ofthe two events and find the Probability by adding theprobabilities of those For two eventsE1andE2in a sample spaceS, we define theirintersectionE1 E2as the set of outcomes that are in bothevents. The intersectionE1 E2is an event by itself, and it occurswhen bothE1and E2occur. The intersection of the heterozygous event and the no-diseaseevent isHM ND={AA}. When possible, we can identify the outcomes in the union ofthe two events and find the Probability by adding theprobabilities of those vs.

7 Marginal Probability We refer to the Probability of the intersection of two events,P(E1 E2), as theirjoint Probability . In contrast, we refer to probabilitiesP(E1) andP(E2) as themarginal probabilitiesof eventsE1andE2. For any two eventsE1andE2, we haveP(E1 E2) =P(E1) +P(E2) P(E1 E2). That is, the Probability of the unionP(E1 E2) is the sum oftheir marginal probabilities minus their joint Probability . The union of the heterozygous and the no-disease events isP(HM ND) =P(HM) +P(ND) P(HM ND)= + = events Two events are calleddisjointormutually exclusiveif theynever occur together: if we know that one of them hasoccurred, we can conclude that the other event has not.

8 Disjoint events have no elements (outcomes) in common, andtheir intersection is the empty set. For the above example, if a person is heterozygous, we knowthat he does not have the disease so the two eventsHTandNDare events For two disjoint eventsE1andE2, the Probability of theirintersection ( , their joint Probability ) is zero:P(E1 E2) =P( ) = 0 Therefore, the Probability of the union of the two disjointevents is simply the sum of their marginal probabilities:P(E1 E2) =P(E1) +P(E2) In general, if we have multiple disjoint events,E1,E2.

9 ,En,then the Probability of their union is the sum of the marginalprobabilities:P(E1 E2 .. En) =P(E1) +P(E2) +..+P(En)Partition When two or more events are disjoint and their union is thesample spaceS, we say that the events form apartitionofthe sample space. Two complementary eventsEandEcalways form a partitionof the sample space since they are disjoint and their union isthe sample Probability Ver often, we need to discuss possible changes in theprobability of one event based on our knowledge regarding theoccurrence of another event. Theconditional Probability , denotedP(E1|E2), is theprobability of eventE1given that another eventE2hasoccurred.

10 The conditional Probability of eventE1given eventE2can becalculated as follows: (assumingP(E2)6= 0)P(E1|E2) =P(E1 E2)P(E2). This is the joint Probability of the two events divided by themarginal Probability of the event on which we are Probability Consider the gene-disease example. Suppose we know that aperson is homozygous and are interested in the probabilitythat this person has the disease,P(D|HM). The Probability of the intersection ofDandHMisP(D HM) =P({aa}) = Therefore, the conditional Probability of having the diseaseknowing that the genotype is homozygous can be obtained asfollows:P(D|HM) =P(D HM)P(HM)= In this case, the Probability of the disease has increased fromP(D) = toP(D|HM) = law of total Probability By rearranging the equation for conditional probabilities, wehaveP(E1 E2) =P(E1|E2)P(E2).