# BDA3 Chapter 1 Exercise 3

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# BDA3 Chapter 1 Exercise 3

Here’s my solution to exercise 3, chapter 1, of Gelman’s *Bayesian Data Analysis* (BDA), 3rd edition. There are solutions to some of the exercises on the book’s webpage.

Suppose a particular gene for eye colour has two alleles: a dominant X and a recessive x allele. Having \(xx\) gives you blue eyes, otherwise you have brown eyes. Suppose also that the proportion of blue-eyed people is \(p^2\), and the proportion of heterozygotes is \(2p(1 – p)\). There are 3 questions to answer:

- What is the probability of a brown-eyed child of brown-eyed parents being a heterozygote?
- If such a heterozygote, Judy, has n brown-eyed children with a random heterozygote, what’s the probability that Judy is a heterozygote?
- Under the conditions of part 2, what is the probability that Judy’s first grandchild has blue eyes?

## Simulation

Let’s first set up some data with which we can verify the results via simulation.

### Data

We’ll simulate a large population of individuals where the probability of the recessive allele is 0.2.

set.seed(11146) N <- 5000000 p <- 0.2 alleles <- c('x', 'X') weights <- c(p, 1 - p) df <- tibble(id = 1:N %>% as.character()) %>% mutate( allele1 = sample(alleles, N, prob = weights, replace = TRUE), allele2 = sample(alleles, N, prob = weights, replace = TRUE), genotype = if_else(allele1 == allele2, 'homozygote', 'heterozygote'), eye_colour = if_else(allele1 == 'x' & allele2 == 'x', 'blue', 'brown') )

id | allele1 | allele2 | genotype | eye_colour |
---|---|---|---|---|

1 | X | X | homozygote | brown |

2 | x | X | heterozygote | brown |

3 | X | x | heterozygote | brown |

4 | X | X | homozygote | brown |

5 | x | X | heterozygote | brown |

6 | X | X | homozygote | brown |

This has the correct distribution of alleles, since \(p^2 \approx\) 0.04 and \((1-p)^2\approx\) 0.64.

allele_distribution <- df %>% group_by(allele1, allele2) %>% tally() %>% mutate(frac = n / sum(n))

allele1 | allele2 | n | frac |
---|---|---|---|

x | x | 200006 | 0.2000018 |

x | X | 800015 | 0.7999982 |

X | x | 800802 | 0.2002016 |

X | X | 3199177 | 0.7997984 |

This also has the correct distribution of eye colours.

eye_colour_distribution <- df %>% group_by(eye_colour) %>% tally() %>% mutate(frac = n / sum(n))

eye_colour | n | frac |
---|---|---|

blue | 200006 | 0.0400012 |

brown | 4799994 | 0.9599988 |

The genotype distribution is also correct, since \(2p(1-p) \approx\) 0.32.

genotype_distribution <- df %>% group_by(genotype) %>% tally() %>% mutate(frac = n / sum(n))

genotype | n | frac |
---|---|---|

heterozygote | 1600817 | 0.3201634 |

homozygote | 3399183 | 0.6798366 |

### Reproduction

Let’s also define a couple of functions to simulate reproduction within our population. The `pair`

function matches up random individuals from the first table with random individuals from the second.

pair <- function(df1, df2) { inner_join( df1 %>% select(-matches('\\.(x|y)$')) %>% select(matches('^(id|allele|genotype|eye)')) %>% ungroup() %>% sample_frac(size = 1) %>% mutate(row = row_number()), df2 %>% select(-matches('\\.(x|y)$')) %>% select(matches('^(id|allele|genotype|eye)')) %>% ungroup() %>% sample_frac(size = 1) %>% mutate(row = row_number()), by = 'row' ) %>% select(-row) %>% return() }

The `reproduce`

function then randomly generates a child from the paired individuals.

