# Using R to explore probability distributions

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## Goals

This lab represents an introduction to the R computing language, which will be useful not only for today’s lab, which explores the most common probability distributions used in Bayesian phylogenetics, but also for Phylogenetics software written as R extensions (i.e. APE).

## Objectives

After completing this lab, you will be able to…

- plot density functions corresponding to common probability distributions used as prior distributions in Bayesian phylogenetics
- use R’s “p*” functions to compute the cumulative probability distribution for common probability distributions
- use R’s “q*” functions to compute quantiles for common probability distributions
- use R’s “r*” functions to draw random samples from common probability distributions
- use R’s “curve” function to plot a density curve
- calculate the correct $\mu$ and $\sigma$ parameters for a Lognormal prior distribution given the desired mean and standard deviation
- explain how an Exponential distribution is a special case of a Gamma distribution
- explain how a Beta distribution is a special case of a Dirichlet distribution
- list 3 probability distributions that would be appropriate for an edge length or kappa parameter
- list 1 probability distribution that would be appropriate for the proportion of invariable sites parameter
- list 1 probability distribution that would be appropriate for equilibrium nucleotide relative frequencies
- list 1 probability distribution that would be appropriate for the GTR exchangeability parameters

## Copy the template

Copy the text below into a text file to record you answers to questions posed in this lab.

```
1. What is the value of beta if alpha is known and the mean must be 1.0?
answer
2. Why is the scale parameter beta never estimated when using the Gamma distribution to model among-site rate heterogeneity?
answer
3. What is the value of the scale parameter beta if mean = 2 and variance = 4?
answer
4. What is the value of the shape parameter alpha if mean = 2 and variance = 4 (hint: use the value of beta you just calculated)?
answer
5. Compute the sample mean of the values you generated from a Gamma(0.5, 2) distribution, does it appear that the mean is correct?
answer
6. How close is the sample mean to the expected mean for a Gamma(0.5, 2) with sample size 10000?
answer
7. How close is the sample mean to the expected mean for a Gamma(0.5, 2) with sample size 1000000?
answer
8. Why are the histograms for x and y so different? (hint: use help(rgamma) to read about the rgamma function)
answer
9. What are the true means of x and y?
answer
10. Calculate the Gamma probability density for the value 0.99 assuming shape alpha=10 and scale beta = 0.1 (use the formula labeled "Density function" in the table above)
answer
11. Use the `pgamma` command to compute the probability that a value drawn from a Gamma(shape=2,scale=0.5) distribution would fall between 0.5 and 4.5? Hint: you will need to either invoke the command twice or provide it with a vector containing two values, then do some subtraction with the results.
answer
12. What value $x_r$ produces an interval (0,$x_r$) that accounts for 95% of the values drawn from a Gamma(shape=2,scale=0.5) distribution?
answer
13. What pair of values ($x_l$,$x_r$) produces an interval that accounts for the middle 95% of the values drawn from a Gamma(shape=2, scale=0.5) distribution (by middle, I mean that 2.5% of the distribution should be less than $x_l$ and 2.5% greater than $x_r$)?
answer
14. Is the height of the density at x=0 related to the rate parameter?
answer
15. Which of these two densities has the larger mean? (Try to answer this question first by looking at the plot, then compute the numbers to check your intuition.)
answer
16. Which of these two densities has the larger variance? (Again, answer the question first by looking at the plot.)
answer
17. What fraction of the Exponential(10) distribution is less than 1? (Hint: use the `pexp` command.)
answer
18. What fraction of the Exponential(0.1) distribution is less than 1?
answer
19. Fill in the blank: 95% of the Exponential(0.1) distribution is less than _____. (Hint: use the `qexp` command.)
answer
20. What is the sample mean of your 1 million Lognormal(1,2) random variates?
answer
21. What is the sample standard deviation of your 1 million Lognormal(1,2) random variates?
answer
22. What is the sample mean of the _logarithm_ of your 1 million Lognormal(1,2) random variates?
answer
23. What is the sample standard deviation of the _logarithm_ of your 1 million Lognormal(1,2) random variates?
answer
24. What is the median of a Lognormal(1,2) distribution? (Hint: look up Lognormal distribution in Wikipedia)
answer
25. What is the sample median of the 1 million Lognormal(1,2) variates that you generated at the beginning of this section?
answer
26. What would mu and sigma be if you wanted the mean to be 0.1 and the standard deviation 0.5?
answer
27. Which distribution -- Beta(1,1) or Beta(0.1, 0.1) -- would be most appropriate as a prior distribution for a parameter representing the chances of a coin coming up heads on any given flip if you believed strongly that the world was full of trick coins (two-headed or two-tailed coins)?
answer
28. What is the effect of increasing the magnitude of all four parameters from 1 to 100 to 1 million?
answer
29. Which of these three Dirichlet distributions comes closest to the assumptions made by the JC69 model?
answer
```

## Installing R and RStudio on your laptop

R computing language is one of two general-purpose programming languages that we will use this semester (the other being Python). While R and Python share many similarities, R excels in tasks that involve statistics and creating plots. While a spreadsheet program like Excel can create charts, it is often easier to create the plot you need in R. The purpose of this lab is to introduce you to R by showing you how to make various plots related to probability distributions. This will also enable you to learn about several important probability distributions that serve as prior distributions in Bayesian phylogenetics. This introduction to R will serve us well later in the course when we make use of phylogenetic packages that are written in the R language.

### Installing R

Install R (if you haven’t already) by going to the R project website and following the download instructions for your platform. **You will be doing this lab on your own laptop, not the cluster**, because much of this lab involves plots, which are most easily generated and displayed on your own laptop. I have installed the latest version of R (which, at this moment, is version 4.3.2, nicknamed “Eye Holes”), but these instructions should work with slightly older versions of R.

### Installing RStudio

We will use R via RStudio, which will allows you to manipulate R source code and see the results within the same application window. RStudio is a separate app that you can download from the RStudio downloads page.

## Creating plots using R

**Prompt:** To use R, type commands into the console window at the R prompt (`>`

). Try the following commands for starters. Note that I’ve started each line with the `>`

character. That `>`

character is the R command prompt: you should **type in only what follows the > character**.

