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Fix documentation syntax bugs
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4 changed files with 151 additions and 163 deletions
7
.vscode/settings.json
vendored
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.vscode/settings.json
vendored
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@ -1,9 +1,10 @@
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{
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"cSpell.words": [
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"beefblup",
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"BLUP",
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"dairyblup",
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"EBVs",
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"EPDs",
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"autob",
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"beefblup",
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"dairyblup"
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]
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}
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}
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@ -15,6 +15,8 @@ makedocs(;
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),
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pages=[
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"Home" => "index.md",
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"How to Calculate EPDs" => "how-to-calculate-epds.md",
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"CLI Reference (WIP)" => "beefblup-cli.md"
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],
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)
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@ -1,7 +1,8 @@
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```@meta
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CurrentModule = BeefBLUP
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```
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beefblup Command Line Interface (CLI) documentation
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# beefblup Command Line Interface (CLI) documentation
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> _A work in progress_
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@ -15,7 +15,7 @@ Since I'm mostly talking to American beef producers, though, we'll stick with
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EPDs for most of this discussion.
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Expected Breeding Values (EBVs) (which are more often halved and published as
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Expected Progeny Differences [EPDs] or Predicted Transmitting Abilities [PTAs]
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Expected Progeny Differences (EPDs) or Predicted Transmitting Abilities (PTAs)
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in the United States) are generally found using Charles Henderson's linear
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mixed-model equations. Great, you say, what is that? I'm glad you asked...
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@ -29,15 +29,17 @@ P = G + E
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Where:
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- _P_ = phenotype
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- _G_ = genotype (think: breeding value)
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- _E_ = environmental factors
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- ``P`` = phenotype
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- ``G`` = genotype (think: breeding value)
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- ``E`` = environmental factors
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Now, we can't identify _every_ environmental factor that affects phenotype, but
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we can identify some of them, so let's substitute _E_ with some absolutes. A
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we can identify some of them, so let's substitute ``E`` with some absolutes. A
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good place to start is the "contemporary group" listings for the trait of
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interest in the [BIF Guidelines], though for the purposes of this example, I'm
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only going to consider sex, and birth year.
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interest in the
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[BIF Guidelines](https://beefimprovement.org/wp-content/uploads/2018/03/BIFGuidelinesFinal_updated0318.pdf),
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though for the purposes of this example, I'm only going to consider sex, and
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birth year.
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```math
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P = G + E_{year} + E_{sex}
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@ -45,61 +47,61 @@ P = G + E_{year} + E_{sex}
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Where:
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- _E<sub>n</sub>_ is the effect of _n_ on the phenotype
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- ``E_n`` is the effect of ``n`` on the phenotype
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Now let's say I want to find the weaning weight breeding value (_G_) of my
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Now let's say I want to find the weaning weight breeding value (``G``) of my
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favorite herd bull. I compile his stats, and then plug them into the equation
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and solve for G, right? Let's try that.
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and solve for ``G``, right? Let's try that.
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### Calf Records
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ID | Birth Year | Sex | YW (kg)
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-- | - | - | -
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1 | 1990 | Male | 354
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| ID | Birth Year | Sex | YW (kg) |
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|:-- | :--------- | :----- |:------- |
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| 1 | 1990 | Male | 354 |
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```math
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354 \textup{kg} &= G_1 + E_{1990} + E_{male}
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354 \ \textup{kg} = G_1 + E_{1990} + E_{male}
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```
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Hmm. I just realized I don't know any of those _E_ values. Come to think of it,
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Hmm. I just realized I don't know any of those ``E`` values. Come to think of it,
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I remember from math class that I will need as many equations as I have
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unknowns, so I will add equations for other animals that I have records for.