reproduce <- function(pairs, n=1) { pairs %>% crossing(child = 1:n) %>% mutate( # the variables x and y indicate the allele taken from parent x and y, respectively x = rbinom(n(), 1, 0.5) + 1, y = rbinom(n(), 1, 0.5) + 1, allele1 = if_else(x == 1, allele1.x, allele2.x), allele2 = if_else(y == 1, allele1.y, allele2.y), genotype = if_else(allele1 == allele2, 'homozygote', 'heterozygote'), eye_colour = if_else(allele1 == 'x' & allele2 == 'x', 'blue', 'brown'), id = paste(id.x, id.y, child, sep = '-') ) %>% return() }

The `kids`

table then represents the next generation from random mating within the entire population.

kids <- df %>% pair(df) %>% reproduce()

allele1.x | allele2.x | allele1.y | allele2.y | allele1 | allele2 | x | y |
---|---|---|---|---|---|---|---|

X | X | X | x | X | X | 2 | 1 |

X | x | X | X | x | X | 2 | 1 |

X | X | X | x | X | x | 1 | 2 |

X | X | X | X | X | X | 1 | 1 |

X | x | X | x | X | X | 1 | 1 |

x | X | X | X | x | X | 1 | 1 |

The parent attributes are contained in the `kids`

table, with the `.x`

suffix for one parent and `.y`

for the other.

## Part 1

We’ll use \(A\) to stand for the allele combination, e.g. \(XX\), or \(Xx = xX\), and \(E\) for eye colour. The subscripts \(i = 1, 2\) will be used for each of the two parents, and the absence of subscripts will indicate the variable for the child. We need to calculate the probability that the child is heterogenous given that they are brown-eyed with brown-eyed parents:

\[ \mathbb P (A = Xx \mid E, E_1, E_2 = B). \]

It will be easier to calculate this if we can rewrite it as a probability conditional only on \(A_\bullet\)-variables. First note that

\[ \begin{align} \mathbb P (A, A_1, A_2) &= \mathbb P (A \mid A_1, A_2) \mathbb P(A_1, A_2) \\ &= \mathbb P (A \mid A_1, A_2) \mathbb P(A_1) \mathbb P (A_2) \end{align} \]

using the chain rule and the assumption of random mating. Therefore,

\[ \begin{align} & P (A = Xx \mid E_\bullet = B) \\ &= \frac{\mathbb P (E_\bullet = B \mid A = Xx) \cdot \mathbb P (A = Xx)}{\mathbb P (E_\bullet = B)} \\ &= \frac{ \sum_{a_1, a_2} \mathbb P (E_\bullet = B \mid A = Xx, A_1 = a_1, A_2 = a_2) \cdot \mathbb P (A = Xx \mid A_1 = a_1, A_2 = a_2) \cdot \mathbb P (A_1 = a_1) \cdot \mathbb P (A_2 = a_2) }{ \sum_{a, a_1, a_2} \mathbb P (E_\bullet = B \mid A = a, A_1 = a_1, A_2 = a_2) \cdot \mathbb P (A = a \mid A_1 = a_1, A_2 = a_2) \cdot \mathbb P (A_1 = a_1) \cdot \mathbb P (A_2 = a_2) }, \end{align} \]

where the numerator is marginalised over possible values of \(A_1\) and \(A_2\), and the denominator additionally over \(A\).

The factors involving \(E_\bullet\) are either 1 or 0, depending only on whether the given combination of alleles can give rise to brown eyes or not, respectively. Moreover, \(\mathbb P (A_i = XX) = (1 – p)^2\) and \(\mathbb P (A_i = Xx) = 2p(1 – p)\), where the case \(A_i = xx\) is impossible conditional on everybody having brown eyes. The only non-trivial calculations now involve \(\mathbb P (A = a \mid A_1 = a_1, A_2 = a_2)\):

\[ \begin{align} \mathbb P (A = Xx \mid A_1 = Xx, A_2 = Xx) &= \frac{1}{2} \\ \mathbb P (A = Xx \mid A_1 = Xx, A_2 = XX) &= \frac{1}{2} \\ \mathbb P (A = Xx \mid A_1 = XX, A_2 = XX) &= 0 \\ \mathbb P (A = XX \mid A_1 = Xx, A_2 = Xx) &= \frac{1}{4} \\ \mathbb P (A = XX \mid A_1 = Xx, A_2 = XX) &= \frac{1}{2} \\ \mathbb P (A = XX \mid A_1 = XX, A_2 = XX) &= 1, \end{align} \]

as can be verified by inspection.

