Statistical FoundationsJinliang YangOct. 3th, 20221 / 43

Population Genetics vs. Quantitative genetics?

Population genetics

Pop-gen is the study of evolution.
The language of pop-gen is Mathematics.

2 / 43

Population Genetics vs. Quantitative genetics?

Population genetics

Pop-gen is the study of evolution.
The language of pop-gen is Mathematics.

Quantitative genetics

Quant-gen is the study of the complex trait, or phenotype.
The language of Quant-gen is Statistics.

3 / 43

Quantitative genetics almost synonymous with statistics

R. A. Fisher is a founder of quantitative genetics but also of analysis of variance and randomization procedures in statistics.

The early geneticist Karl Pearson originated the concepts of regression and correlation.

4 / 43

Quantitative genetics almost synonymous with statistics

R. A. Fisher is a founder of quantitative genetics but also of analysis of variance and randomization procedures in statistics.

The early geneticist Karl Pearson originated the concepts of regression and correlation.

In the next couple weeks, we will be deeply involved with the statistical evalution of the basic quantiative genetic models.

5 / 43

Quantitative genetics vs. statistics

Many genetic factors
- Determining the quantitative traits are almost always normally distributed

6 / 43

Quantitative genetics vs. statistics

Many genetic factors
- Determining the quantitative traits are almost always normally distributed
Genetic factors act in pairs (two alleles per locus)
- Explanatory variables
- Each with two or three levels of variation

7 / 43

Quantitative genetics vs. statistics

Many genetic factors
- Determining the quantitative traits are almost always normally distributed
Genetic factors act in pairs (two alleles per locus)
- Explanatory variables
- Each with two or three levels of variation
Passed on to progeny at random
- Random events

8 / 43

Quantitative genetics vs. statistics

Many genetic factors
- Determining the quantitative traits are almost always normally distributed
Genetic factors act in pairs (two alleles per locus)
- Explanatory variables
- Each with two or three levels of variation
Passed on to progeny at random
- Random events
Genetic factors sometimes show independent assortment
- Independence

9 / 43

Quantitative traits: statistical notation

Conceptual Notation

+ error

10 / 43

Quantitative traits: statistical notation

Conceptual Notation

+ error

Matrix Notation

$\begin{align*} \underbrace{\begin{bmatrix} Y_1\\ Y_2\\ \vdots \\ Y_n\end{bmatrix}}_{n \times 1} &= \underbrace{\begin{bmatrix} X_{11} & X_{12} & \cdots & X_{1m} \\ X_{21} & X_{22} & \cdots & X_{2m} \\ \vdots & \vdots & \ddots & \vdots \\ X_{n1} & X_{n2} & \cdots & X_{nm} \end{bmatrix}}_{n \times m} \quad \underbrace{\begin{bmatrix} a_1\\ a_2\\ \vdots \\ a_m\end{bmatrix}}_{m \times 1} +\underbrace{\begin{bmatrix} \epsilon_1\\ \epsilon_2\\ \vdots \\ \epsilon_m\end{bmatrix}}_{n \times 1} \end{align*}$

11 / 43

Quantitative traits: statistical notation

Conceptual Notation

+ error

Matrix Notation

Statistical Notation

$\begin{align*} Y_{i} = \sum\limits_{j=1}^{j=m} X_{ij} \alpha_{j} + \epsilon_i \end{align*}$

12 / 43

Why the normal distribution?

13 / 43

Why the normal distribution?

The Central Limit Theorem (CLT)

The CLT states that the sums of a set of random variables $(X_1, X_2, X_3, ..., X_n)$ is normally distributed no matter the distribution the individual X's were sampled from, as long as they were sampled from identical distributions.

14 / 43

A simulation experiment

$\begin{align*} Y_{i} = \sum\limits_{j=1}^{j=m} X_{ij} \alpha_{j} + \epsilon_i \end{align*}$

For a given individual ( $i=1$ ) with a number of loci ( $m=1,000$ )
Each allele is $X_j \in (A, a)$ , with the probability of $p$ or $q=1-p$

15 / 43

A simulation experiment

$\begin{align*} Y_{i} = \sum\limits_{j=1}^{j=m} X_{ij} \alpha_{j} + \epsilon_i \end{align*}$

For a given individual ( $i=1$ ) with a number of loci ( $m=1,000$ )
Each allele is $X_j \in (A, a)$ , with the probability of $p$ or $q=1-p$
The effect of $j$ th allele ( $\alpha_j$ ) can be samples from any distribution (e.g., uniform distribution)

According to the CLT, if $m$ is sufficiently large, the sum is normally distributed.