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### Calf Records
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ID | Birth Year | Sex | YW (kg)
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-- | - | - | -
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1 | 1990 | Male | 354
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2 | 1990 | Female | 251
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3 | 1991 | Male | 327
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4 | 1991 | Female | 328
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5 | 1991 | Male | 301
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6 | 1991 | Female | 270
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7 | 1992 | Male | 330
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| ID | Birth Year | Sex | YW (kg) |
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|:--- |:---------- |:------ |:------- |
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| 1 | 1990 | Male | 354 |
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| 2 | 1990 | Female | 251 |
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| 3 | 1991 | Male | 327 |
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| 4 | 1991 | Female | 328 |
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| 5 | 1991 | Male | 301 |
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| 6 | 1991 | Female | 270 |
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| 7 | 1992 | Male | 330 |
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```math
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\begin{aligned}
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251 \textup{kg} &= G_2 + E_{1990} + E_{female} \\
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327 \textup{kg} &= G_3 + E_{1991} + E_{male} \\
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328 \textup{kg} &= G_4 + E_{1991} + E_{female} \\
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301 \textup{kg} &= G_5 + E_{1991} + E_{male} \\
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270 \textup{kg} &= G_6 + E_{1991} + E_{female} \\
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330 \textup{kg} &= G_7 + E_{1992} + E_{male}
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251 \ \textup{kg} &= G_2 + E_{1990} + E_{female} \\
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327 \ \textup{kg} &= G_3 + E_{1991} + E_{male} \\
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328 \ \textup{kg} &= G_4 + E_{1991} + E_{female} \\
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301 \ \textup{kg} &= G_5 + E_{1991} + E_{male} \\
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270 \ \textup{kg} &= G_6 + E_{1991} + E_{female} \\
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330 \ \textup{kg} &= G_7 + E_{1992} + E_{male}
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\end{aligned}
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```
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Drat! Every animal I added brings more variables into the system than it
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eliminates! In fact, since each cow brings in _at least_ one term
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(_G<sub>n</sub>_), I will never be able to write enough equations to solve for
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_G_ numerically. I will have to use a different approach.
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(``G_n``), I will never be able to write enough equations to solve for
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``G`` numerically. I will have to use a different approach.
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## The statistical model: the setup
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Since I can never solve for _G_ directly, I will have to find some way to
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estimate it. I can switch to a statistical model and solve for _G_ that way. The
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Since I can never solve for ``G`` directly, I will have to find some way to
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estimate it. I can switch to a statistical model and solve for ``G`` that way. The
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caveat with a statistical model is that there will be some level of error, but
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so long as we know and can control the level of error, that will be better than
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not knowing _G_ at all.
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not knowing ``G`` at all.
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Since we're switching into a statistical space, we should also switch the
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variables we're using. I'll rewrite the first equation as
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@ -110,12 +112,12 @@ y = b + u + e
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Where:
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- _y_ = Phenotype
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- _b_ = Environment
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- _u_ = Genotype
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- _e_ = Error
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- ``y`` = Phenotype
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- ``b`` = Environment
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- ``u`` = Genotype
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- ``e`` = Error
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It's not as easy as simply substituting _b_ for every _E_ that we had above,
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It's not as easy as simply substituting ``b`` for every ``E`` that we had above,
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however. The reason for that is that we must make the assumption that
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environment is a **fixed effect** and that genotype is a **random effect**. I'll
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go over why that is later, but for now, understand that we need to transform the
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@ -126,20 +128,20 @@ We'll start with the environment terms.
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## The statistical model: environment as fixed effects
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To properly transform the equations, I will have to introduce
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_b<sub>mean</sub>_ terms in each animal's equation. This is part of the fixed
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``b_{mean}`` terms in each animal's equation. This is part of the fixed
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effect statistical assumption, and it will let us obtain a solution.