**Comments:** Also, you need not type in the `#`

character or any characters that follow `#`

on the same line. These represent comments in R, and will be ignored completely by the R interpreter.

```
> help()
```

You can get help on a particular topic like this:

```
> help(plot)
```

To search for all help topics that mention the word “gamma”:

```
> help.search("gamma")
```

That last one will generate a confusing list of topics, some of which do not appear to have anything to do with the word “gamma”! Nevertheless, this search feature is good to know about if you are desperate for information about some topic for which the basic help command turns up nothing. (Google is also an excellent help search tool: start your Google search with a capital letter R followed by a keyword.)

## Creating a histogram

Drawing 1000 standard normal variates and creating a histogram is as easy as:

```
> y <- rnorm(1000, 0, 1)
```

In this case, y will be a list of 1000 random normal (mean 0, std. dev. 1) variates (a **variate** is just one draw from the distribution). Note that `<-`

means “assign to”. You can also use an equal sign here, but the `<-`

is preferred, saving `=`

for specifying values for arguments in function calls (you will see examples of this below).

To view the values in the variable `y`

, just type the name of the variable:

```
> y
```

To create a histogram of the values stored in `y`

:

```
> hist(y)
```

Important time-saving hint: you can save a lot of typing by using your up-arrow to recall previously-issued commands. You can then modify the previously-issued command and hit the Enter key to submit the modified version.

## Creating a plot

Let’s begin by making a simple scatterplot using these data:

```
x y
1 1
2 1
1 3
4 5
2 4
3 5
4 7
5 8
```

In the R console window, enter the data and plot it as follows (remember that you should **not** enter the initial `>`

on each line; that’s the R prompt):

```
> x <- c(1,2,1,4,2,3,4,5)
> y <- c(1,1,3,5,4,5,7,8)
> plot(x,y)
```

The `c`

in `c(...)`

means “combine”, so the `c`

function’s purpose is to combine its arguments. Thus, both `x`

and `y`

are not single values, but instead each is a collection of 8 numbers. Collections of numbers like this are known as **vectors** in mathematics, or **arrays** or **lists** in many computer programming languages.

## Modifying a plot

R does a fairly nice job of selecting defaults for plots, but I find myself wanting to tweak things. This section explains how to modify various aspects of plots.

### Changing the plotting symbol

Use the `pch`

parameter to change the symbol R uses to plot points:

```
> plot(x, y, pch=19)
```

Common values for `pch`

include 19 (solid dot), 20 (bullet), 21 (circle), 22 (square), 23 (diamond), 24 (up-pointing triangle), and 25 (down-pointing triangle). The stroke and fill color for symbols 21-25 can be changed using the `col`

and `bg`

parameters. For example, to use red squares with yellow fill, try this:

```
> plot(x, y, pch=22, col="red", bg="yellow")
```

See the help topic on `points`

for more info:

```
> help(points)
```

### Plot type

The default plot type is `p`

for points, but other plot types can be used (use the command `help(plot)`

for details). Let’s use `"l"`

(that’s a lower-case L) type to connect our points with lines:

```
> plot(x, y, type="l")
```

Clearly, this is not something that you would want to do with these data, but we will use `type="l"`

extensively later in this tutorial. To use both points and lines, set the type to both (`b`

):

```
> plot(x, y, type="b")
```

### Line width

To make the line connecting the points thicker, use the `lwd`

option (`lwd=1`

is the default line width):

```
> plot(x,y,type="l",lwd=3)
```

### Plot and axis labels

The `xlab`

, `ylab`

, and `main`

parameters can be used to change the x-axis label, the y-axis label and the main plot title, respectively:

```
> plot(x, y, type="l", xlab="width", ylab="height", main="Leaf width vs. height")
```

### Losing the box

The `bty`

parameter can be set to `"n"`

to remove the box around the plot:

```
> plot(x, y, type="l", bty="n")
```

### Losing an axis

The `xaxt`

and `yaxt`

parameters can be set to `"n"`

to remove the x- and y-axis, respectively, from the plot. Here for example is a plot with no y-axis (note that I’ve also set the y-axis label to an empty string with `ylab=""`

and removed the box around the plot with `bty`

):

```
> plot(x, y, type="l", yaxt="n",ylab="",bty="n")
```

### Changing axis labeling

It is often desirable to change the values used on the x- or y-axis. You can use the `ylim`

parameter to make the y-axis extend from 0 to 10 (instead of the default, which is 1 to 8 in this case because that is the observed range of y values):

```
> plot(x, y, type="l", ylim=c(0,10))
```

Note that there is a little overhang at both ends on the y-axis (the 0 tick mark is slightly above the bottom of the plot and the tick mark for 10 is slightly below the top of the plot). This is because, by default, the y-axis style parameter `yaxs`

is set to `"r"`

, which causes R to extend the range you defined (using `ylim`

) by 4% at each end. You can set `yaxs="i"`

to make R strictly honor the limits you set:

```
> plot(x, y, type="l", ylim=c(0,10), yaxs="i")
```

You can change the x-axis in the same way using `xlim`

and `xaxs`

instead of `ylim`

and `yaxs`

.

## Data frames

The `c(...)`

notation used above is fine for small examples, but if you have a lot of data, you will prefer to read in the data from a file. R uses the concept of a data frame to store a collection of data. For example, save the following data in a text file on your hard drive named *myfile.txt*. Be sure to use a plain text editor such as BBEdit (Mac) or Notepad++ (Windows), **not** Word (which will add a lot of formatting information that R will not understand):

```
one two three four
A 0 1 2 3
B 4 5 6 7
C 8 9 10 11
D 12 13 14 15
```

Note that the first row has only 4 elements whereas all others have 5. If the first row contains one fewer elements than the other rows, then R assumes that the first row contains column names.

Use this command to read these data into a data frame:

```
> d <- read.table("myfile.txt")
```

If you get an error, it is probably because the current working directory used by R is not the directory in which you stored your file. You can ask R to tell you the current working directory:

```
> getwd()
```

and you can set the current working directory as described below. Assuming that you are using Windows and *myfile.txt* is stored in *C:\phylogenetics\rlab*:

```
> setwd("C:\\phylogenetics\\rlab") # Windows version
```

Note that (for rather esoteric reasons) the single backslash characters in the actual path must be replaced by double backslashes in R commands.

