Probability and random variables

A breeder can describe the phenotypes already recorded, but selection decisions concern offspring and candidates whose outcomes are not yet known. Probability supplies a model for that uncertainty. It does not guarantee the genotype of one offspring. Instead, it describes the pattern expected when the same inheritance process is repeated under stated assumptions.

Outcomes and events in a Mendelian cross

Consider one autosomal locus with alleles \(A\) and \(a\). Two heterozygous parents are crossed:

\[ Aa \times Aa. \]

Under ordinary Mendelian segregation, each parent contributes \(A\) with probability \(1/2\) and \(a\) with probability \(1/2\). Assume that transmission from one parent is independent of transmission from the other, and that the four gamete combinations have equal fertilisation and survival. These assumptions define the probability model. Segregation distortion, differential survival, or genotyping error would require a different model.

An outcome records one possible result. The sample space \(\Omega\) is the set of all outcomes considered by the model:

\[ \Omega=\{(A,A),(A,a),(a,A),(a,a)\}. \]

The first entry is the sire’s contribution and the second is the dam’s. Each ordered pair has probability \(1/4\). Keeping the pairs ordered matters: \((A,a)\) and \((a,A)\) are different transmission outcomes even though both produce genotype \(Aa\).

An event is a set of outcomes. If \(H\) denotes the event that an offspring is heterozygous, then

\[ H=\{(A,a),(a,A)\}, \qquad P(H)=\frac{2}{4}=\frac{1}{2}. \]

The genotype probabilities are therefore \(P(AA)=1/4\), \(P(Aa)=1/2\), and \(P(aa)=1/4\). These values describe proportions over repeated crosses. A family of four offspring need not contain exactly one \(AA\), two \(Aa\), and one \(aa\).

Conditional probability and independence

Conditional probability asks how the probability of one event changes after another event is known. For events \(B\) and \(C\), with \(P(C)>0\),

\[ P(B\mid C)=\frac{P(B\cap C)}{P(C)}. \]

Let \(B\) be the event that the offspring is \(AA\), and let \(C\) be the event that it carries at least one \(A\) allele. Three of the four outcomes belong to \(C\), and one of those three belongs to \(B\). Consequently, \(P(B\mid C)=1/3\), not \(1/4\). Learning that the offspring carries \(A\) has changed the relevant set of possible outcomes.

Two events are independent when knowing that one occurred does not change the probability of the other. Let \(S\) mean that the sire contributes \(A\), and \(D\) mean that the dam contributes \(A\). In this model,

\[ P(S\cap D)=\frac{1}{4}=P(S)P(D) =\frac{1}{2}\times\frac{1}{2}. \]

This equality expresses the independence assumption for the two parental transmissions. Independence is not a general claim about all genetic events. Nearby loci can be inherited together because of linkage, and relatives can have correlated genetic values.

A random variable turns outcomes into numbers

A random variable maps each outcome to a numerical value. Define \(G\) as the number of \(A\) alleles carried by an offspring. This is the usual 0, 1, 2 allele-dosage coding:

Transmission outcome Genotype \(G\) Probability
\((a,a)\) \(aa\) 0 \(1/4\)
\((A,a)\) or \((a,A)\) \(Aa\) 1 \(1/2\)
\((A,A)\) \(AA\) 2 \(1/4\)

The outcome is a pair of transmitted alleles, while \(G\) is a function of that outcome. Several outcomes can map to the same value. This distinction lets a statistical model work with numerical marker dosage without claiming that the number itself contains the complete biological history of the alleles.

The expected value of a discrete random variable is its probability-weighted centre:

\[ E(G)=\sum_g gP(G=g) =0\left(\frac14\right)+1\left(\frac12\right) +2\left(\frac14\right)=1. \]

An expectation is a long-run average under the model. Here it happens to be a possible dosage, but an expectation need not be a value that any individual can take. For example, the expected genotype count in a family need not be an integer.

The variance of \(G\) measures probability-weighted squared distance from its expectation:

\[ \operatorname{Var}(G) =\sum_g \{g-E(G)\}^2P(G=g) =1\left(\frac14\right)+0\left(\frac12\right) +1\left(\frac14\right)=0.5. \]

This is a variance of a random variable under a probability model. It differs from the sample variance calculated from realised observations on the previous page, although a sufficiently informative random sample can be used to estimate a population variance.