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Here are the transformed equations:
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```math
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\begin{aligned}
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354 \textup{kg} &= u_1 + b_{mean} + b_{1990} + b_{male} + e_1 \\
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251 \textup{kg} &= u_2 + b_{mean} + b_{1990} + b_{female} + e_2 \\
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327 \textup{kg} &= u_3 + b_{mean} + b_{1991} + b_{male} + e_3 \\
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328 \textup{kg} &= u_4 + b_{mean} + b_{1991} + b_{female} +e_4 \\
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301 \textup{kg} &= u_5 + b_{mean} + b_{1991} + b_{male} + e_5 \\
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270 \textup{kg} &= u_6 + b_{mean} + b_{1991} + b_{female} + e_6 \\
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330 \textup{kg} &= u_7 + b_{mean} + b_{1992} + b_{male} + e_7
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354 \ \textup{kg} &= u_1 + b_{mean} + b_{1990} + b_{male} + e_1 \\
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251 \ \textup{kg} &= u_2 + b_{mean} + b_{1990} + b_{female} + e_2 \\
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327 \ \textup{kg} &= u_3 + b_{mean} + b_{1991} + b_{male} + e_3 \\
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328 \ \textup{kg} &= u_4 + b_{mean} + b_{1991} + b_{female} +e_4 \\
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301 \ \textup{kg} &= u_5 + b_{mean} + b_{1991} + b_{male} + e_5 \\
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270 \ \textup{kg} &= u_6 + b_{mean} + b_{1991} + b_{female} + e_6 \\
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330 \ \textup{kg} &= u_7 + b_{mean} + b_{1992} + b_{male} + e_7
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\end{aligned}
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```
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@ -148,13 +150,13 @@ equations above to a single matrix equation that means the exact same thing.
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```math
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\begin{bmatrix}
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354 \textup{kg} \\
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251 \textup{kg} \\
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327 \textup{kg} \\
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328 \textup{kg} \\
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301 \textup{kg} \\
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270 \textup{kg} \\
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330 \textup{kg}
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354 \ \textup{kg} \\
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251 \ \textup{kg} \\
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327 \ \textup{kg} \\
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328 \ \textup{kg} \\
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301 \ \textup{kg} \\
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270 \ \textup{kg} \\
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330 \ \textup{kg}
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\end{bmatrix}
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=
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\begin{bmatrix}
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@ -200,9 +202,10 @@ e_7
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\end{bmatrix}
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```
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That's a nice equation, but now my hand is getting tired writing all those _b_
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terms over and over again, so I'm going to use [the dot product] to condense
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this down.
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That's a nice equation, but now my hand is getting tired writing all those ``b``
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terms over and over again, so I'm going to use
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[the dot product](https://www.khanacademy.org/math/precalculus/x9e81a4f98389efdf:matrices/x9e81a4f98389efdf:multiplying-matrices-by-matrices/v/matrix-multiplication-intro)
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to condense this down.
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```math
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\begin{bmatrix}
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@ -233,7 +236,6 @@ u_7
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1 & 0 & 1 & 0 & 1 & 0 \\
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1 & 0 & 0 & 1 & 1 & 0
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\end{bmatrix}
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+
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\begin{bmatrix}
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b_{mean} \\
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b_{1990} \\
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@ -257,26 +259,27 @@ e_7
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That matrix in the middle with all the zeros and ones is called the **incidence
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matrix**, and essentially reads like a table with each row corresponding to an
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animal, and each column corresponding to a fixed effect. For brevity, we'll just
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call it _**X**_, though. One indicates that the animal and effect go together,
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call it ``X``, though. One indicates that the animal and effect go together,
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and zero means they don't. For our record, we could write a table to go with
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_**X**_, and it would look like this:
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``X``, and it would look like this:
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Animal | mean | 1990 | 1991 | 1992 | male | female
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-- | - | - | - | - | - | -
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1 | yes | yes | no | no | yes | no
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2 | yes | yes | no | no | no | yes
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3 | yes | no | yes | no | yes | no
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4 | yes | no | yes | no | no | yes
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5 | yes | no | yes | no | yes | no
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6 | yes | no | yes | no | no | yes
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7 | yes | no | no | yes | yes | no
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| Animal | mean | 1990 | 1991 | 1992 | male | female |
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|:------ |:---- |:---- |:---- |:---- |:---- |:------ |
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| 1 | yes | yes | no | no | yes | no |
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| 2 | yes | yes | no | no | no | yes |
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| 3 | yes | no | yes | no | yes | no |
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| 4 | yes | no | yes | no | no | yes |
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| 5 | yes | no | yes | no | yes | no |
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| 6 | yes | no | yes | no | no | yes |
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| 7 | yes | no | no | yes | yes | no |
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Now that we have _**X**_, we have the ability to start making changes to allow
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us to solve for _u_. Immediately, we see that _**X**_ is **singular**, meaning
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Now that we have ``X``, we have the ability to start making changes to allow
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us to solve for ``u``. Immediately, we see that ``X`` is **singular**, meaning
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it can't be solved directly. We kind of already knew that, but now we can
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quantify it. We calculate the [rank of _**X**_], and find that there is only
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enough information contained in it to solve for 4 variables, which means we need
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to eliminate two columns.