If instead you were using a Mac and *myfile.txt* was stored on your desktop, then you would do this:

```
> setwd("~/Desktop") # Mac version
```

To ensure that your data were read correctly, simply type `d`

to see the contents of the variable you created to hold the data:

```
> d
one two three four
A 0 1 2 3
B 4 5 6 7
C 8 9 10 11
D 12 13 14 15
```

To extract one column of your stored data frame, use a `$`

followed by the column name:

```
> col1 <- d$one
> col1
[1] 0 4 8 12
```

The `[1]`

that starts this line tells you that this is the first line of output; it should not be confused with the actual output, which comprises the numbers 0, 4, 8, and 12.

You can also extract single values, although this is not commonly needed:

```
> x <- d[2,3]
> x
[1] 6
```

**Important!** Note that, in R, indices begin with 1. This differs from Python and almost every other programming language, which start counting at 0.

Finally, if you have not named your columns, you can still get, for example, the third column as follows:

```
> d[,3]
[1] 2 6 10 14
```

Note that I’ve left the row part of the specification empty to tell R that I want to retrieve data from all rows for column 3.

To get the data for the third row, just leave the column specification empty:

```
> d[3,]
one two three four
C 8 9 10 11
```

## Exploring probability distributions using R

R’s plotting abilities and its built-in knowledge of many probability distributions used in statistics make it useful for exploring these distributions. The distributions described here are used extensively as prior distributions for parameters in Bayesian phylogenetic analyses, so it is worthwhile to familiarize yourself with their basic properties. With R we can examine them graphically.

### Gamma distribution

The Gamma distribution has two parameters that govern its shape (parameter $\alpha$) and scale (parameter $\beta$). Gamma distributions range from 0.0 to $\infty$, so they provide excellent prior distributions for parameters such as relative rates (e.g. kappa, GTR relative rates, omega) and branch lengths that can be arbitrarily large but not negative.

Here are some basic facts about Gamma distributions:

Shape parameter | $\alpha$ |

Scale parameter | $\beta$ |

Mean | $\alpha \; \beta$ |

Variance | $\alpha \; \beta^2$ |

Density function | $p(x) = \frac{x^{\alpha - 1} e^{-x/\beta}}{\beta^{\alpha} \Gamma(\alpha)}$ |

Note that $\Gamma(\alpha)$ in the density function refers to the gamma *function* (which is distinct from the Gamma *distribution*). The gamma function is easy to calculate when its argument is a positive integer:

$\Gamma(x) = (x-1)!$

For example,

$\Gamma(10) = 9! = (9)(8)(7)(6)(5)(4)(3)(2)(1) = 362880$

(you will use this fact a few minutes from now). When, however, the argument is an arbitrary positive real number (which is the usual situation for $\alpha$), then computing it gets more complicated. Fortunately, R handles details like this for us!

What is the value of $\beta$ if $\alpha$ is known and the mean must be 1.0?

Why is the scale parameter $\beta$ never estimated when using the Gamma distribution to model among-site rate heterogeneity?

What is the value of the scale parameter $\beta$ if mean = 2 and variance = 4?

What is the value of the shape parameter $\alpha$ if mean = 2 and variance = 4 (hint: use the value of $\beta$ you just calculated)?

### Creating a Histogram

Type the following command into the R console to generate 1000 random numbers (variates) from a Gamma distribution with shape=0.5 and scale=2.0 and store them in the variable `x`

:

```
> x <- rgamma(1000, shape=0.5, scale=2.0)
```

The `r`

in `rgamma`

stands for “random”. There is a corresponding “r” command for all the common probability distributions (e.g. rbeta, rbinom, rchisq, rexp, rnorm, runif) and in each case it allows you to draw a sample of values from that distribution (e.g. the beta, binomial, chi-square, exponential, normal and uniform distributions, respectively).

It may appear that nothing has happened, but you can view the 1000 values stored in the variable `x`

by typing `x`

at the prompt.

You can get the sample mean, sample variance, and sample standard deviation using:

```
> mean(x)
> var(x)
> sd(x)
```

You can get more summary statistics using:

```
> summary(x)
```

Creating a basic histogram from these 1000 values is easy:

```
> hist(x)
```

Now refine the histogram by asking R to give you 50 bars:

```
> hist(x, breaks=50)
```

Note that breaks represents a suggestion only. R will often be passive-aggressive and not give you exactly as many bars as you want.

Create a histogram with approximately 40 bars from 10000 variates from a Gamma(shape=0.5, scale=2) distribution.

Compute the sample mean of the values you generated from a Gamma(0.5, 2) distribution, does it appear that the mean is correct?

Create a histogram with 30 bars from 10000 variates from a Gamma distribution in which the mean is 2 and the variance is 1 Assuming you have stored the 10000 variates in the variable `x`

, type `summary(x)`

to get the sample mean, median, etc.

How close is the sample mean to the expected mean 2 with sample size 10000?

Draw another sample with 1 million random variates and summarize.

How close is the sample mean to the expected mean 2 with sample size 1000000?

The results above make sense: the variance of one Gamma random variate is $\alpha \beta^2$, but the variance of the *mean* of $n$ Gamma random variates is $\alpha \beta^2/n$. The variance of the mean is thus inversely proportional to the sample size, so the mean of 1 million Gamma variates was much closer to the expected value 2 than the mean of only 10000 variates.

### Don’t confuse rate with scale!

Only 2 parameters are required to fully specify a Gamma distribution, but the second parameter can be either the scale (what we’ve been using) or the rate (which is the inverse of the scale parameter). Try this experiment, which will plot two histograms on the same plot (by using add=TRUE for the second invocation of hist):

```
> x <- rgamma(1000000, shape=5, scale=5)
> y <- rgamma(1000000, 5, 5)
> hist(x, breaks=50, main="", xlab="", xlim=c(0,80))
> hist(y, col="navy", border=NA, add=TRUE)
```

Why are the histograms for x and y so different? (hint: use help(rgamma) to read about the rgamma function)

What are the true means of x and y?

### Plotting the density function

The “d” commands (e.g. dgamma, dbeta, dexp, dnorm, dunif) can be used to compute the probability density function for a series of values. In this part of the tutorial, you will use the dgamma function to compute the Gamma density function for all 501 values in the series 0.0, 0.01, 0.02, …, 5.0. First, generate the values in the series and store them in a variable named x:

```
> x <- seq(0.0, 5.0, 0.01)
```

The `seq`

function generates a series of values starting at the first supplied value (0.0 in this case), stopping at the second supplied value (5.0 in this case), with spacing between values in the sequence determined by the third supplied value (0.01 in this case). To see if 501 values were generated, type

```
> length(x)
```

You can list all the values in `x`

by simply typing `x`

at the console.

