05-RandomVar.Rmd

# Random Variables {#ChapRandomVar}

## Student Learning Objective

This section introduces some important examples of random variables. The
distributions of these random variables emerge as mathematical models of
real-life settings. In two of the examples the sample space is composed
of integers. In the other two examples the sample space is made of
continuum of values. For random variables of the latter type one may use
the density, which is a type of a histogram, in order to describe the
distribution.

By the end of the chapter the student should:

-   Identify the Binomial, Poisson, Uniform, and Exponential random
    variables, relate them to real life situations, and memorize their
    expectations and variances.

-   Relate the plot of the density/probability function and the
    cumulative probability function to the distribution of a random
    variable.

-   Become familiar with the `R` functions that produce the
    density/probability of these random variables and their cumulative
    probabilities.

-   Plot the density and the cumulative probability function of a random
    variable and compute probabilities associated with random variables.

## Discrete Random Variables

In the previous chapter we introduced the notion of a random variable. A
random variable corresponds to the outcome of an observation or a
measurement prior to the actual making of the measurement. In this
context one can talk of all the values that the measurement may
potentially obtain. This collection of values is called the *sample
space*. To each value in the sample space one may associate the
*probability* of obtaining this particular value. Probabilities are like
relative frequencies. All probabilities are positive and the sum of the
probabilities that are associated with all the values in the sample
space is equal to one.

A random variable is defined by the identification of its sample space
and the probabilities that are associated with the values in the sample
space. For each type of random variable we will identify first the
sample space — the values it may obtain — and then describe the
probabilities of the values. Examples of situations in which each type
of random variable may serve as a model of a measurement will be
provided. The `R` system provides functions for the computation of
probabilities associated with specific types of random variables. We
will use these functions in this and in proceeding chapters in order to
carry out computations associated with the random variables and in order
to plot their distributions.

The distribution of a random variable, just like the distribution of
data, can be characterized using numerical summaries. For the latter we
used summaries such as the mean and the sample variance and standard
deviation. The mean is used to describe the central location of the
distribution and the variance and standard deviation are used to
characterize the total spread. Parallel summaries are used for random
variable. In the case of a random variable the name *expectation* is
used for the central location of the distribution and the *variance* and
the *standard deviation* (the square root of the variation) are used to
summarize the spread. In all the examples of random variables we will
identify the expectation and the variance (and, thereby, also the
standard deviation).

Random variables are used as probabilistic models of measurements.
Theoretical considerations are used in many cases in order to define
random variables and their distribution. A random variable for which the
values in the sample space are separated from each other, say the values
are integers, is called a *discrete random variable*. In this section we
introduce two important integer-valued random variables: The *Binomial*
and the *Poisson* random variables. These random variables may emerge as
models in contexts where the measurement involves counting the number of
occurrences of some phenomena.

Many other models, apart from the Binomial and Poisson, exist for
discrete random variables. An example of such model, the
Negative-Binomial model, will be considered in
Section \@ref(sec:RVarExercises). Depending on the specific context that
involves measurements with discrete values, one may select the Binomial,
the Poisson, or one of these other models to serve as a theoretical
approximation of the distribution of the measurement.

### The Binomial Random Variable

The Binomial random variable is used in settings in which a trial that
has two possible outcomes is repeated several times. Let us designate
one of the outcomes as “Success" and the other as “Failure". Assume that
the probability of success in each trial is given by some number $p$
that is larger than 0 and smaller than 1. Given a number $n$ of repeats
of the trial and given the probability of success, the actual number of
trials that will produce “Success" as their outcome is a random
variable. We call such random variable *Binomial*. The fact that a
random variable $X$ has such a distribution is marked by the expression:
“$X \sim \mathrm{Binomial}(n,p)$".

As an example consider tossing 10 coins. Designate “Head" as success and
“Tail" as failure. For fair coins the probability of “Head" is $1/2$.
Consequently, if $X$ is the total number of “Heads" then
$X \sim \mathrm{Binomial}(10,0.5)$, where $n=10$ is the number of trials
and $p=0.5$ is the probability of success in each trial.