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quantify it. We calculate the
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[rank of ``X``](https://math.stackexchange.com/a/2080577),
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and find that there is only enough information contained in it to solve for 4
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variables, which means we need to eliminate two columns.
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There are several ways to effectively eliminate fixed effects in this type of
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system, but one of the simplest and the most common methods is to declare a
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@ -288,27 +291,22 @@ last occuring form of each variable.
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### Base population
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<dl>
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<dt>Year</dt>
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<dd>1992</dd>
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<dt>Sex</dt>
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<dd>Male</dd>
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</dl>
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- **Year**: 1992
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- **Sex**: Female
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Now in order to use the base population, we simply drop the columns representing
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conformity with the traits in the base population from _**X**_. Our new
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conformity with the traits in the base population from ``X````. Our new
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equation looks like
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```math
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\begin{bmatrix}
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354 \textup{kg} \\
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251 \textup{kg} \\
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327 \textup{kg} \\
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328 \textup{kg} \\
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301 \textup{kg} \\
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270 \textup{kg} \\
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330 \textup{kg}
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354 \ \textup{kg} \\
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251 \ \textup{kg} \\
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327 \ \textup{kg} \\
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328 \ \textup{kg} \\
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301 \ \textup{kg} \\
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270 \ \textup{kg} \\
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330 \ \textup{kg}
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\end{bmatrix}
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=
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\begin{bmatrix}
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@ -350,24 +348,23 @@ e_7
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And the table for humans to understand:
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Animal | mean | 1990 | 1991 | female
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-- | - | - | - | -
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1 | yes | yes | no | no
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2 | yes | yes | no | yes
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3 | yes | no | yes | no
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4 | yes | no | yes | yes
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5 | yes | no | yes | no
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6 | yes | no | yes | yes
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7 | yes | no | no | no
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| Animal | mean | 1990 | 1991 | female |
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|:------ |:---- |:---- |:---- |:------ |
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| 1 | yes | yes | no | no |
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| 2 | yes | yes | no | yes |
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| 3 | yes | no | yes | no |
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| 4 | yes | no | yes | yes |
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| 5 | yes | no | yes | no |
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| 6 | yes | no | yes | yes |
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| 7 | yes | no | no | no |
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Even though each animal is said to participate in the mean, the result for the
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mean will now actually be the average of the base population. Math is weird
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sometimes.
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Double-checking, the rank of _**X**_ is still 4, so we can solve for the average
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Double-checking, the rank of ``X`` is still 4, so we can solve for the average
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of the base population, and the effect of being born in 1990, the effect of
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being born in 1991, and the effect of being female (although I think [Calvin
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already has an idea about that one]).
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being born in 1991, and the effect of being male.
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Whew! That was some transformation. We still haven't constrained this model
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enough to solve it, though. Now on to the genotype.