R makes it easy to compute a quantity for every number in a list. Here is how to compute the Gamma density for each value stored in the variable `x`

:

```
> y <- dgamma(x, shape=10, scale=0.1)
```

Before plotting the density function, you should convince yourself that R is computing the density correctly by calculating one value yourself by hand. First print out the 100th `x`

value as follows:

```
> x[100]
```

The value displayed should be 0.99.

Calculate the Gamma probability density for the value 0.99 assuming shape $\alpha=10$ and scale $\beta = 0.1$ (use the formula labeled “Density function” in the table above)

You can either use a calculator, or use Python or R to do the calculation (just don’t use the R function dgamma to do the calculation). Compare your value to the value computed by R:

```
> y[100]
```

Now plot the density function as follows:

```
> plot(x,y)
```

This command says to plot all possible pairs of points, where one value is taken from `x`

and the other from `y`

. There are 501 values in both `x`

and `y`

, so 501 points will be plotted. This plot looks pretty ugly, with lots of overlapping points. Instead, let’s connect the points with a line but not print all 501 points themselves. You can do this by adding `type="l"`

to your plot command:

```
> plot(x,y,type="l")
```

The type is specified to be a lower case L character here (L for line). For other plot types, see the online help by typing:

```
> help(plot)
```

I nearly always want the lines in my plots to be thicker. Here is how to triple the thickness of the plotted line using the `lwd`

(line width) setting:

```
> plot(x,y,type="l", lwd=3)
```

Finally, make the plotted line blue using the `col`

(color) setting:

```
> plot(x,y,type="l", lwd=3, col="blue")
```

Here is a very helpful PDF file showing many named colors.

Also, there are many other settings that affect plots. Get help on the `par`

command to see these (just about any setting you can specify for `par`

also works for `plot`

):

```
> help(par)
```

### Using the distribution function

We’ve seen the “d” and “r” commands (for density function and random number generation), but there are two additional standard commands defined for probability distributions. The “p” command computes the cumulative distribution function for a distribution. For the Gamma distribution, you can use the `pgamma`

command to compute the integral of the density function up to a specified value. First, plot a Gamma density with shape 2 and scale 0.5:

```
> x <- seq(0.0, 5.0, 0.01)
> y <- dgamma(x, shape=2.0, scale=0.5)
> plot(x, y, type="l")
```

The total area under this curve is 1.0. What is your guess as to the area of the curve from 0 to 1?
Compute the area under the density from 0 to 1 as follows using the `pgamma`

command:

```
> pgamma(1, shape=2, scale=0.5)
[1] 0.5939942
```

Thus, nearly 60% of random draws from a Gamma(2,0.5) distribution would be less than 1.0. You can compute several cumulative probabilities by supplying a vector to the pgamma command. Suppose you wanted to know the cumulative probability for each of the values on the x-axis (i.e. 1, 2, 3, 4 and 5):

```
> pgamma(c(1,2,3,4,5), shape=2, scale=0.5)
[1] 0.5939942 0.9084218 0.9826487 0.9969808 0.9995006
```

Note that 5 values are produced by this command, corresponding to the 5 values supplied in the vector `c(1,2,3,4,5)`

. The last one (0.9995006) means that 99.95% of random draws from this distribution would be less than 5.0.

Use the

`pgamma`

command to compute the probability that a value drawn from a Gamma(shape=2,scale=0.5) distribution would fall between 0.5 and 4.5? Hint: you will need to either invoke the command twice or provide it with a vector containing two values, then do some subtraction with the results.

### Quantiles

A quantile is a point along the x-axis of a density plot that corresponds to a particular cumulative probability. For example, the 60% quantile corresponds to a cumulative probability of 0.6. If you computed the area under the density curve up to some point `x`

and the cumulative probability was 0.6, then that value of `x`

is the 60% quantile.

For example, in the previous section, the value 1 is very close to the 60% quantile. Suppose we wanted to divide up a Gamma(2, 0.5) distribution into four equal-area pieces, and needed to find the points along the x-axis corresponding to the boundaries between these pieces (does this problem sound familiar?). We could do this in R as follows:

```
> qgamma(c(0.25, 0.5, 0.75, 1), shape=2, scale=0.5)
[1] 0.4806394 0.8391735 1.3463173
```

Note that we have supplied 4 values to the `qgamma`

function via the vector `c(0.25, 0.5, 0.75, 1)`

. We did not actually need to supply the fourth value (1.0) because we know that the 100% quantile is equal to $\infty$. In case you haven’t guessed, these would be the boundaries used if we applied discrete Gamma rate heterogeneity to a model, specifying 4 categories and shape=2.0.

You could check this in PAUP* using PAUP*’s `gammaplot`

command:

```
gammaplot shape=2 ncat=4;
```

Since we’re not using the cluster today, I’ll just show you the results of issuing this command in PAUP*:

```
Cut-points and category rates for discrete gamma approximation
------ cut-points ------
category lower upper rate (mean)
-------------------------------------------------
1 0.00000000 0.48063938 0.29327472
2 0.48063938 0.83917350 0.65501368
3 0.83917350 1.34631726 1.06998966
4 1.34631726 infinity 1.98172195
```

So the `pgamma`

command can be used to ask questions like “What are the chances that a Gamma(2,0.5) variate would be less than 1.0?” whereas the `qgamma`

command is used to ask the related question “What interval accounts for 95% of the values drawn from a Gamma(2,0.5) distribution?”

What value x produces an interval (0,x) that accounts for 95% of the values drawn from a Gamma(shape=2,scale=0.5) distribution?

What pair of values (x,y) produces an interval that accounts for the middle 95% of the values drawn from a Gamma(shape=2, scale=0.5) distribution (by middle, I mean that 2.5% of the distribution should be less than x and 2.5% greater than y)?