16 / 43

A simulation experiment

$\begin{align*} Y_{i} = \sum\limits_{j=1}^{j=m} X_{ij} \alpha_{j} + \epsilon_i \end{align*}$

For a given individual ( $i=1$ ) with a number of loci ( $m=1,000$ )
Each allele is $X_j \in (A, a)$ , with the probability of $p$ or $q=1-p$
The effect of $j$ th allele ( $\alpha_j$ ) can be samples from any distribution (e.g., uniform distribution)

According to the CLT, if $m$ is sufficiently large, the sum is normally distributed.

m <- 1000
## for each allele, the chance of A or a is equal to 0.5
x <- rbinom(n=m, size=1, prob=0.5)
## sample effect from a uniform distribution:
a <- runif(n=m)
y <- sum(x*a) + 0
y

## [1] 243.8146

17 / 43

A simulation experiment

$\begin{align*} Y_{i} = \sum\limits_{j=1}^{j=m} X_{ij} \alpha_{j} + \epsilon_i \end{align*}$

set.seed(1234) # seed for random number generator
m <- 1000
n = 2000 # simulate a population of 2,000 individuals
out <- c() # create an empty vector
for(i in 1:n){
  x <- rbinom(n=m, size=1, prob=0.5) ## for each allele, the chance of A = 0.5
  a <- runif(n=m) ## sample effect from a uniform distribution:
  y <- sum(x*a)
  out <- c(out, y)
} 
#shapiro.test(out) # W = 0.99928, p-value = 0.6622
hist(out, breaks=50, col="#b8860b", main="Phenotype Distribution", xlab="")

18 / 43

A simulation experiment

$\begin{align*} Y_{i} = \sum\limits_{j=1}^{j=m} X_{ij} \alpha_{j} + \epsilon_i \end{align*}$

set.seed(1234) # seed for random number generator
m <- 2
n = 2000 # simulate a population of 2,000 individuals
out <- c()
for(i in 1:n){
  x <- rbinom(n=m, size=1, prob=0.5) ## for each allele, the chance of A = 0.5
  a <- runif(n=m) ## sample effect from a uniform distribution:
  y <- sum(x*a)
  out <- c(out, y)
} 
#shapiro.test(out) # W = 0.91117, p-value < 2.2e-16
hist(out, breaks=50, col="#b8860b", main="Phenotype Distribution", xlab="")

19 / 43

Probability Density

For a continuous trait, i.e., kernel number per ear (raning from 0 to 1,000), what is the $Pr(Y=100)$ ?

20 / 43

Probability Density

For a continuous trait, i.e., kernel number per ear (raning from 0 to 1,000), what is the $Pr(Y=100)$ ?

Assuming wheat plant height in a population is normally distributed, with m = 30 inch, sd=5. Question: $Pr(30 < Y \leq 50)$ ?

21 / 43

Probability Density

For a continuous trait, i.e., kernel number per ear (raning from 0 to 1,000), what is the $Pr(Y=100)$ ?

Assuming wheat plant height in a population is normally distributed, with m = 30 inch, sd=5. Question: $Pr(30 < Y \leq 50)$ ?

Using the probability density function (or pdf) and integration, we can calculate the probability that Y is contained in a certain bracket as:

$\begin{align*} Pr(30 < Y \leq 50) = \int_{30}^{50} f(y)d_y \end{align*}$

22 / 43

Probability Density

For a continuous trait, i.e., kernel number per ear (raning from 0 to 1,000), what is the $Pr(Y=100)$ ?

Assuming wheat plant height in a population is normally distributed, with m = 30 inch, sd=5. Question: $Pr(30 < Y \leq 50)$ ?

Using the probability density function (or pdf) and integration, we can calculate the probability that Y is contained in a certain bracket as:

$\begin{align*} Pr(30 < Y \leq 50) = \int_{30}^{50} f(y)d_y \end{align*}$

23 / 43

Expectation and variance

Define the random variable $X$ (i.e. for genotype) which counts the number of allele $A$ . $\begin{align*} X &= \begin{cases} 2 & \text{if } AA \text{ with frequency } p^2 \\ 1 & \text{if } Aa \text{ with frequency } 2p(1-p) \\ 0 & \text{if } aa \text{ with frequency } (1-p)^2 \end{cases} \\ \end{align*}$ where $p$ is the allele frequency of $A$ .

24 / 43

Expectation and variance

Then, according to Formula (1) in the Note:

$\begin{align*} E(f(X)) = \sum\limits_{i=1}^kf(x_i)Pr(X = x_i) \end{align*}$

25 / 43

Expectation and variance

Then, according to Formula (1) in the Note:

$\begin{align*} E(f(X)) = \sum\limits_{i=1}^kf(x_i)Pr(X = x_i) \end{align*}$

Expected value of $X$ :

26 / 43

Expectation and variance

Then, according to Formula (1) in the Note:

$\begin{align*} E(f(X)) = \sum\limits_{i=1}^kf(x_i)Pr(X = x_i) \end{align*}$

Expected value of $X$ :

$\begin{align*} E[X] &= 0 \times (1 - p)^2 + 1 \times [2p(1-p)] + 2 \times p^2 = 2p \\ \end{align*}$

27 / 43

Expectation and variance

Define the random variable $X$ which counts the number of allele $A$ . $\begin{align*} X &= \begin{cases} 2 & \text{if } AA \text{ with frequency } p^2 \\ 1 & \text{if } Aa \text{ with frequency } 2p(1-p) \\ 0 & \text{if } aa \text{ with frequency } (1-p)^2 \end{cases} \\ \end{align*}$ where $p$ is the allele frequency of $A$ .