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@ -389,35 +386,35 @@ We'll need a pedigree for our animals:
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### Calf Records
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ID | Sire | Dam | Birth Year | Sex | YW (kg)
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-- | - | - | - | - | -
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1 | NA | NA | 1990 | Male | 354
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2 | NA | NA | 1990 | Female | 251
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3 | 1 | NA | 1991 | Male | 327
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4 | 1 | NA | 1991 | Female | 328
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5 | 1 | 2 | 1991 | Male | 301
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6 | NA | 2 | 1991 | Female | 270
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7 | NA | NA | 1992 | Male | 330
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| ID | Sire | Dam | Birth Year | Sex | YW (kg) |
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|:-- |:---- |:--- |:---------- |:------ |:------- |
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| 1 | NA | NA | 1990 | Male | 354 |
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| 2 | NA | NA | 1990 | Female | 251 |
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| 3 | 1 | NA | 1991 | Male | 327 |
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| 4 | 1 | NA | 1991 | Female | 328 |
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| 5 | 1 | 2 | 1991 | Male | 301 |
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| 6 | NA | 2 | 1991 | Female | 270 |
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| 7 | NA | NA | 1992 | Male | 330 |
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Now, because cows sexually reproduce, the genotype of one animal is halfway the
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same as that of either parent.<sup>[a](#a)</sup> It should go without saying
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that each animal's genotype is identical to that of itself. From this we can
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then find the numerical multiplier for any relative (grandparent = 1/4, full
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sibling = 1, half sibling = 1/2, etc.). Let's write those values down in a
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table.
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same as that of either parent (exception: inbreeding, see below). It should go
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without saying that each animal's genotype is identical to that of itself. From
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this we can then find the numerical multiplier for any relative (grandparent =
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1/4, full sibling = 1, half sibling = 1/2, etc.). Let's write those values down
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in a table.
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ID | 1 | 2 | 3 | 4 | 5 | 6 | 7
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-- | - | - | - | - | - | - | -
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1 | 1 | 0 | 1/2 | 1/2 | 1/2 | 0 | 0
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2 | 0 | 1 | 0 | 0 | 1/2 | 1/2 | 0
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3 | 1/2 | 0 | 1 | 1/4 | 1/4 | 0 | 0
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4 | 1/2 | 0 | 1/4 | 1 | 1/4 | 0 | 0
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||||
5 | 1/2 | 1/2 | 1/4 | 1/4 | 1 | 1/4 | 0
|
||||
6 | 0 | 1/2 | 0 | 0 | 1/4 | 1 | 0
|
||||
7 | 0 | 0 | 0 | 0 | 0 | 0 | 1
|
||||
| ID | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
|
||||
|:-- |:--- |:--- |:--- |:--- |:--- |:--- |:-- |
|
||||
| 1 | 1 | 0 | 1/2 | 1/2 | 1/2 | 0 | 0 |
|
||||
| 2 | 0 | 1 | 0 | 0 | 1/2 | 1/2 | 0 |
|
||||
| 3 | 1/2 | 0 | 1 | 1/4 | 1/4 | 0 | 0 |
|
||||
| 4 | 1/2 | 0 | 1/4 | 1 | 1/4 | 0 | 0 |
|
||||
| 5 | 1/2 | 1/2 | 1/4 | 1/4 | 1 | 1/4 | 0 |
|
||||
| 6 | 0 | 1/2 | 0 | 0 | 1/4 | 1 | 0 |
|
||||
| 7 | 0 | 0 | 0 | 0 | 0 | 0 | 1 |
|
||||
|
||||
Hmm. All those numbers look suspiciously like a matrix. Why don't I put them
|
||||
into a matrix called _**A**_?
|
||||
into a matrix called ``A``?
|
||||
|
||||
```math
|
||||
\begin{bmatrix}
|
||||
|
@ -431,25 +428,25 @@ into a matrix called _**A**_?