It may happen that all 10 coins turn up “Tail". In this case $X$ is
equal to 0. It may also be the case that one of the coins turns up
“Head" and the others turn up “Tail". The random variable $X$ will
obtain the value 1 in such a case. Likewise, for any integer between 0
and 10 it may be the case that the number of “Heads" that turn up is
equal to that integer with the other coins turning up “Tail". Hence, the
sample space of $X$ is the set of integers $\{0, 1, 2, \ldots, 10\}$.
The probability of each outcome may be computed by an appropriate
mathematical formula that will not be discussed here[^5_1].

The probabilities of the various possible values of a Binomial random
variable may be computed with the aid of the `R` function “`dbinom`"
(that uses the mathematical formula for the computation). The input to
this function is a sequence of values, the value of $n$, and the value
of $p$. The output is the sequence of probabilities associated with each
of the values in the first input.

```{r RVar2, fig.cap='The Binomial(10,0.5) Distribution', echo=FALSE, message=FALSE,warning=FALSE, out.width = '60%', fig.align = "center"}
plot(0:10, dbinom(0:10, 10,.5), type="h", ylab ="Probability", xlab="x", lwd=3)
```

For example, let us use the function in order to compute the probability
that the given Binomial obtains an odd value. A sequence that contains
the odd values in the Binomial sample space can be created with the
expression “`c(1,3,5,7,9)`". This sequence can serve as the input in the
first argument of the function “`dbinom`". The other arguments are
“`10`" and “`0.5`", respectively:

```{r}
dbinom(c(1,3,5,7,9),10,0.5)
```

Observe that the output of the function is a sequence of the same length
as the first argument. This output contains the Binomial probabilities
of the values in the first argument. In order to obtain the probability
of the event $\{\mbox{X is odd}\}$ we should sum up these probabilities,
which we can do by applying the function “`sum`" to the output of the
function that computes the Binomial probabilities:

```{r}
sum(dbinom(c(1,3,5,7,9),10,0.5))
```

Observe that the probability of obtaining an odd value in this specific
case is equal to one half.

Another example is to compute all the probabilities of all the potential
values of a $\mathrm{Binomial}(10,0.5)$ random variable:

```{r}
x <- 0:10
dbinom(x,10,0.5)
```

The expression “`start.value:end.value`" produces a sequence of numbers
that initiate with the number “`start.value`" and proceeds in jumps of
size one until reaching the number “`end.value`". In this example,
“`0:10`" produces the sequence of integers between 0 and 10, which is
the sample space of the current Binomial example. Entering this sequence
as the first argument to the function “`dbinom`" produces the
probabilities of all the values in the sample space.

One may display the distribution of a discrete random variable with a
bar plot similar to the one used to describe the distribution of data.
In this plot a vertical bar representing the probability is placed above
each value of the sample space. The hight of the bar is equal to the
probability. A bar plot of the $\mathrm{Binomial}(10,0.5)$ distribution
is provided in Figure \@ref(fig:RVar2).

Another useful function is “`pbinom`", which produces the cumulative
probability of the Binomial:

```{r}
pbinom(x,10,0.5)
cumsum(dbinom(x,10,0.5))
```

The output of the function “`pbinom`" is the cumulative probability
$\Prob(X \leq x)$ that the random variable is less than or equal to the
input value. Observe that this cumulative probability is obtained by
summing all the probabilities associated with values that are less than
or equal to the input value. Specifically, the cumulative probability at
$x=3$ is obtained by the summation of the probabilities at $x=0$, $x=1$,
$x=2$, and $x=3$:

$$\Prob(X \leq 3) = 0.0009765625 + 0.009765625 + 0.0439453125 + 0.1171875 = 0.171875$$
The numbers in the sum are the first 4 values from the output of the
function “`dbinom(x,10,0.5)`", which computes the probabilities of the
values of the sample space.

In principle, the expectation of the Binomial random variable, like the
expectation of any other (discrete) random variable is obtained from the
application of the general formulae:

$$\Expec(X) = \sum_x \big(x \times \Prob(X = x)\big)\;,\quad \Var(X) = \sum_x\big( (x-\Expec(X))^2 \times \Prob(x)\big)\;.$$
However, in the specific case of the Binomial random variable, in which
the probability $\Prob(X = x)$ obeys the specific mathematical formula
of the Binomial distribution, the expectation and the variance reduce to
the specific formulae: 

$$\Expec(X) = n p\;,\quad \Var(X) = n p(1-p)\;.$$
Hence, the expectation is the product of the number of trials $n$ with
the probability of success in each trial $p$. In the variance the number
of trials is multiplied by the product of a probability of success ($p$)
with the probability of a failure ($1-p$).