## Exponential distribution

The Exponential distribution is a special case of the Gamma distribution. Exponential distributions are Gamma distributions in which the shape parameter equals 1.0. The single parameter of an Exponential distribution is called the rate parameter and is the inverse of the scale parameter of the corresponding Gamma distribution. Like all other Gamma distributions, Exponential distributions range from 0.0 to infinity, so they provide excellent prior distributions for all the parameters for which Gamma distributions are appropriate (various relative rates and branch lengths). Exponential distributions are popular as priors because they force you to only come up with one number (the rate parameter) rather than the two (shape and scale) needed for Gamma distributions, but this does not in any way justify using Exponential distributions over Gamma distributions!

Here are some basic facts about the Exponential distribution:

Rate parameter | $\lambda$ |

Mean | $\frac{1}{\lambda}$ |

Variance | $\frac{1}{\lambda^2}$ |

Density function | $p(x) = \lambda e^{-\lambda x}$ |

### Exponential densities

Let’s plot a couple of Exponential densities and compare them. First, plot an Exponential(10) density function as follows:

```
> x <- seq(0.0,5.0,0.01)
> y <- dexp(x, rate=10)
> plot(x, y, type="l", col="blue", lwd=3, ylim=c(0, 10), xaxs="i")
```

Note that I’ve colored the line blue (`col="blue"`

), made it thick (`lwd=3`

), set the y-axis limits to 0.0 and 10.0 (`ylim=c(0,10)`

), and specified that the x-axis limits should be adhered to strictly (no added 4% to the left and right ends).

Now let’s add an Exponential(0.1) density (in red) to the plot using the `lines`

function:

```
> y2 <- dexp(x, rate=0.1)
> lines(x, y2, col="red", lwd=3, ylim=c(0, 10), xaxs="i")
```

The `lines`

command is like the `plot`

command except that it adds a line to an existing plot rather than creating a new plot (also, notice that `type="l"`

is neither necessary nor allowed because the `lines`

command only generates lines). (As you might imagine, there is also a `points`

command that adds points to an existing plot.)

Is the height of the density at x=0 related to the rate parameter?

Which of these two densities has the larger mean? (Try to answer this question first by looking at the plot, then compute the numbers to check your intuition.)

Which of these two densities has the larger variance? (Again, answer the question first by looking at the plot.)

What fraction of the Exponential(10) distribution is less than 1? (Hint: use the

`pexp`

command.)

What fraction of the Exponential(0.1) distribution is less than 1?

Fill in the blank: 95% of the Exponential(0.1) distribution is less than _____. (Hint: use the

`qexp`

command.)

## Lognormal distribution

The Lognormal distribution fills a similar niche as the Gamma distribution. Like the Gamma distribution, it has **support** (the interval in which its density is greater than 0) from 0 to $\infty$, making it suitable as a prior for edge lengths, rate ratios, and other parameters that must necessarily be positive but do not have an upper bound on their possible values. One difference between Lognormal and Gamma distributions is that Lognormal distributions always have density 0.0 at the value 0.0, whereas Gamma distributions with shape parameter less than 1 have infinite density at the value 0.0. So if zero is really not an option, a Lognormal prior may be a better choice than a Gamma distribution.

The Lognormal distribution has two parameters: $\mu$ and $\sigma$. The names of these parameters may be familiar to you as the mean and standard deviation parameters of a Normal distribution. Let’s generate 1 million random variates from a Lognormal distribution and ask what the sample mean and standard deviation are:

```
library(stats)
z <- rlnorm(1000000, 1, 2)
mean(z)
sd(z)
```

What is the sample mean of your 1 million Lognormal(1,2) random variates?

What is the sample standard deviation of your 1 million Lognormal(1,2) random variates?

If the mean and standard deviation were not what you expected, it is because $\mu$ and $\sigma$ do **not** represent the mean and standard deviation of a Lognormal distribution! Let me show you what these parameters actually represent:

```
logz <- log(z)
mean(logz)
sd(logz)
```

What is the sample mean of the

logarithmof your 1 million Lognormal(1,2) random variates?

What is the sample standard deviation of the

logarithmof your 1 million Lognormal(1,2) random variates?

Create a density plot of the `logz`

values:

```
plot(density(logz), type="l", col="blue", lwd=2, xlim=c(-10,10))
```

and compare that to a Normal(1,2) density:

```
x <- seq(-10, 10, 0.1)
lines(x, dnorm(x, 1, 2), col="red", lwd=5, lty="dotted", xlim=c(-10,10))
```

So it appears that $\mu$ and $\sigma$ parameters actually represent the mean and standard deviation of the *normal distribution* that results when you take the *logarithm* of a Lognormal random variate! Now you know why it is called a Lognormal distribution.

What is the mean of a Lognormal distribution? The mean equals $e^{\mu + \sigma^2/2}$. For the example above, the mean would thus equal $e^{1 + 2^2/2} = e^{3} = 20.09$, which should be close to the sample mean you obtained.

What is the median of a Lognormal(1,2) distribution? (Hint: look up Lognormal distribution in Wikipedia)

What is the sample median of the 1 million Lognormal(1,2) variates that you generated at the beginning of this section?

To see what this Lognormal(1,2) density function looks like, plot it as follows:

```
x <- seq(0, 5, 0.1)
plot(x, dlnorm(x, 1, 2), type="l", col="navy", lwd=2, xlim=c(0, 5))
```

Note that a Lognormal distribution looks nothing like a Normal distribution. While a Normal distribution is symmetric, the Lognormal distribution is very asymmetric. Normal distributions have a left tail that extends out to $-\infty$ (and also a right tail that extends out to $+\infty$), whereas Lognormal variates are strictly positive.

### Calculating $\mu$ and $\sigma$ from mean and standard deviation

If you want to use a Lognormal distribution as a prior, you probably know the mean (`m`

) and standard devation (`sd`

) that you want to use. How do you obtain the $\mu$ and $\sigma$ parameters needed to specify the Lognormal prior distribution? First calculate the coefficient of variation (`cv`

), which is the ratio of the standard deviation to the mean, then use that to calculate `mu`

and `sigma`

:

```
cv <- sd/m
variance <- log(1 + cv^2)
sigma <- sqrt(variance)
mu <- log(m) - variance/2
```

What would mu and sigma be if you wanted the mean to be 0.1 and the standard deviation 0.5?