|
|||
\end{bmatrix}
|
||||
```
|
||||
|
||||
Now I'm going to take the matrix with all of the _u_ values, and call it
|
||||
_**μ**_. To quantify the idea of genetic relationship, I will then say that
|
||||
Now I'm going to take the matrix with all of the ``u`` values, and call it
|
||||
``μ``. To quantify the idea of genetic relationship, I will then say that
|
||||
|
||||
```math
|
||||
\textup{var}(μ) = \mathbf{A}σ_μ^2
|
||||
\textup{var}(μ) = A σ_μ^2
|
||||
```
|
||||
|
||||
Where:
|
||||
|
||||
- _**A**_ = the relationship matrix defined above
|
||||
- _σ<sub>μ</sub><sup>2</sup>_ = the standard deviation of all the genotypes
|
||||
- ``A`` = the relationship matrix defined above
|
||||
- ``σ_μ^2`` = the standard deviation of all the genotypes
|
||||
|
||||
To fully constrain the system, I have to make two more assumptions: 1) that the
|
||||
error term in each animal's equation is independent from all other error terms,
|
||||
and 2) that the error term for each animal is independent from the value of the
|
||||
genotype. I will call the matrix holding the _e_ values _**ε**_ and then say
|
||||
genotype. I will call the matrix holding the ``e`` values ``ε`` and then say
|
||||
|
||||
```math
|
||||
\textup{var}(ϵ) = \mathbf{I}σ_ϵ^2
|
||||
\textup{var}(ϵ) = I σ_ϵ^2
|
||||
```
|
||||
|
||||
```math
|
||||
|
@ -458,9 +455,6 @@ genotype. I will call the matrix holding the _e_ values _**ε**_ and then say
|
|||
|
||||
Substituting in the matrix names, our equation now looks like
|
||||
|
||||
![\Large Figure 25. Nearly complete mixed-model
|
||||
equation](https://latex.codecogs.com/svg.latex?%5Cinline%20%5Cbegin%7Bbmatrix%7D%20354%5Ctextup%7B%20kg%7D%5C%5C%20251%5Ctextup%7B%20kg%7D%5C%5C%20327%5Ctextup%7B%20kg%7D%5C%5C%20328%5Ctextup%7B%20kg%7D%5C%5C%20301%5Ctextup%7B%20kg%7D%5C%5C%20270%5Ctextup%7B%20kg%7D%5C%5C%20330%5Ctextup%7B%20kg%7D%20%5Cend%7Bbmatrix%7D%20%3D%20%5Cmu%20+%20X%20%5Cbegin%7Bbmatrix%7D%20b_%7Bmean%7D%5C%5C%20b_%7B1990%7D%5C%5C%20b_%7B1991%7D%5C%5C%20b_%7Bfemale%7D%5C%5C%20%5Cend%7Bbmatrix%7D%20+%20%5Cvarepsilon)
|
||||
|
||||
```math
|
||||
\begin{bmatrix}
|
||||
354 \textup{kg} \\
|
||||
|
@ -484,14 +478,14 @@ b_{male} \\
|
|||
We are going to make three changes to this equation before we are ready to solve
|
||||
it, but they are cosmetic details for this example.
|
||||
|
||||
1. Call the matrix on the left side of the equation _**Y**_ (sometimes it's
|
||||
1. Call the matrix on the left side of the equation ``Y`` (sometimes it's
|
||||
called the **matrix of observations**)
|
||||
2. Multiply _**μ**_ by an identity matrix called _**Z**_. Multiplying by the
|
||||
2. Multiply ``μ`` by an identity matrix called ``Z``. Multiplying by the
|
||||
identity matrix is the matrix form of multiplying by one, so nothing changes,
|
||||
but if we later want to find one animal's genetic effect on another animal's
|
||||
performance (e.g. a **maternal effects model**), we can alter _**Z**_ to
|
||||
allow that/
|
||||
3. Call the matrix with all the _b_ values _**β**_.
|
||||
performance (e.g. a **maternal effects model**), we can alter ``Z`` to
|
||||
allow that
|
||||
3. Call the matrix with all the ``b`` values ``β``.
|
||||
|
||||
With all these changes, we now have
|
||||
|
||||
|
@ -506,9 +500,6 @@ Charles Henderson used to first predict breeding values of livestock.
