# Lab 4: probability distributions, averaging, and Jensen's inequality

## 1  Random distributions in R

R knows about lots of probability distributions. For each, it can generate random numbers drawn from the distribution ("deviates"); compute the cumulative distribution function and the probability distribution function; and compute the quantile function, which gives the x value such that ò0x P(x)  dx (area under the curve from 0 to x) is a specified value, such as 0.95 (think about "tail areas" from standard statistics).
Figure 1: R functions for an arbitrary distribution dist, showing density function (ddist), cumulative distribution function (pdist), quantile function (qdist), and random-deviate function (rdist).
Let's take the binomial distribution (yet again) as an example.
• rbinom(n,size,p) gives n random draws from the binomial distribution with parameters size (total number of draws) and p (probability of success on each draw). You can give different parameters for each draw. For example:
```> rbinom(10, size = 8, p = 0.5)

```
``` [1] 3 6 3 4 6 7 4 3 2 2

```
```> rbinom(3, size = 8, p = c(0.2, 0.4, 0.6))

```
```[1] 2 4 7

```
Figure 2 shows the result of drawing 200 values from a binomial distribution with N=12 and p=0.1 and plotting the results as a factor (with 200 draws we don't have to worry about any of the 13 possible outcomes getting missed and excluded from the plot):
```> plot(factor(rbinom(200, size = 12, p = 0.1)), xlab = "# of successes",
+     ylab = "# of trials out of 200")

```
Figure 2: Results of rbinom
• dbinom(x,size,p) gives the value of the probability distribution function (pdf) at x (for a continous distribution, the analogous function would compute the probability density function). Since the binomial is discrete, x has to be an integer, and the pdf is just the probability of getting that many successes; if you try dbinom with a non-integer x, you'll get zero and a warning.
• pbinom(q,size,p) gives the value of the cumulative distribution function (cdf) at q (e.g. pbinom(7,size=10,prob=0.4));
• qbinom(p,size,prob) gives the quantile function, where p is a number between 0 and 1 (an area under the pdf, or value of the cdf) and qbinom is the value such that P(X £ p)=q. The quantile function Q is the inverse of the cumulative distribution function C: if Q(p)=q then C(q)=p. Example: qbinom(0.95,size=10,prob=0.4).
These four functions exist for each of the distributions R has built in: e.g. for the normal distribution they're rnorm(), pnorm(), dnorm(), qnorm(). Each distribution has its own set of parameters (so e.g. pnorm() is pnorm(x,mean=0,sd=1)).
Exercise 1: For the binomial distribution with 10 trials and a success probability of 0.2:
• Pick 8 random values and sort them into increasing order (if you set.seed(1001) beforehand, you should get X=0 (twice), X=2 (5 times), and X=4 and X=5 (once each)).
• Calculate the probabilities of getting 3, 4, or 5 successes. Answer:
```[1] 0.20132659 0.08808038 0.02642412

```
• Calculate the probability of getting 5 or more successes. Answer:
```[1] 0.0327935

```
• What tail values would you use to test against the (two-sided) null hypothesis that p=0.2? (Use qbinom() to get the answer, and use pbinom(0:10,size=10,prob=0.2) and pbinom(0:10,size=10,prob=0.2,lower.tail=FALSE) to check that your answer makes sense.
You can use the R functions to test your understanding of a distribution and make sure that random draws match up with the theoretical distributions as they should. This procedure is particularly valuable when you're developing new probability distributions by combining simpler ones, e.g. by zero-inflating or compounding distributions.
The results of a large number of random draws should have the correct moments (mean and variance), and a histogram of those random draws (with freq=FALSE or prob=TRUE) should match up with the theoretical distribution. For example, draws from a binomial distribution with p=0.2 and N=20 should have a mean of approximately Np=4 and a variance of Np(1-p)=3.2:
```> set.seed(1001)
> N = 20
> p = 0.2
> x = rbinom(10000, prob = p, size = N)
> c(mean(x), var(x))