You can confirm that you calculated $\mu$ and $\sigma$ correctly by simulating, say, 10 million values from a Lognormal(mu, sigma) distribution and checking that the mean and standard deviation are 0.1 and 0.5, respectively:

```
x <- rlnorm(1000000, mu, sigma)
mean(x)
sd(x)
```

(Note: even with a sample size of 10 million, you will not be able to nail the standard deviation with as much accuracy as you might expect, but your sample standard deviation should be within about 0.02-0.05 from the true value.)

## Beta distribution

The Beta distribution differs from the Gamma distribution in that Beta random variables have an upper bound (1.0) as well as a lower bound (0.0). Two parameters (a and b, but often referred to as $\alpha$ and, perversely, $\beta$) govern the shape of the Beta distribution. Beta distributions have support on the interval [0,1] and thus are natural priors for parameters such as the proportion of invariable sites (and the proportion of heads parameter in coin flipping examples), which are restricted to the [0,1] interval.

Here are some basic facts about Beta distributions:

Shape parameter 1 | $a$ |

Shape parameter 2 | $b$ |

Mean | $\frac{a}{a+b}$ |

Variance | $\frac{ab}{(a+b)^2 (a+b+1)}$ |

Density function | $p(x) = \frac{\Gamma(a+b)}{\Gamma(a) \Gamma(b)} x^{a-1} (1-x)^{b-1}$ |

The $\Gamma$ function above is the same one used in the Gamma distribution.

### Symmetrical Beta distributions

Beta densities are symmetrical about 0.5 if $a = b$. Try plotting a Beta(3,3) distribution in R as follows:

```
> x <- seq(0.0, 1.0, 0.01)
> y <- dbeta(x, 3, 3)
> plot(x, y, type="l")
```

A somewhat easier way to make a plot (that we will use from this point onward) involves using the `curve`

function. The `curve`

function handles details like generating a list of `x`

values to plot - you need only specify a “from” and “to” value for the x-axis. If you use a variable name not equal to `x`

for the x-axis, then you also need to specify `xname`

. Below I am pretending that the plotted Beta distribution is my prior for the pinvar (proportion of invariable sites) parameter:

```
> curve(dbeta(pinvar, 3, 3), from=0, to=1, xname="pinvar")
```

As $a$ and $b$ get larger (still constraining $a = b$), the density becomes more and more sharply peaked at 0.5. Generate the plot again, this time for a Beta(100,100) density. Let’s also plot the Beta(3,3) density for comparison. Note the addition of `add=TRUE`

to the second curve, which causes the second curve to be added to the first one (very handy for comparing two curves!):

```
> curve(dbeta(x, 100, 100), from=0, to=1)
> curve(dbeta(x, 3, 3), from=0, to=1, col="blue",lty="dotted", add=TRUE)
```

I did not need to specify `xname="x"`

in either curve above because I used `x`

as my x-axis variable (which is the default). I also made the second curve have a dotted, blue line by adding the line type (`lty`

) and color (`col`

) attributes to the command.

Note that if $a = b$, the variance formula given above simplifies to just $1/(8a + 4)$, so the variance of a Beta(3,3) distribution is 1/28 = 0.0357, whereas the variance of a Beta(100,100) distribution is 1/804 = 0.00124, which is 28.8 times smaller. My point here is just that “sharper” densities (i.e. narrower peaks) have smaller variances.

If $a=b=1$, the Beta distribution is identical to a Uniform(0,1) distribution, and the density function is simply a straight line at height 1.0:

```
> curve(dbeta(x, 1, 1), from=0, to=1)
```

Plot a Beta density when $a=b=0.1$.

Which distribution – Beta(1,1) or Beta(0.1, 0.1) – would be most appropriate as a prior distribution for a parameter representing the chances of a coin coming up heads on any given flip if you believed strongly that the world was full of trick coins (two-headed or two-tailed coins)?

### Asymmetrical Beta distributions

If $a$ is not equal to $b$, Beta densities are skewed to the right or left and are not symmetrical. Take a look at the formula for the mean: $a/(a+b)$. If $a < b$, the mean is less than 0.5 and the Beta distribution leans to the left. If $a > b$, the Beta distribution leans to the right.

Try plotting these Beta densities: Beta(1,100), Beta(100,1), Beta(2,5), and Beta(5,2).

```
curve(dbeta(x, 1, 100))
curve(dbeta(x, 100, 1))
curve(dbeta(x, 2, 5))
curve(dbeta(x, 5, 2))
```

Plot a Beta(20,50) density in blue, then a Beta(2,5) density in red on top of the Beta(20,50) density.

```
curve(dbeta(x, 20, 50), col="blue")
curve(dbeta(x, 2, 5), col="red", add=TRUE)
```

Calculate the mean and variance of a Beta(2,5) distribution using the formulas provided at the beginning of this section. Now sample 10000 values from a Beta(2,5) distribution and compute the sample mean and variance as follows:

```
> x <- rbeta(10000, 2, 5)
> mean(x)
[1] 0.2850546
> var(x)
[1] 0.02490558
```

Create a histogram of the values in x:

```
> hist(x)
```

Save the histogram as a PDF file named *beta25a.pdf* (we’ll be using this file again in a minute). In the Plots window of RStudio, use the Export dropdown and choose “Save as PDF…” to save the file.

### Relationship to the Gamma distribution

Beta distributions are related to Gamma distributions. If `x`

is a Gamma($a$,1) random variable, and `y`

is a Gamma($b$,1) random variable, then $x/(x+y)$ is distributed as Beta($a$,$b$). Now that you have just generated 10000 Beta(2,5) random variables, let’s compare that to 10000 random variables generated as described above:

```
> xx <- rgamma(10000, 2, 1)
> yy <- rgamma(10000, 5, 1)
> z <- xx/(xx+yy)
> mean(z)
> var(z)
```

Note in line 3 that R automatically applies formulas repeatedly to each value of a list. Both `xx`

and `yy`

are lists of 10000 values. Line 3 produces 10000 values stored in `z`

where `z[i] = xx[i]/(xx[i] + yy[i])`

for i = 1, 2, …, 10000.

Create a histogram of the values in `z`

, and save as a PDF file named *beta25b.pdf* (the procedure is described in the previous section). Compare *beta25a.pdf* with *beta25b.pdf* (they should be very similar because we have simply generated 10000 Beta(2,5) random variables in two different ways).