|
|||
|
||||
Henderson proved that the mixed-model equation can be solved by the following:
|
||||
|
||||
![\Large Figure 27. Solution to mixed-model
|
||||
equation](https://latex.codecogs.com/svg.latex?%5Cinline%20%5Cbegin%7Bbmatrix%7D%20%5Chat%7B%5Cbeta%7D%5C%5C%20%5Chat%7B%5Cmu%7D%20%5Cend%7Bbmatrix%7D%20%3D%5Cbegin%7Bbmatrix%7D%20X%27X%26X%27Z%5C%5C%20Z%27X%26Z%27Z+A%5E%7B-1%7D%5Clambda%20%5Cend%7Bbmatrix%7D%5E%7B-1%7D%20%5Cbegin%7Bbmatrix%7D%20X%27Y%5C%5C%20Z%27Y%20%5Cend%7Bbmatrix%7D)
|
||||
|
||||
```math
|
||||
\begin{bmatrix}
|
||||
\hat{β} \\
|
||||
|
@ -517,7 +508,7 @@ equation](https://latex.codecogs.com/svg.latex?%5Cinline%20%5Cbegin%7Bbmatrix%7D
|
|||
=
|
||||
\begin{bmatrix}
|
||||
X'X & X'Z \\
|
||||
Z'X & Z'Z+A^{-1}\lambda
|
||||
Z'X & Z'Z+A^{-1}λ
|
||||
\end{bmatrix}^{-1}
|
||||
\begin{bmatrix}
|
||||
X'Y \\
|
||||
|
@ -529,30 +520,23 @@ Where
|
|||
|
||||
- The variables with hats are the statistical estimates of their mixed-model
|
||||
counterparts
|
||||
- The predicted value of _**μ**_ is called the _Best Linear Unbiased
|
||||
- The predicted value of ``μ`` is called the _Best Linear Unbiased
|
||||
Predictor_ or _BLUP_
|
||||
- The estimated value of _**β**_ is called the _Best Linear Unbiased Estimate_
|
||||
- The estimated value of ``β`` is called the _Best Linear Unbiased Estimate_
|
||||
or _BLUE_
|
||||
- ' is the transpose operator
|
||||
- λ is a single real number that is a function of the heritability for the trait
|
||||
being predicted. It can be left out in many cases (λ = 1).
|
||||
- λ = (1-h<sup>2</sup>)/h<sup>2</sup>
|
||||
- ``λ`` is a single real number that is a function of the heritability for the trait
|
||||
being predicted. It can be left out in many cases (``λ = 1``).
|
||||
- ``λ = \frac{1-h^2}{h^2}``
|
||||
|
||||
What happened to
|
||||
|
||||
## Footnotes
|
||||
|
||||
### a
|
||||
### Exception
|
||||
|
||||
An animal **can** share its genome with itself by a factor of more than one:
|
||||
that's called inbreeding! We can account for this, and `beefblup` does as it
|
||||
calculates _**A**_. This is an area that actually merits a good deal of study:
|
||||
calculates ``A``. This is an area that actually merits a good deal of study:
|
||||
see chapter 2 of _Linear Models for the Prediction of Animal Breeding Values_ by
|
||||
Raphael A. Mrode (ISBN 978 1 78064 391 5).
|
||||
|
||||
[BIF Guidelines]:
|
||||
https://beefimprovement.org/wp-content/uploads/2018/03/BIFGuidelinesFinal_updated0318.pdf
|
||||
[The Dot Product]:
|
||||
https://www.khanacademy.org/math/precalculus/x9e81a4f98389efdf:matrices/x9e81a4f98389efdf:multiplying-matrices-by-matrices/v/matrix-multiplication-intro
|
||||
[rank of _**X**_]: https://math.stackexchange.com/a/2080577 [Calvin already has
|
||||
an idea about that one]: https://www.gocomics.com/calvinandhobbes/1992/12/02
|
||||
|
|
Loading…
Reference in a new issue