```
```[1] 4.001200 3.144913

```
The mean is very close, the variance is a little bit farther off. Just for the heck of it, we can use the replicate() function to re-do this command many times and see how close we get:
```> var_dist = replicate(1000, var(rbinom(10000, prob = p, size = N)))

```
(this may take a little while; if it takes too long, lower the number of replicates to 100).
Looking at the summary statistics and at the 2.5% and 97.5% quantiles of the distribution of variances:
```> summary(var_dist)

```
```   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
3.052   3.169   3.199   3.199   3.229   3.340

```
```> quantile(var_dist, c(0.025, 0.975))

```
```    2.5%    97.5%
3.114357 3.285333

```
(Try a histogram too.) Even though there's some variation (of the variance) around the theoretical value, we seem to be doing the right thing since the 95% confidence limits include the theoretical value. (Lab 5 will go into more detail on running simulations to check the expected variation of different measurement as a function of parameters and sample size.)
Finally, Figure 3 shows the entire simulated frequency distribution along with the theoretical values. The steps in R are:
1. pick 10,000 random deviates:
```> x = rbinom(10000, prob = p, size = N)

```
2. Tabulate the values, and divide by the number of samples to get a probability distribution:
```> tx = table(factor(x, levels = 0:12))/10000

```
(The levels command is necessary in this case because the probability of x=12 with p=0.2 and N=12 is actually so low (8.7×10-5) that there's a reasonable chance that a sample of 10,000 won't include any samples with 12 successes.)
3. Draw a barplot of the values, extending the y-limits a bit to make room for the theoretical values and saving the x locations at which the bars are drawn:
```> b1 = barplot(tx, ylim = c(0, 0.23), ylab = "Probability")

```
4. Add the theoretical values, plotting them at the same x-locations as the centers of the bars:
```> points(b1, dbinom(0:12, prob = p, size = N), pch = 16)

```
(barplot() doesn't put the bars at x locations corresponding to their numerical values, so you have to save those values as b1 and re-use them to make sure the theoretical values end up in the right place.)
A few alternative ways to do this plot would be:
1. ```> plot(factor(x))
> points(b1, 10000 * dbinom(0:12, prob = p, size = N))

```
(plots the number of observations without rescaling and scales the probability distribution to match);
2. ```> plot(table(x)/10000)
> points(0:12, dbinom(0:12, prob = p, size = N))

```
Plotting a table does a plot with type="h" (high density), which plots a vertical line for each value. I think it's not quite as pretty as the barplot, but it works. Unlike factors, tables can be scaled numerically, and the lines end up at the right numerical locations, so we can just use 0:12 as the x locations for the theoretical values.
3. You could also draw a histogram: since histograms were really designed continuous data you have to make sure the breaks occur in the right place (halfway between counts):
```> h = hist(x, breaks = seq(-0.5, 12.5, by = 1), col = "gray", prob = TRUE)
> points(0:12, dbinom(0:12, prob = p, size = N))

```
Figure 3: Checking binomial deviates against theoretical values.
Doing the equivalent plot for continuous distributions is actually somewhat easier, since you don't have to deal with the complications of a discrete distribution: just use hist(...,prob=TRUE) to show the sampled distribution (possibly with ylim adjusted for the maximum of the theoretical density distribution) and ddist(x,[parameters],add=TRUE) to add the theoretical curve (e.g.: curve(dgamma(x,shape=2,scale=1,add=FALSE))).
Exercise 2*: Pick 10,000 negative binomial deviates with m = 2, k=0.5. Pick one of the ways above to draw the distribution. Check that the mean and variance agree reasonably well with the theoretical values. Add points representing the theoretical distribution to the plot. Now translate the m and k parameters into their p and nn equivalents (the coin-flipping derivation of the negative binomial), and add those points to the plot [use a different plotting symbol to make sure you can see that they overlap with the theoretical values based on the m, k parameterization].