Now let’s plug in these values into the formula for the desired probability. The numerator is

\[ \begin{align} & P (A = Xx \mid E_\bullet = B) \\ &= \frac{ \frac{1}{2} \cdot (2p(1 – p))^2 + \frac{1}{2} \cdot 2 \cdot 2p(1 – p)(1 – p)^2 + 0 \cdot (1 – p)^4 }{ (\frac{1}{2} + \frac{1}{4}) \cdot 4p^2(1 – p)^2 + (\frac{1}{2} + \frac{1}{2}) \cdot 4p(1 – p)^3 + (0 + 1) \cdot (1 – p)^4 } \\ &= \frac{(1 – p)^2}{(1 – p)^2} \frac{ 2p^2 + 2p(1 – p) }{ 3p^2 + 4p(1 – p) + (1 – p)^2 } \\ &= \frac{ 2p^2 + 2p – 2p^2 }{ 3p^2 + 4p – 4p^2 + 1 + p^2 – 2p } \\ &= \frac{ 2p }{ 1 + 2p }, \end{align} \]

as required. This is approximately \(2p\) for small \(p\), and is approximatily \(\frac{1}{2}\) for large \(p\).

## Part 1 simulation

To condition on brown-eyed children from brown-eyed parents, we can just filter the `kids`

table. Such a child is called `judy`

in this exercise.

judy <- kids %>% filter( eye_colour.x == 'brown', eye_colour.y == 'brown', eye_colour == 'brown' ) judy_genotypes <- judy %>% group_by(genotype) %>% tally() %>% mutate(frac = n / sum(n))

genotype | n | frac |
---|---|---|

heterozygote | 1280529 | 0.2858268 |

homozygote | 3199559 | 0.7141732 |

This is very close to the theoretical value of \(\frac{2p}{1 + 2p}\approx\) 0.286.

## Part 2

Denote by \(E_{C_\bullet} = B\) the condition that all of Judy’s children have brown eyes, and by \(A^p = a\) the condition that Judy’s partner has allele combination \(a\). Then

\[ \begin{align} & \mathbb P (A = Xx \mid E_\bullet = B = E_{C_\bullet}, A^p = Xx) \\ &= \frac{ \mathbb P (A = Xx \mid E_\bullet = B, A^p = Xx) \cdot \mathbb P (E_{C_\bullet} = B \mid E_\bullet = B, A^p = Xx = A) }{ \mathbb P (E_{C_\bullet} = B \mid E_\bullet = B, A^p = Xx) } \\ &= \frac{ \mathbb P (A = Xx \mid E_\bullet = B) \cdot \mathbb P (E_{C_\bullet} = B \mid A^p = Xx = A) }{ \sum_a \mathbb P (E_{C_\bullet} = B \mid E_\bullet = B, A^p = Xx, A = a) \cdot \mathbb P (A = a \mid E_\bullet = B, A^p = Xx) } \\ &= \frac{ \mathbb P (A = Xx \mid E_\bullet = B) \cdot \mathbb P (E_{C_\bullet} = B \mid A^p = Xx = A) }{ \sum_a \mathbb P (E_{C_\bullet} = B \mid A^p = Xx, A = a) \cdot \mathbb P (A = a \mid E_\bullet = B) } \\ &= \frac{ \frac{2p}{1 + 2p} \cdot (\frac{3}{4})^n }{ \mathbb P (E_{C_\bullet} = B \mid A^p = Xx = A) \cdot \mathbb P (A = Xx \mid E_\bullet = B) + \mathbb P (E_{C_\bullet} = B \mid A^p = Xx, A = XX) \cdot \mathbb P (A = XX \mid E_\bullet = B) } \\ &= \frac{ \frac{2p}{1 + 2p} \cdot (\frac{3}{4})^n }{ \frac{2p}{1 + 2p} \cdot (\frac{3}{4})^n + \frac{1}{1 + 2p} } \\ &= \frac{2p \cdot (\frac{3}{4})^n}{2p \cdot (\frac{3}{4})^n + 1} \\ &= \frac{2p \cdot 3^n}{2p \cdot 3^n + 4^n} , \end{align} \]

