Random Data
6 Random Data
Although Einstein said that god does not play dice,
R can. For example
> sample(1:6,10,replace=T)
[1] 6 4 4 3 5 2 3 3 5 4
or with a function
> RollDie = function(n) sample(1:6,n,replace=T)
> RollDie(5)
[1] 3 6 1 2 2
In fact,
R can create lots of different types of random
numbers ranging from familiar families of distributions to specialized
ones.
6.1 Random number generators in R the ``r'' functions.
As we know, random numbers are described by a distribution. That is,
some function which specifies the probability that a random number
is in some range. For example
P(
a <
X £ b). Often this is
given by a probability density (in the continuous case) or by a
function
P(
X=
k) =
f(
k) in the discrete case.
R will give
numbers drawn from lots of different distributions. In order to use
them, you only need familiarize yourselves with the parameters that
are given to the functions such as a mean, or a rate. Here are
examples of the most common ones. For each, a histogram is given
for a random sample of size 100, and density (using the ``d''
functions) is superimposed as appropriate.

Uniform.
 Uniform numbers are ones that are "equally likely" to
be in the specified range. Often these numbers are in [0,1] for
computers, but in practice can be between [a,b] where a,b depend
upon the problem. An example might be the time you wait at a traffic
light. This might be uniform on [0,2].
> runif(1,0,2) # time at light
[1] 1.490857 # also runif(1,min=0,max=2)
> runif(5,0,2) # time at 5 lights
[1] 0.07076444 0.01870595 0.50100158 0.61309213 0.77972391
> runif(5) # 5 random numbers in [0,1]
[1] 0.1705696 0.8001335 0.9218580 0.1200221 0.1836119
The general form is runif(n,min=0,max=1) which allows you to
decide how many uniform random numbers you want (n), and the
range they are chosen from ([min,max])
To see the distribution with min=0 and max=1 (the
default) we have
> x=runif(100) # get the random numbers
> hist(x,probability=TRUE,col=gray(.9),main="uniform on [0,1]")
> curve(dunif(x,0,1),add=T)
Figure 25: 100 uniformly random numbers on [0,1]
The only tricky thing was plotting the histogram with a background
``color''. Notice how the dunif function was used with the
curve function.
 Normal.
 Normal numbers are the backbone of classical statistical theory
due to the central limit theorem The normal
distribution has two parameters a mean µ and a standard deviation
s. These are the location and spread parameters. For example,
IQs may be normally distributed with mean 100 and standard deviation
16, Human gestation may be normal with mean 280 and
standard deviation about 10 (approximately). The family of normals can be standardized
to normal with mean 0 (centered) and variance 1. This is achieved by
"standardizing" the numbers, i.e. Z=(Xµ)/s.
Here are some examples
> rnorm(1,100,16) # an IQ score
[1] 94.1719
> rnorm(1,mean=280,sd=10)
[1] 270.4325 # how long for a baby (10 days early)
Here the function is called as rnorm(n,mean=0,sd=1) where
one specifies the mean and the standard deviation.
To see the shape for the defaults (mean 0, standard deviation 1) we
have (figure 26)
> x=rnorm(100)
> hist(x,probability=TRUE,col=gray(.9),main="normal mu=0,sigma=1")
> curve(dnorm(x),add=T)
## also for IQs using rnorm(100,mean=100,sd=16)
Figure 26: Normal(0,1) and normal(100,16)
 Binomial.
 The binomial random numbers are discrete random
numbers. They have the distribution of the number of successes in
n independent Bernoulli trials where a Bernoulli trial results in
success or failure, success with probability p.
A single Bernoulli trial is given with n=1 in the binomial
> n=1, p=.5 # set the probability
> rbinom(1,n,p) # different each time
[1] 1
> rbinom(10,n,p) # 10 different such numbers
[1] 0 1 1 0 1 0 1 0 1 0
A binomially distributed number is the same as the number of 1's in
n such Bernoulli numbers. For the last example, this would be 5.
There are then two parameters n (the number of Bernoulli trials) and
p (the success probability).
To generate binomial numbers, we simply change the value of n
from 1 to the desired number of trials. For example, with 10 trials:
> n = 10; p=.5
> rbinom(1,n,p) # 6 successes in 10 trials
[1] 6
> rbinom(5,n,p) # 5 binomial number
[1] 6 6 4 5 4
The number of successes is of course discrete, but as n gets large,
the number starts to look quite normal. This is a case of the central limit theorem
which states in general that (X^{}  µ)/s is normal in the
limit (note this is standardized as above) and in our specific case
that
is approximately normal, where p^{^} = (number of successes)/n.