## 2  Averaging across discrete and continuous distributions

Suppose we have a (tiny) data set; we can organize it in two different ways, in standard long format or in tabular form:
```> dat = c(5, 6, 5, 7, 5, 8)
> dat

```
```[1] 5 6 5 7 5 8

```
```> tabdat = table(dat)
> tabdat

```
```dat
5 6 7 8
3 1 1 1

```
To get the (sample) probability distribution of the data, just scale by the total sample size:
```> prob = tabdat/length(dat)
> prob

```
```dat
5         6         7         8
0.5000000 0.1666667 0.1666667 0.1666667

```
(dividing by sum(tabdat) would be equivalent).
In the long format, we can take the mean with mean(dat) or, replicating the formula åxi/N exactly, sum(dat)/length(dat).
In the tabular format, we can calculate the mean with the formula åP(x) x, which in R would be sum(prob*5:8) or more generally
```> vals = as.numeric(names(prob))
> sum(prob * vals)

```
```[1] 6

```
(you could also get the values by as.numeric(levels(prob)), or by sort(unique(dat))).
However, mean(prob) or mean(tabdat) is just plain wrong (at least, I can't think of a situation where you would want to calculate this value).
Exercise 3: figure out what it means that mean(tabdat) equals 1.5.
Going back the other way, from a table to raw values, we can use the rep() function to repeat values an appropriate number of times. In its simplest form, rep(x,n) just creates a vector repeats x (which may be either a single value or a vector) n times, but if n is a vector as well then each element of x is repeated the corresponding number of times: for example,
```> rep(c(1, 2, 3), c(2, 1, 5))

```
```[1] 1 1 2 3 3 3 3 3

```
gives two copies of 1, one copy of 2, and five copies of 3.
Therefore,
```> rep(vals, tabdat)

```
```[1] 5 5 5 6 7 8

```
will recover our original data (although not in the original order) by repeating each element of vals the correct number of times.

### 2.1  Jensen's inequality

Looking at Schmitt et al's data, the recruitment level very nearly fits an exponential distribution with a mean of 24.5 (so l = 1/24.5). Schmitt et al. also say that recruitment (R) as a function of settlement (S) is R = aS/(1+(a/b)S), with a = 0.696 (initial slope, recruits per 0.1 m2 patch reef per recruit) and b = 9.79 (asymptote, recruits per 0.1 m2 patch reef).
Let's see how strong Jensen's inequality is for this population. We'll figure out the average by approximating an integral by a sum: ò0¥ f(S) P(S)  dS » åf(Si) P(Si) DS. We need to set the range big enough to get most of the probability of the distribution, and the DS small enough to get most of the variation in the distribution; we'll try 0-200 in steps of 0.1. (If I set the range too small or the DS too big, I'll miss a piece of the distribution or the function. If I try to be too precise, I'll waste time computing.)
In R:
```> a = 0.696
> b = 9.79
> dS = 0.1
> S = seq(0, 200, by = dS)
> pS = dexp(S, rate = 1/24.5)
> fS = a * S/(1 + (a/b) * S)
> sum(pS * fS * dS)

```
```[1] 5.008049

```
R also knows how to integrate functions numerically: it can even approximate an integral from 0 to ¥. First we have to define a (vectorizable) function:
```> tmpf = function(S) {
+     dexp(S, rate = 1/24.5) * a * S/(1 + (a/b) * S)
+ }

```
Then we can just ask R to integrate it:
```> i1 = integrate(tmpf, lower = 0, upper = Inf)
> i1

```
```5.010691 with absolute error < 5.5e-05

```
(Use adapt(), in the adapt package, if you need to do multidimensional integrals.)
This integral shows that we were pretty close with our first approximation. However, numerical integration will always imply some level of approximation; be careful with functions with sharp spikes, because it can be easy to miss important parts of the function.
Now to try out the delta function approximation:
```> d1 = D(expression(a * x/(1 + (a/b) * x)), "x")
> d2 = D(d1, "x")

```
As stated above, the mean value of the distribution is about 24.5. The variance of the exponential distribution is equal to the mean squared, or 600.25.
```> Smean = 24.5
> Svar = Smean^2
> d2_num = eval(d2, list(a = 0.696, b = 9.79, x = Smean))
> mval = a * Smean/(1 + (a/b) * Smean)
> dapprox = mval + 1/2 * Svar * d2_num
> merr = (mval - i1\$value)/i1\$value
> merr