Verify the cross in R

The following code constructs the four equally likely ordered transmissions. Logical values are added as 0 and 1, which gives the \(A\)-allele dosage.

Code
cross <- expand.grid(sire = c("A", "a"), dam = c("A", "a"))
cross$dosage <- (cross$sire == "A") + (cross$dam == "A")
probability <- prop.table(table(cross$dosage))
expectation <- sum(as.numeric(names(probability)) * probability)
variance <- sum(
  (as.numeric(names(probability)) - expectation)^2 * probability
)

cross
  sire dam dosage
1    A   A      2
2    a   A      1
3    A   a      1
4    a   a      0
Code
probability

   0    1    2 
0.25 0.50 0.25
Code
c(expectation = expectation, variance = variance)
expectation    variance 
        1.0         0.5

The result gives dosage probabilities 0.25, 0.50, and 0.25, expectation 1, and variance 0.5. The calculation verifies the stated model; it is not an empirical test of segregation in the project dataset.

Probability notation used later

The expression \(P(A)\) denotes the probability of event \(A\). The union \(A\cup B\) means that at least one of two events occurs, while the intersection \(A\cap B\) means that both occur. For mutually exclusive events, \(P(A\cup B)=P(A)+P(B)\). For general events, the shared intersection must be subtracted once:

\[ P(A\cup B)=P(A)+P(B)-P(A\cap B). \]

The complement \(A^c\) contains outcomes outside \(A\), so \(P(A^c)=1-P(A)\). These rules will reappear when genotype classes, binary phenotypes, and Bayesian mixture classes are assigned probabilities.

Exercises

  1. In a hypothetical transmission model, the sire contributes \(A\) with probability 0.6 and the dam contributes \(A\) with probability 0.3. Assuming independent parental transmissions, what is the probability that an offspring carries exactly one \(A\) allele? What is the probability that it carries at least one \(A\) allele?

Exactly one \(A\) can arise when the sire contributes \(A\) and the dam contributes \(a\), or when the sire contributes \(a\) and the dam contributes \(A\). Its probability is \(0.6(1-0.3)+(1-0.6)0.3=0.42+0.12=0.54\). The probability of at least one \(A\) is one minus the probability that both contribute \(a\): \(1-(1-0.6)(1-0.3)=1-0.28=0.72\).

  1. Suppose an \(Aa\) parent is crossed with an \(aa\) parent. Construct the ordered sample space and calculate the expectation and variance of the offspring’s \(A\)-allele dosage.

Writing the heterozygous parent’s contribution first, the ordered sample space is \(\{(A,a),(a,a)\}\). The \(aa\) parent always contributes \(a\), while the \(Aa\) parent contributes \(A\) or \(a\) with equal probability. The two outcomes give dosage 1 with probability \(1/2\) and dosage 0 with probability \(1/2\). Thus \(E(G)=0.5\), and \(\operatorname{Var}(G)=(0-0.5)^2(0.5)+(1-0.5)^2(0.5)=0.25\).

  1. For the original \(Aa\times Aa\) cross, calculate \(P(AA\mid\text{offspring is homozygous})\).

The homozygous outcomes are \((A,A)\) and \((a,a)\), each with probability \(1/4\). Conditional on this two-outcome set, one outcome is \(AA\), so the conditional probability is \((1/4)/(1/2)=1/2\).

  1. For the \(Aa\times Aa\) model, list the ordered outcomes in the event that an offspring carries at least one \(A\) allele. Why are \((A,a)\) and \((a,A)\) separate outcomes but the same genotype?

The event is \(\{(A,A),(A,a),(a,A)\}\). The ordered pair records which parent contributed each allele, so the two heterozygous transmissions differ as outcomes. Both contain one \(A\) and one \(a\), so both give genotype \(Aa\).

  1. Let \(A\) and \(B\) be events with \(P(A)=0.6\), \(P(B)=0.5\), and \(P(A\cap B)=0.2\). Calculate \(P(A\cup B)\) and \(P(A^c)\).

\(P(A\cup B)=0.6+0.5-0.2=0.9\), because the intersection is subtracted once. The complement probability is \(P(A^c)=1-0.6=0.4\).

A random variable is fully described only after probabilities have been assigned to its possible values. The next page develops probability distributions, which provide reusable models for genotype counts, binary traits, residuals, and marker effects.