You can also plot the histograms on top of each other as follows (I’ve used the `rgb`

command (which stands for red, green, blue) for specifying colors so that I could add some transparency: the fourth argument to `rgb`

after red, green, and blue is 0.2 (20% of fully opaque):

```
> hist(x,col=rgb(1,0,0,.2), breaks=30)
> hist(z, col=rgb(0,0,1,.2), breaks=30,add=TRUE)
```

The area in common will now be purplish (red plus blue), while areas that represent one or the other will be reddish or bluish in color.

## Dirichlet distribution

Gamma distributions can be used as priors for branch lengths, relative rates, the among-site rate heterogeneity shape parameter and other parameters whose range of valid values is 0 to $\infty$. Beta distributions are good for proportions, such as pinvar, and parameters whose range is 0 to 1. What about base frequencies? The four base frequencies all lie between 0 and 1, but any prior distribution used for base frequencies must take account of the fact that the four base frequencies must add up to 1. The Dirichlet distribution is perfect for constrained proportions like this.

The Dirichlet distribution behaves in many ways like a Beta distribution. In fact, the Beta distribution is a special case of the Dirichlet distribution or, said another way, the Dirichlet distribution is a multivariate version of the Beta distribution. Think of the Beta distribution as a univariate distribution governing the proportion, $p$, where $p$ and $1-p$ are subject to the same constraint as base frequencies (that they must add to 1). A Beta($a$,$b$) distribution is thus equivalent to a Dirichlet($a$,$b$) distribution. There are four base frequency parameters, so four Dirichlet parameters are needed (i.e., a Dirichlet($a$,$b$,$c$,$d$) distribution) to create a prior appropriate for a vector of four base frequencies: $\pi_A$, $\pi_C$, $\pi_G$, $\pi_T$.

Here are some basic facts about the Dirichlet($a$,$b$,$c$,$d$) distribution, specifically when used as a prior distribution for base frequencies:

Shape parameters 1,2,3,4 | $a$,$b$,$c$,$d$ |

Sum of shape parameters | $s = a + b + c + d$ |

Mean of $\pi_A$, $\pi_C$, $\pi_G$, and $\pi_T$ | $\frac{a}{s}$, $\frac{b}{s}$, $\frac{c}{s}$, and $\frac{d}{s}$ |

Variance of $\pi_A$, $\pi_C$, $\pi_G$, and $\pi_T$ | $\frac{a(s-a)}{s^2 (s+1)}$, $\frac{b(s-b)}{s^2 (s+1)}$, $\frac{c(s-c)}{s^2 (s+1)}$, and $\frac{d(s-d)}{s^2 (s+1)}$ |

Density function | $p(\pi_A, \pi_C, \pi_G, \pi_T) = \frac{\Gamma(s)}{\Gamma(a) \Gamma(b) \Gamma(c) \Gamma(d)} \pi_A^{a-1} \pi_C^{b-1} \pi_G^{c-1} \pi_T^{d-1}$ |

### Relationship to the Gamma distribution

There are no `rdirichlet`

, `ddirichlet`

, `qdirichlet`

, or `pdirichlet`

functions in R, but it is easy to sample from a Dirichlet using the fact that, like the Beta distribution, the Dirichlet distribution is related to the Gamma distribution. If

$x_1 \sim \mbox{Gamma}(a,1)$

$x_2 \sim \mbox{Gamma}(b,1)$

$x_3 \sim \mbox{Gamma}(c,1)$

$x_4 \sim \mbox{Gamma}(d,1)$

(where $x_1 \sim \mbox{Gamma}(a, 1)$ is shorthand for “$x_1$ is a random number drawn from a Gamma distribution with shape $a$ and scale 1”), then a Dirichlet($a$,$b$,$c$,$d$) random variable (a vector of 4 values) can be created by dividing each $x_i$ by the sum of all four $x_i$ values. Create 10 random Dirichlet(1,1,1,1) variables using this method and list them:

```
> x1 <- rgamma(10, 1, 1)
> x2 <- rgamma(10, 1, 1)
> x3 <- rgamma(10, 1, 1)
> x4 <- rgamma(10, 1, 1)
> s <- x1 + x2 + x3 + x4
> d <- c(x1/s, x2/s, x3/s, x4/s)
```

Note the use of the `c(...)`

function to construct a vector of four values, each of which is one of the $x_i$ values divided by $s$, which is the sum of all four $x_i$ values.

Now display the 10 Dirichlet vectors:

```
> d
[1] 0.09234290 0.37789542 0.08640752 0.28049539 0.14485492 0.05511074 0.14355295
[8] 0.05358819 0.03350763 0.19362171 0.39036151 0.25134264 0.29663099 0.43909681
[15] 0.13981216 0.29531499 0.34553287 0.76487943 0.13311580 0.23832980 0.11153418
[22] 0.35743685 0.03021077 0.12139093 0.63609473 0.17125022 0.27907977 0.09122259
[29] 0.44937431 0.19609418 0.40576141 0.01332508 0.58675072 0.15901687 0.07923820
[36] 0.47832406 0.23183441 0.09030979 0.38400226 0.37195431
```

Unfortunately, R dumped all 10 vectors together, creating a single vector of 40 values! We can coerce R into displaying these 4 at a time by using the `dim`

command to redimension `d`

into a matrix with 10 rows and 4 columns:

```
> dim(d) <- c(10,4)
> d
[,1] [,2] [,3] [,4]
[1,] 0.09234290 0.3903615 0.11153418 0.40576141
[2,] 0.37789542 0.2513426 0.35743685 0.01332508
[3,] 0.08640752 0.2966310 0.03021077 0.58675072
[4,] 0.28049539 0.4390968 0.12139093 0.15901687
[5,] 0.14485492 0.1398122 0.63609473 0.07923820
[6,] 0.05511074 0.2953150 0.17125022 0.47832406
[7,] 0.14355295 0.3455329 0.27907977 0.23183441
[8,] 0.05358819 0.7648794 0.09122259 0.09030979
[9,] 0.03350763 0.1331158 0.44937431 0.38400226
[10,] 0.19362171 0.2383298 0.19609418 0.37195431
```

Each row now represents a random draw from a Dirichlet(1,1,1,1) distribution. If we were using this Dirichlet distribution as a prior for base frequencies, each row would represent one representative vector of base frequencies from our prior. Given the impossibility of plotting a 4-dimensional density function, it would thus be useful to use the method above to produce samples from various Dirichlet distributions to see what they imply about the distribution of base frequencies.