The graphs (figure 27) show 100
binomially distributed random numbers for 3 values of n and for
p=.25. Notice in the graph, as n increases the shape becomes more
and more bellshaped. These graphs were made with the commands
> n=5;p=.25 # change as appropriate
> x=rbinom(100,n,p) # 100 random numbers
> hist(x,probability=TRUE,)
## use points, not curve as dbinom wants integers only for x
> xvals=0:n;points(xvals,dbinom(xvals,n,p),type="h",lwd=3)
> points(xvals,dbinom(xvals,n,p),type="p",lwd=3)
... repeat with n=15, n=50
Figure 27: Random binomial data with the theoretical distribution
 Exponential
 The exponential distribution is important for
theoretical work. It is used to describe lifetimes of
electrical components (to first order). For example, if the mean
life of a light bulb is 2500 hours one may think its lifetime is
random with exponential distribution having mean 2500. The one
parameter is the rate = 1/mean. We specify it as follows
rexp(n,rate=1). Here is an example with the rate being
1/2500 (figure 28).
> x=rexp(100,1/2500)
> hist(x,probability=TRUE,col=gray(.9),main="exponential mean=2500")
> curve(dexp(x,1/2500),add=T)
Figure 28: Random exponential data with theoretical density
There are others of interest in statistics. Common ones are the Poisson, the
Student
tdistribution, the
F distribution, the beta distribution and the
c^{2} (chi
squared) distribution.
6.2 Sampling with and without replacement using
sample
R has the ability to sample with and without replacement. That
is, choose at random from a collection of things such as the numbers
1 through 6 in the dice rolling example. The sampling can be done
with replacement (like dice rolling) or without replacement (like a
lottery). By default
sample samples without replacement each object
having equal chance of being picked. You need to specify
replace=TRUE if you want to sample with
replacement. Furthermore, you can specify separate probabilities for each if desired.
Here are some examples
## Roll a die
> sample(1:6,10,replace=TRUE)
[1] 5 1 5 3 3 4 5 4 2 1 # no sixes!
## toss a coin
> sample(c("H","T"),10,replace=TRUE)
[1] "H" "H" "T" "T" "T" "T" "H" "H" "T" "T"
## pick 6 of 54 (a lottery)
> sample(1:54,6) # no replacement
[1] 6 39 23 35 25 26
## pick a card. (Fancy! Uses paste, rep)
> cards = paste(rep(c("A",2:10,"J","Q","K"),4),c("H","D","S","C"))
> sample(cards,5) # a pair of jacks, no replacement
[1] "J D" "5 C" "A S" "2 D" "J H"
## roll 2 die. Even fancier
> dice = as.vector(outer(1:6,1:6,paste))
> sample(dice,5,replace=TRUE) # replace when rolling dice
[1] "1 1" "4 1" "6 3" "4 4" "2 6"
The last two illustrate things that can be done with a little typing
and a lot of thinking using the fun commands
paste for pasting
together strings,
rep for repeating things and
outer for
generating all possible products.
6.3 A bootstrap sample
Bootstrapping is a method of sampling from a data set to make
statistical inference. The intuitive idea is that by sampling, one
can get an idea of the variability in the data. The process involves
repeatedly selecting samples and then forming a statistic. Here is a
simple illustration on obtaining a sample.
The built in data set
faithful has a variable ``eruptions''
that measures the time between eruptions at Old Faithful. It has an
unusual distribution. A bootstrap sample is just a sample with
replacement from the given values. It can be found as follows
> data(faithful) # part of R's base
> names(faithful) # find the names for faithful
[1] "eruptions" "waiting"
> eruptions = faithful[['eruptions']] # or attach and detach faithful
> sample(eruptions,10,replace=TRUE)
[1] 2.03 4.37 4.80 1.98 4.32 2.18 4.80 4.90 4.03 4.70
> hist(eruptions,breaks=25) # the dataset
## the bootstrap sample
> hist(sample(eruptions,100,replace=TRUE),breaks=25)
Figure 29: Bootstrap sample
Notice that the bootstrap sample
has a similar histogram, but
it is different (figure
29).