As illustration, let us compute for the given example the expectation
and the variance according to the general formulae for the computation
of the expectation and variance in random variables and compare the
outcome to the specific formulae for the expectation and variance in the
Binomial distribution:

```{r}
X.val <- 0:10
P.val <- dbinom(X.val,10,0.5)
EX <- sum(X.val*P.val)
EX
sum((X.val-EX)^2*P.val)
```

This agrees with the specific formulae for Binomial variables, since
$10 \times 0.5 = 5$ and $10 \times 0.5 \times(1-0.5) = 2.5$.

Recall that the general formula for the computation of the expectation
calls for the multiplication of each value in the sample space with the
probability of that value, followed by the summation of all the
products. The object “`X.val`" contains all the values of the random
variable and the object “`P.val`" contains the probabilities of these
values. Hence, the expression “`X.val*P.val`" produces the product of
each value of the random variable times the probability of that value.
Summation of these products with the function “`sum`" gives the
expectation, which is saved in an object that is called “`EX`".

The general formula for the computation of the variance of a random
variable involves the product of the squared deviation associated with
each value with the probability of that value, followed by the summation
of all products. The expression “`(X.val-EX)^ 2`" produces the sequence
of squared deviations from the expectation for all the values of the
random variable. Summation of the product of these squared deviations
with the probabilities of the values (the outcome of
“`(X.val-EX)^2*P.val`") gives the variance.

When the value of $p$ changes (without changing the number of trials
$n$) then the probabilities that are assigned to each of the values of
the sample space of the Binomial random variable change, but the sample
space itself does not. For example, consider rolling a die 10 times and
counting the number of times that the face 3 was obtained. Having the
face 3 turning up is a “Success". The probability $p$ of a success in
this example is 1/6, since the given face is one out of 6 equally likely
faces. The resulting random variable that counts the total number of
success in 10 trials has a $\mathrm{Binomial}(10,1/6)$ distribution. The
sample space is yet again equal to the set of integers
$\{0,1, \ldots,10\}$. However, the probabilities of values are
different. These probabilities can again be computes with the aid of the
function “`dbinom`":

```{r}
dbinom(x,10,1/6)
```

In this case smaller values of the random variable are assigned higher
probabilities and larger values are assigned lower probabilities..

```{r RVar3, fig.cap='The Binomial Distribution for Various Probability of "Success"" $p$', echo=FALSE, message=FALSE,warning=FALSE, out.width = '60%', fig.align = "center"}
plot(0:10, dbinom(0:10, 10,.5), type="h", ylab ="Probability", xlab="x", lwd=c(3), ylim=c(0,.4))
lines((0:10)-.15, dbinom(0:10, 10,1/6), col="red", type="h", lwd=c(3))
lines((0:10)+.15, dbinom(0:10, 10,.6), col="green", type="h", lwd=c(3))
legend(7.5, .39, c("p=0.5", "p=1/6", "p=0.6"), col = c(1, 2, 3), lty = c(1, 1, 1), lwd=c(3,3,3))
```

In Figure \@ref(fig:RVar3) the probabilities for
$\mathrm{Binomial}(10,1/6)$, the $\mathrm{Binomial}(10,1/2)$, and the
$\mathrm{Binomial}(10,0.6)$ distributions are plotted side by side. In
all these 3 distributions the sample space is the same, the integers
between 0 and 10. However, the probabilities of the different values
differ. (Note that all bars should be placed on top of the integers. For
clarity of the presentation, the bars associated with the
$\mathrm{Binomial}(10,1/6)$ are shifted slightly to the left and the
bars associated with the $\mathrm{Binomial}(10,0.6)$ are shifted
slightly to the right.)

The expectation of the $\mathrm{Binomial}(10,0.5)$ distribution is equal
to $10 \times 0.5 = 5$. Compare this to the expectation of the
$\mathrm{Binomial}(10,1/6)$ distribution, which is
$10 \times (1/6) = 1.666667$ and to the expectation of the
$\mathrm{Binomial}(10,0.6)$ distribution which equals
$10 \times 0.6 = 6$.