where we have used conditional independence several times for the probability of the child’s alleles given the parents’ alleles. As \(n \rightarrow \infty\), this probability shrinks to 0.

## Part 2 simulation

To simulate part 2, we need to pair `judy`

with heterozygotes from the general population, then filter for those children with brown eyes.

judy_kids <- df %>% filter(genotype == 'heterozygote') %>% pair(judy) %>% reproduce() %>% ungroup() %>% filter(eye_colour == 'brown')

Amongst `judy_kids`

, Judy’s attributes have the `.y`

suffix. Given the above conditions, the probability of her possible genotypes are then:

judy_genotypes_posterior <- judy_kids %>% group_by(genotype.y) %>% tally() %>% mutate(frac = n / sum(n))

genotype.y | n | frac |
---|---|---|

heterozygote | 343962 | 0.2313911 |

homozygote | 1142534 | 0.7686089 |

This is close to the theoretical value of \(\frac{6p}{6p + 4}\approx\) 23.1%.

## Part 3

Let’s introduce some notation. Let \(A_g\) be the alleles of Judy’s first grandchild, the child of \(c\) with alleles \(A_c\) whose partner has alleles \(A_c^p\). We wish to calculate \(\mathbb P (A_g = xx \mid E_\bullet = B = E_{C_\bullet}, A^p = Xx)\).

First note that

$$ \[\begin{align} & \mathbb P (A_c = Xx \mid E_\bullet = E_{C_\bullet} = B, A^p = Xx, A = a) \\ &= \frac{ \mathbb P (E_{C_\bullet} = B \mid E_\bullet = B, A_c = Xx = A^p, A = a) \cdot \mathbb P (A_c = Xx \mid E_\bullet = B, A^p = Xx, A = a, A) }{ \mathbb P (E_{C_\bullet} = B \mid E_\bullet = B, A^p = Xx, A = a) } \\ &= \frac{ \mathbb P (E_{C_\bullet} = B \mid E_\bullet = B, A_c = Xx = A^p, A = a) \cdot 0.5 }{ \mathbb P (E_{C_\bullet} = B \mid E_\bullet = B, A^p = Xx, A = a) } \\ &= \begin{cases} \frac{\left(\frac{3}{4}\right)^{n-1} \cdot \frac{1}{2}}{\left(\frac{3}{4}\right)^n} &\text{if } A = Xx \\ 1 \cdot \frac{1}{2} / 1 &\text{othewrise} \end{cases} \\ &= \begin{cases} \frac{2}{3} &\text{if } A = Xx \\ \frac{1}{2} &\text{othewrise} \end{cases} . \end{align}\]$$

Thus,

$$ \[\begin{align} & \mathbb P (A_c = Xx \mid E_\bullet = B = E_{C_\bullet}, A^p = Xx) \\ &= \sum_a \mathbb P (A_c = Xx \mid E_\bullet = B = E_{C_\bullet}, A^p = Xx, A = a) \cdot \mathbb P (A = a \mid E_\bullet = B = E_{C_\bullet}, A^p = Xx) \\ &= \frac{2}{3} \cdot \frac{2p \cdot (\frac{3}{4})^n}{2p \cdot (\frac{3}{4})^n + 1} + \frac{1}{2} \cdot \frac{1}{2p \cdot (\frac{3}{4})^n + 1} \\ &= \frac{p\left( \frac{3}{4} \right)^{n-1} + 0.5}{2p \cdot \left(\frac{3}{4}\right)^n + 1} , \end{align}\]$$

which converges to \(\frac{1}{2}\) as \(n \rightarrow \infty\).