28 / 43

Expectation and variance

Expected value of $X^2$ :

29 / 43

Expectation and variance

Expected value of $X^2$ :

$\begin{align*} E[X^2] &= 0^2 \times (1 - p)^2 + 1^2 \times [2p(1-p)] + 2^2 \times p^2 \\ &= 2p(1-p) + 4p^2 \\ \end{align*}$

30 / 43

Expectation and variance

Expected value of $X^2$ :

$\begin{align*} E[X^2] &= 0^2 \times (1 - p)^2 + 1^2 \times [2p(1-p)] + 2^2 \times p^2 \\ &= 2p(1-p) + 4p^2 \\ \end{align*}$

Thus, the variance of allelic counts is

31 / 43

Expectation and variance

Expected value of $X^2$ :

$\begin{align*} E[X^2] &= 0^2 \times (1 - p)^2 + 1^2 \times [2p(1-p)] + 2^2 \times p^2 \\ &= 2p(1-p) + 4p^2 \\ \end{align*}$

Thus, the variance of allelic counts is

$\begin{align*} Var(X) &= E[X^2] - E[X]^2 \\ &= 2p(1-p) + 4p^2 - (2p)^2\\ &= 2p(1-p) \end{align*}$

32 / 43

Alternative coding

Define the random variable $X$ as below:

$\begin{align*} X &= \begin{cases} 1 & \text{if } AA \text{ with frequency } p^2 \\ 0 & \text{if } Aa \text{ with frequency } 2p(1-p) \\ -1 & \text{if } aa \text{ with frequency } (1-p)^2 \end{cases} \\ \end{align*}$ where $p$ is the allele frequency of $A$ .

33 / 43

Alternative coding

Define the random variable $X$ as below:

Then, $\begin{align*} E[X] &= -1 \times (1 - p)^2 + 0 \times [2p(1-p)] + 1 \times p^2 \\ &= −(1 − 2p + p^2) + p^2 = 2p-1 \\ E[X^2] &= (-1)^2 \times (1 - p)^2 + 0^2 \times [2p(1-p)] + 1^2 \times p^2 \\ &= 1 − 2p + p^2 +p^2 = 2p^2 − 2p + 1 \\ \end{align*}$ Thus, the variance of allelic counts is $\begin{align*} Var(X) &= E[X^2] - E[X]^2 \\ &= 2p^2 − 2p + 1 − (4p^2 − 4p + 1)\\ &= -2p^2 + 2p = 2p(1-p) \end{align*}$

34 / 43

Examples for probabilities

Two variables: Genotype and Milk Yield (MY)

Genotype (G)	$MY \leq 100$	$100 < MY \leq 300$	$MY > 300$	Marginal $Pr(G)$
aa	0.10	0.04	0.02	0.16
Aa	0.14	0.18	0.16	0.48
AA	0.06	0.10	0.20	0.36
Marg. Prob.	0.30	0.32	0.38	1.00

35 / 43

Examples for probabilities

Two variables: Genotype and Milk Yield (MY)

Genotype (G)	$MY \leq 100$	$100 < MY \leq 300$	$MY > 300$	Marginal $Pr(G)$
aa	0.10	0.04	0.02	0.16
Aa	0.14	0.18	0.16	0.48
AA	0.06	0.10	0.20	0.36
Marg. Prob.	0.30	0.32	0.38	1.00

Joint Probability

Two random variables to occur together.

36 / 43

Examples for probabilities

Two variables: Genotype and Milk Yield (MY)

Genotype (G)	$MY \leq 100$	$100 < MY \leq 300$	$MY > 300$	Marginal $Pr(G)$
aa	0.10	0.04	0.02	0.16
Aa	0.14	0.18	0.16	0.48
AA	0.06	0.10	0.20	0.36
Marg. Prob.	0.30	0.32	0.38	1.00

Joint Probability

Two random variables to occur together.

What is the joint probability of $Pr(G=aa, MY > 300)$ ?