6.4 d, p and q functions
The
d functions were used to plot the theoretical densities above.
As with the ``
r'' functions, you need to specify the parameters, but
differently, you need to specify the
x values (not the number of
random numbers
n).
Figure 30: Illustration of 'p' and 'q' functions
The
p and
q functions are for the cumulative
distribution functions and the quantiles. As mentioned, the
distribution of a random number is specified by the probability that
the number is between
a and
b for arbitrary
a and
b,
P(
a <
X £ b). In fact, the value
F(
x) =
P(
X £ b) is enough.
The
p functions answer what is the probability that a random
variable is less than
x. Such as for a standard normal, what is the
probability it is less than .7?
> pnorm(.7) # standard normal
[1] 0.7580363
> pnorm(.7,1,1) # normal mean 1, std 1
[1] 0.3820886
Notationally, these answer
P(
Z £ .7) where
Z is a standard normal
or normal(1,1). To answer
P(
Z > .7) is also easy. You can do the
work by noting this is 1 
P(
Z £ .7) or let
R do the work,
by specifying
lower.tail=F as in:
> pnorm(.7,lower.tail=F)
[1] 0.2419637
The
q function are inverse to this. They ask, what value corresponds
to a given probability. This the quantile or point in the data that
splits it accordingly. For example, what value of
z has .75 of the
area to the right for a standard normal? (This is
Q_{3})
> qnorm(.75)
[1] 0.6744898
Notationally, this is finding
z which solves 0.75 =
P(
Z £ z).
6.5 Standardizing, scale and z scores
To standardize a random variable you subtract the mean and then
divide by the standard deviation. That is
To do so requires knowledge of the mean and standard deviation.
You can also standardize a sample. There is a convenient function
scale that will do this for you. This will make your
sample have mean 0 and standard deviation 1. This is useful for comparing
random variables which live on different scales.
Normal random variables are often standardized as the distribution
of the standardized normal variable is again normal with mean 0 and
variance 1. (The ``standard'' normal.) The
zscore of a normal
number is the value of it after standardizing.
If we have normal data with mean 100 and standard deviation 16 then the
following will find the
zscores
> x = rnorm(5,100,16)
>
> x
[1] 93.45616 83.20455 64.07261 90.85523 63.55869
> z = (x100)/16
> z
[1] 0.4089897 1.0497155 2.2454620 0.5715479 2.2775819
The
zscore is used to look up the probability of being to the
right of the value of
x for the given random variable. This way
only one table of normal numbers is needed. With
R, this is not
necessary. We can use the
pnorm function directly
> pnorm(z)
[1] 0.34127360 0.14692447 0.01236925 0.28381416 0.01137575
> pnorm(x,100,16) # enter in parameters
[1] 0.34127360 0.14692447 0.01236925 0.28381416 0.01137575
6.6 Problems

6.1
 Generate 10 random numbers from a uniform distribution on [0,10].
Use R to find the maximum and minimum values.x
 6.2
 Generate 10 random normal numbers with mean 5 and standard deviation
5 (normal(5,5)). How many are less than 0? (Use R)
 6.3
 Generate 100 random normal numbers with mean 100 and standard
deviation 10. How many are 2 standard deviations from the mean
(smaller than 80 or bigger than 120)?
 6.4
 Toss a fair coin 50 times (using R). How many heads do you have?
 6.5
 Roll a ``die'' 100 times. How many 6's did you see?
 6.6
 Select 6 numbers from a lottery containing 49 balls. What is the
largest number? What is the smallest? Answer these using R.
 6.7
 For normal(0,1), find a number z^{*} solving P(Z £ z^{*}) =
.05 (use qnorm).
 6.8
 For normal(0,1), find a number z^{*} solving P(z^{*} £ Z £ z^{*}) =
.05 (use qnorm and symmetry).
 6.9
 How much area (probability) is to the right of 1.5 for a
normal(0,2)?
 6.10
 Make a histogram of 100 exponential numbers with mean 10.
Estimate the median. Is it more or less than the mean?
 6.11
 Can you figure out what this R command does?
> rnorm(5,mean=0,sd=1:5)
 6.12
 Use R to pick 5 cards from a deck of 52. Did you get a pair
or better? Repeat until you do. How long did it take?
Copyright © John Verzani, 20012. All rights reserved.