The probability that Judy’s first grandchild is a homozygote can then be calculated by marginalising over the allele combinations of the child and their partner:

$$ \[\begin{align} & \mathbb P (A_g = xx \mid E_\bullet = B = E_{C_\bullet}, A^p = Xx) \\ &= \sum_{a_c, a_c^p} \mathbb P (A_g = xx \mid E_\bullet = B = E_{C_\bullet}, A^p = Xx, A_c = a_c, A_c^p = a_c^p) \cdot \mathbb P (A_c = a_c, A_c^p = a_c^p \mid E_\bullet = B = E_{C_\bullet}, A^p = Xx) \\ &= \sum_{a_c, a_c^p} \mathbb P (A_g = xx \mid A_c = a_c, A_c^p = a_c^p) \cdot \mathbb P (A_c^p = a_c^p ) \cdot \mathbb P (A_c = a_c \mid E_\bullet = B = E_{C_\bullet}, A^p = Xx) \\ &= \sum_{a_c^p} \mathbb P (A_g = xx \mid A_c = Xx, A_c^p = a_c^p) \cdot \mathbb P (A_c^p = a_c^p ) \cdot \mathbb P (A_c = Xx \mid E_\bullet = B = E_{C_\bullet}, A^p = Xx) \\ &= \frac{p\left( \frac{3}{4} \right)^{n-1} + 0.5}{2p \cdot \left(\frac{3}{4}\right)^n + 1} \cdot \sum_{a_c^p} \mathbb P (A_g = xx \mid A_c = Xx, A_c^p = a_c^p) \cdot \mathbb P (A_c^p = a_c^p ) \\ &= \frac{p\left( \frac{3}{4} \right)^{n-1} + 0.5}{2p \cdot \left(\frac{3}{4}\right)^n + 1} \cdot \left( \mathbb P (A_g = xx \mid A_c = Xx, A_c^p = Xx) \cdot \mathbb P (A_c^p = Xx ) + \mathbb P (A_g = xx \mid A_c = Xx, A_c^p = xx) \cdot \mathbb P (A_c^p = xx ) \right) \\ &= \frac{p\left( \frac{3}{4} \right)^{n-1} + 0.5}{2p \cdot \left(\frac{3}{4}\right)^n + 1} \cdot \left( \frac{1}{4} \cdot 2p(1 – p) + \frac{1}{2} \cdot p^2 \right) \\ &= \frac{p\left( \frac{3}{4} \right)^{n-1} + 0.5}{2p \cdot \left(\frac{3}{4}\right)^n + 1} \cdot \frac{p}{2} , \end{align}\]$$

since

the grandchild can only be blue-eyed if the (brown-eyed) child has at least one x-allele, i.e. the child is \(Xx\);

\(A\) and \(A^p\) are independent by the random mating assumption; and

\(A_c\) and \(A_c^p\) are independent by the random mating assumption.

As \(n \rightarrow \infty\), this probability converges to \(\frac{p}{4}\).

## Part 3 simulation

To simulate Judy’s grandkids, we pair up `judy_kids`

with members of the general population.

judy_grandkids <- judy_kids %>% pair(df) %>% reproduce() judy_grandkids %>% summarise(mean(eye_colour == 'blue')) %>% pull() %>% signif(3) [1] 0.054

The above fraction of grandkids with blue eyes is consistent with the theoretical value of \(\frac{4p + 0.5}{6p + 4}\frac{p}{2} \approx\) 0.0538.

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**Brian Callander**.

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