37 / 43

Examples for probabilities

Two variables: Genotype and Milk Yield (MY)

Genotype (G)	$MY \leq 100$	$100 < MY \leq 300$	$MY > 300$	Marginal $Pr(G)$
aa	0.10	0.04	0.02	0.16
Aa	0.14	0.18	0.16	0.48
AA	0.06	0.10	0.20	0.36
Marg. Prob.	0.30	0.32	0.38	1.00

38 / 43

Examples for probabilities

Two variables: Genotype and Milk Yield (MY)

Genotype (G)	$MY \leq 100$	$100 < MY \leq 300$	$MY > 300$	Marginal $Pr(G)$
aa	0.10	0.04	0.02	0.16
Aa	0.14	0.18	0.16	0.48
AA	0.06	0.10	0.20	0.36
Marg. Prob.	0.30	0.32	0.38	1.00

Marginal Probability

A sum of mutually exclusive and exhaustive set of events.

39 / 43

Examples for probabilities

Two variables: Genotype and Milk Yield (MY)

Genotype (G)	$MY \leq 100$	$100 < MY \leq 300$	$MY > 300$	Marginal $Pr(G)$
aa	0.10	0.04	0.02	0.16
Aa	0.14	0.18	0.16	0.48
AA	0.06	0.10	0.20	0.36
Marg. Prob.	0.30	0.32	0.38	1.00

Marginal Probability

A sum of mutually exclusive and exhaustive set of events.

What is the marginal probability of $Pr(G=Aa)$ ?

40 / 43

Examples for probabilities

Two variables: Genotype and Milk Yield (MY)

Genotype (G)	$MY \leq 100$	$100 < MY \leq 300$	$MY > 300$	Marginal $Pr(G)$
aa	0.10	0.04	0.02	0.16
Aa	0.14	0.18	0.16	0.48
AA	0.06	0.10	0.20	0.36
Marg. Prob.	0.30	0.32	0.38	1.00

41 / 43

Examples for probabilities

Two variables: Genotype and Milk Yield (MY)

Genotype (G)	$MY \leq 100$	$100 < MY \leq 300$	$MY > 300$	Marginal $Pr(G)$
aa	0.10	0.04	0.02	0.16
Aa	0.14	0.18	0.16	0.48
AA	0.06	0.10	0.20	0.36
Marg. Prob.	0.30	0.32	0.38	1.00

Conditional Probability

$Pr(X = x | Y = y) = \frac{Pr(X = x, Y = y)}{ Pr(Y = y)}$

42 / 43

Examples for probabilities

Two variables: Genotype and Milk Yield (MY)

Genotype (G)	$MY \leq 100$	$100 < MY \leq 300$	$MY > 300$	Marginal $Pr(G)$
aa	0.10	0.04	0.02	0.16
Aa	0.14	0.18	0.16	0.48
AA	0.06	0.10	0.20	0.36
Marg. Prob.	0.30	0.32	0.38	1.00

Conditional Probability

$Pr(X = x | Y = y) = \frac{Pr(X = x, Y = y)}{ Pr(Y = y)}$

What is the conditional probability of $Pr(MY \leq 100 | G=Aa)$ ?

43 / 43

↑, ←, Pg Up, k	Go to previous slide
↓, →, Pg Dn, Space, j	Go to next slide
Home	Go to first slide
End	Go to last slide
Number + Return	Go to specific slide
b / m / f	Toggle blackout / mirrored / fullscreen mode
c	Clone slideshow
p	Toggle presenter mode
t	Restart the presentation timer
?, h	Toggle this help

Statistical Foundations

Jinliang Yang

Oct. 3th, 2022

Population Genetics vs. Quantitative genetics?

Population genetics

Population Genetics vs. Quantitative genetics?

Population genetics

Quantitative genetics

Quantitative genetics almost synonymous with statistics

Quantitative genetics almost synonymous with statistics

Quantitative genetics vs. statistics

Quantitative genetics vs. statistics

Quantitative genetics vs. statistics

Quantitative genetics vs. statistics

Quantitative traits: statistical notation

Conceptual Notation

Quantitative traits: statistical notation

Conceptual Notation

Matrix Notation

Quantitative traits: statistical notation

Conceptual Notation

Matrix Notation

Statistical Notation

Why the normal distribution?

Why the normal distribution?

The Central Limit Theorem (CLT)

A simulation experiment

A simulation experiment

A simulation experiment

A simulation experiment

A simulation experiment

Probability Density

Probability Density

Probability Density

Probability Density

Expectation and variance

Expectation and variance

Expectation and variance

Expectation and variance

Expectation and variance

Expectation and variance

Expectation and variance

Expectation and variance

Expectation and variance

Alternative coding

Alternative coding

Examples for probabilities

Examples for probabilities

Joint Probability

Examples for probabilities

Joint Probability

Examples for probabilities

Examples for probabilities

Marginal Probability

Examples for probabilities

Marginal Probability

Examples for probabilities

Examples for probabilities

Conditional Probability

Examples for probabilities

Conditional Probability

Population Genetics vs. Quantitative genetics?

Population genetics

Help