```
```[1] 0.2412107

```
```> err = (dapprox - i1\$value)/i1\$value
> err

```
```[1] -0.04637931

```
The answer from the delta method (f([`x])+(s2 f"([`x])/2)) is only about 5% below the true value, as opposed to the naive answer (f([`x])) which is about 25% high. (We have to say i1\$value to extract the actual value of the integral from the variable i1; try str(i1) if you want to see all the information that R is storing about the integral.)
Exercise 4*: try the above exercise again, but this time with a gamma distribution instead of an exponential. Keep the mean equal to 24.5 and change the variance to 100, 25, and 1 (use the information that the mean of the gamma distribution is shape*scale and the variance is shape*scale^2). Including the results for the exponential (which is a gamma with shape=1), make a table showing how the (1) true value of mean recruitment [calculated by numerical integration in R either using integrate() or summing over small DS] (2) value of recruitment at the mean settlement (3) delta-method approximation (4,5) proportional error in #2 and #3 change with the variance.

## 3  The method of moments: reparameterizing distributions

In the chapter, I showed how to use the method of moments to estimate the parameters of a distribution by setting the sample mean and variance ([`x], s2) equal to the theoretical mean and variance of a distribution and solving for the parameters. For the negative binomial, in particular, I found m = [`x] and k=([`x])/(s2/[`x] -1).
You can also define your own functions that use your own parameterizations: call them my_function rather than just replacing the standard R functions (which will lead to insanity in the long run).
For example, defining
```> my_dnbinom = function(x, mean, var, ...) {
+     mu = mean
+     k = mean/(var/mean - 1)
+     dnbinom(x, mu = mu, size = k, ...)
+ }
> my_rnbinom = function(n, mean, var, ...) {
+     mu = mean
+     k = mean/(var/mean - 1)
+     rnbinom(n, mu = mu, size = k, ...)
+ }

```
(the ... in the function takes any other arguments you give to my_dnbinom and just passes them through, unchanged, to dnbinom).
Defining your own functions can be handy if you need to work on a regular basis with a distribution that uses a different parameterization than the one built into the standard R function.
You can use the kinds of histograms shown above to test your results (remembering that the method of moments estimates may be slightly biased especially for small samples - but they shouldn't cause errors as large as those caused by typical algebra mistakes).
```> x = my_rnbinom(1e+05, mean = 1, var = 4)
> mean(x)

```
```[1] 0.9999

```
```> var(x)

```
```[1] 4.00574

```
```> tx = table(factor(x, levels = 0:max(x)))/1e+05
> b1 = barplot(tx, ylab = "Probability")
> points(b1, my_dnbinom(0:max(x), mean = 1, var = 4), pch = 16)
> abline(v = 1)

```
Exercise 5*: Morris (1997) gives a definition of the beta function that is different from the standard statistical parameterization. The standard parameterization is
 Beta(x|a,b) = G(a+b) G(a)G(b) xa-1(1-x)b-1
whereas Morris uses
 Beta(x|P,q) = G(q) G(qP)G(q(1-P)) xqP-1 (1-x)q(1-P)-1.
Find expressions for P and q in terms of a and b and vice versa. Explain why you might prefer Morris's parameterization. Define a new set of functions that generate random deviates from the beta distribution (my_rbeta) and calculate the density function (my_dbeta) in terms of P and q. Generate a histogram from this distribution and draw a vertical line showing the mean of the distribution.

## 4  Creating new distributions

### 4.1  Zero-inflated distributions

The general formula for the probability distribution of the a zero-inflated distribution, with an underlying distribution P(x) and a zero-inflation probability of pz, is:
 Prob(0)
 =
 pz + (1-pz) P(0)
 Prob(x > 0)
 =
 (1-pz) P(x)
So, for example, we could define a probability distribution for a zero-inflated negative binomial as follows:
```> dzinbinom = function(x, mu, size, zprob) {
+     ifelse(x == 0, zprob + (1 - zprob) * dnbinom(0, mu = mu,
+         size = size), (1 - zprob) * dnbinom(x, mu = mu, size = size))
+ }