To make it easier to explore the properties of Dirichlet distributions, let’s put the commands you just entered into a function that can be called with different combinations of parameters. Copy the following into your clipboard and paste it into the R console window.

```
Dirichlet <- function(a,b,c,d) {
x1 <- rgamma(10, a, 1)
x2 <- rgamma(10, b, 1)
x3 <- rgamma(10, c, 1)
x4 <- rgamma(10, d, 1)
s <- x1 + x2 + x3 + x4
d <- c(x1/s, x2/s, x3/s, x4/s)
dim(d) <- c(10,4)
d
}
```

### Symmetrical Dirichlet($a$,$b$,$c$,$d$) distributions

Like Beta distributions, Dirichlet distributions are symmetrical if all parameters are equal. Use your new R function, which I’ve arbitrarily (but appropriately!) named “Dirichlet”, to explore a series of symmetrical Dirichlet distributions:

```
> Dirichlet(1,1,1,1)
> Dirichlet(100,100,100,100)
> Dirichlet(1000000,1000000,1000000,1000000) # yes, that's 1 million!
```

Can you see a trend?

What is the effect of increasing the magnitude of all four parameters from 1 to 100 to 1 million?

Which of these three Dirichlet distributions comes closest to the assumptions made by the JC69 model?

In the case of a Dirichlet($a$,$b$,$c$,$d$) distribution, as the parameters become larger (assuming $a=b=c=d$), the density becomes more and more sharply peaked at (0.25, 0.25, 0.25, 0.25). The magnitude of the Dirichlet parameters thus determines the *informativeness* of the prior: if all four parameters equal 1, the Dirichlet is flat or uninformative (any possible combination of base frequencies has the same probability density as any other combination), whereas if all four parameters equal 1 million, then approximately equal base frequencies (e.g. 0.250241, 0.250133, 0.249963, 0.249663) get much more weight than very unequal base frequencies (e.g., 0.000404, 0.531065, 0.000002, 0.468529) and the distribution would be considered informative.

We can visualize these samples better by plotting them. Below is an R function that plots the four base frequencies in different colors (note that it draws 1000 samples rather than just 10):

```
PlotDirichlet <- function(a,b,c,d) {
x1 <- rgamma(1000, a, 1)
x2 <- rgamma(1000, b, 1)
x3 <- rgamma(1000, c, 1)
x4 <- rgamma(1000, d, 1)
s <- x1 + x2 + x3 + x4
d <- c(x1/s, x2/s, x3/s, x4/s)
dim(d) <- c(1000,4)
plot(1:1000, d[,1], type="l", col="red", ylim=c(0,1), ylab="frequencies")
lines(1:1000, d[,2],col="blue")
lines(1:1000, d[,3],col="green")
lines(1:1000, d[,4],col="black")
}
```

The `1:1000`

in the `plot`

and `lines`

commands is shorthand for `seq(1, 1000, 1)`

(i.e., a sequence of values starting at 1, ending at 1000, with a step size of 1. Also, `d[,1]`

means “the first column of `d`

”, `d[,2]`

means “the second column of `d`

”, etc.

Now plot a few Dirichlet distributions:

```
> PlotDirichlet(1,1,1,1)
> PlotDirichlet(10,10,10,10)
> PlotDirichlet(100,100,100,100)
```

The x-axis in these plots doesn’t have any meaning (it is just the row number of the matrix `d`

), but looking at the plot it is easy to see which Dirichlet distributions have higher variance (and hence are less informative) and which have lower variance (and thus are more informative).

### Asymmetrical Dirichlet distributions

If some of the four parameters differ from the others, Dirichlet($a$,$b$,$c$,$d$) densities are not symmetrical. It might be useful to consider using asymmetrical Dirichlet distributions if you know your sequences are GC rich, for example. Here are three GC-rich Dirichlet distributions varying in informativeness but all having mean ($\pi_A$, $\pi_C$, $\pi_G$, $\pi_T$) = (0.2, 0.3, 0.3, 0.2):

```
> PlotDirichlet(2, 3, 3, 2)
> PlotDirichlet(20, 30, 30, 20)
> PlotDirichlet(200, 300, 300, 200)
```

### Another way to view Dirichlet samples

Check out this applet, which allows you to view 4-parameter Dirichlet distributions suitable for modeling nucleotide frequencies:

Dirichlet nucleotide frequencies prior applet

### Dirichlet prior for GTR relative rates

As you know, the GTR model specifies six relative rates (sometimes called exchangeability parameters) corresponding to the six possible types of substitutions in the rate matrix (actually there are 12, but the GTR model is symmetrical in that the relative rate for A to G substitutions is identical to that for G to A substitutions). Bayesian programs typically use a 6-parameter Dirichlet($a$,$b$,$c$,$d$,$e$,$f$) distribution as the prior distribution of these GTR relative rates.

Below is a revised Dirichlet function that takes 6 values rather than 4. Copy this into your R console:

```
Dirichlet <- function(a,b,c,d,e,f) {
x1 <- rgamma(10, a, 1)
x2 <- rgamma(10, b, 1)
x3 <- rgamma(10, c, 1)
x4 <- rgamma(10, d, 1)
x5 <- rgamma(10, e, 1)
x6 <- rgamma(10, f, 1)
s <- x1 + x2 + x3 + x4 + x5 + x6
d <- c(x1/s, x2/s, x3/s, x4/s, x5/s, x6/s)
dim(d) <- c(10,6)
d
}
```

Now apply the new function:

```
> Dirichlet(1,1,1,1,1,1)
> Dirichlet(1000,4000,1000,1000,4000,1000)
```

Now each row represents a set of GTR relative rates. The first case is a flat prior: it says that we have no idea which of six substitution classes we expect to be evolving at a relatively fast rate compared to the other substitution classes. The second case is an informative prior that says we expect the transition rate to be 4 times that of the transversion rate (assuming the ordering is $r_{AC}$, $r_{AG}$, $r_{AT}$, $r_{CG}$, $r_{CT}$, $r_{GT}$).

Try modifying the `PlotDirichlet`

function to plot 6-parameter Dirichlet distributions in the same way we plotted 4-parameter Dirichlet distributions above.