```
(the name, dzinbinom, follows the R convention for a probability distribution function: a d followed by the abbreviated name of the distribution, in this case zinbinom for "zero-inflated negative binomial").
The ifelse() command checks every element of x to see whether it is zero or not and fills in the appropriate value depending on the answer.
A random deviate generator would look like this:
```> rzinbinom = function(n, mu, size, zprob) {
+     ifelse(runif(n) < zprob, 0, rnbinom(n, mu = mu, size = size))
+ }

```
The command runif(n) picks n random values between 0 and 1; the ifelse command compares them with the value of zprob. If an individual value is less than zprob (which happens with probability zprob=pz), then the corresponding random number is zero; otherwise it is a value picked out of the appropriate negative binomial distribution.
Exercise 6: check graphically that these functions actually work. For an extra challenge, calculate the mean and variance of the zero-inflated negative binomial and compare it to the results of rzinbinom(10000,mu=4,size=0.5,zprob=0.2).

## 5  Compounding distributions

The key to compounding distributions in R is that the functions that generate random deviates can all take a vector of different parameters rather than a single parameter. For example, if you were simulating the number of hatchlings surviving (with individual probability 0.8) from a series of 8 clutches, all of size 10, you would say
```> rbinom(8, size = 10, prob = 0.8)

```
```[1] 7 6 9 7 9 9 6 8

```
but if you had a series of clutches of different sizes, you could still pick all the random values at the same time:
```> clutch_size = c(10, 9, 9, 12, 10, 10, 8, 11)
> rbinom(8, size = clutch_size, prob = 0.8)

```
```[1] 10  7  8  8  6  7  6  8

```
Taking this a step farther, the clutch size itself could be a random variable:
```> clutch_size = rpois(8, lambda = 10)
> rbinom(8, size = clutch_size, prob = 0.8)

```
```[1]  7 10  7 11 13  4  5  4

```
We've just generated a Poisson-binomial random deviate ...
As a second example, I'll follow Clark et al. in constructing a distribution that is a compounding of normal distributions, with 1/variance of each sample drawn from a gamma distribution.
First pick the variances as the reciprocals of 10,000 values from a gamma distribution with shape 5 (setting the scale equal to 1/5 so the mean will be 1):
```> var_vals = 1/rgamma(10000, shape = 5, scale = 1/5)

```
Take the square root, since dnorm uses the standard deviation and not the variance as a parameter:
```> sd_vals = sqrt(var_vals)

```
Generate 10,000 normal deviates using this range of standard deviations:
```> x = rnorm(10000, mean = 0, sd = sd_vals)

```
Figure 4 shows a histogram of the following commands:
```> hist(x, prob = TRUE, breaks = 100, col = "gray")
> curve(dt(x, df = 11), add = TRUE, lwd = 2)

```
The superimposed curve is a t distribution with 11 degrees of freedom; it turns out that if the underlying gamma distribution has shape parameter p, the resulting t distribution has df=2p+1. (Figuring out the analytical form of the compounded probability distribution or density function, or its equivalence to some existing distribution, is the hard part; for the most part, though, you can find these answers in the ecological and statistical literature if you search hard enough.
```> hist(x, prob = TRUE, breaks = 100, col = "gray")
> curve(dt(x, df = 11), add = TRUE, lwd = 2)

```
Figure 4: Clark model: inverse gamma compounded with normal, equivalent to the Student t distribution
Exercise 7*: generate 10,000 values from a gamma-Poisson compounded distribution with parameters shape=k=0.5, scale=m/k=4/0.5=8 and demonstrate that it's equivalent to a negative binomial with the appropriate m and k parameters.
Extra credit: generate 10,000 values from a lognormal-Poisson distribution with the same expected mean and variance (the variance of the lognormal should equal the variance of the gamma distribution you used as a compounding distribution; you will have to do some algebra to figure out the values of meanlog and sdlog needed to produce a lognormal with a specified mean and variance. Plot the distribution and superimpose the theoretical distribution of the negative binomial with the same mean and variance to see how different the shapes of the distributions are.

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On 28 Sep 2005, 13:38.