The variance of the $\mathrm{Binomial}(10,0.5)$ distribution is
$10 \times 0.5 \times 0.5 = 2.5$. The variance when $p=1/6$ is
$10 \times (1/6) \times (5/6) = 1.388889$ and the variance when $p=0.6$
is $10 \times 0.6 \times 0.4 = 2.4$.

```{example, exrvar1}
As an application of the Binomial distribution
consider a pre-election poll. A candidate is running for office and is
interested in knowing the percentage of support in the general
population in its candidacy. Denote the probability of support by $p$.
In order to estimate the percentage a sample of size 300 is selected
from the population. Let $X$ be the count of supporters in the sample. A
natural model for the distribution of $X$ is the
$\mathrm{Binomial}(300,p)$ distribution, since each subject in the
sample may be a supporter (“Success") or may not be a supporter
(“Failure"). The probability that a subject supports the candidate is
$p$ and there are $n=300$ subjects in the sample.
```

```{example, exrvar2}
As another example consider the procedure for
quality control that is described in Discussion Forum of
Chapter \@ref(ChapProbability). According to the procedure 20 items are
tested and the number of faulty items is recorded. If $p$ is the
probability that an item is identified as faulty then the distribution
of the total number of faulty items may be modeled by the
$\mathrm{Binomial}(20,p)$ distribution.
```

In both examples one may be interested in making statements on the
probability $p$ based on the sample. Statistical inference relates the
actual count obtained in the sample to the theoretical Binomial
distribution in order to make such statements.

### The Poisson Random Variable

The *Poisson* distribution is used as an approximation of the total
number of occurrences of rare events. Consider, for example, the
Binomial setting that involves $n$ trials with $p$ as the probability of
success of each trial. Then, if $p$ is small but $n$ is large then the
number of successes $X$ has, approximately, the Poisson distribution.

The sample space of the Poisson random variable is the unbounded
collection of integers: $\{0,1,2, \ldots\}$. Any integer value is
assigned a positive probability. Hence, the Poisson random variable is a
convenient model when the maximal number of occurrences of the events in
a-priori unknown or is very large. For example, one may use the Poisson
distribution to model the number of phone calls that enter a switchboard
in a given interval of time or the number of malfunctioning components
in a shipment of some product.

```{r RVar4, fig.cap='The Poisson(2) Distribution', echo=FALSE, message=FALSE,warning=FALSE, out.width = '60%', fig.align = "center"}
plot(0:10, dpois(0:10, 2), type="h", ylab ="Probability", xlab="x", lwd=3)
```

The Binomial distribution was specified by the number of trials $n$ and
probability of success in each trial $p$. The Poisson distribution is
specified by its expectation, which we denote by $\lambda$. The
expression “$X \sim \mathrm{Poisson}(\lambda)$" states that the random
variable $X$ has a Poisson distribution[^5_2] with expectation
$\Expec(X) = \lambda$. The function “`dpois`" computes the probability,
according to the Poisson distribution, of values that are entered as the
first argument to the function. The expectation of the distribution is
entered in the second argument. The function “`ppois`" computes the
cumulative probability. Consequently, we can compute the probabilities
and the cumulative probabilities of the values between 0 and 10 for the
$\mathrm{Poisson}(2)$ distribution via:

```{r}
x <- 0:10
dpois(x,2)
ppois(x,2)
```

The probability function of the Poisson distribution with $\lambda = 2$,
in the range between 0 and 10, is plotted in
Figure \@ref(fig:RVar4). Observe that in this example probabilities
of the values 8 and beyond are very small. As a matter of fact, the
cumulative probability at $x=7$ (the 8th value in the output of
“`ppois(x,2)`") is approximately 0.999, out of the total cumulative
probability of 1.000, leaving a total probability of about 0.001 to be
distributed among all the values larger than 7.

```{r RVar5, fig.cap='The Poisson Distribution for Various Values of $\\lambda$', echo=FALSE, message=FALSE,warning=FALSE, out.width = '60%', fig.align = "center"}
plot(0:10, dpois(0:10,1), type="h", ylab ="Probability", xlab="x", lwd=c(3),ylim=c(0, .6))
lines((0:10)-.15, dpois(0:10,.5), col=2, type="h", lwd=c(3))
lines((0:10)+.15, dpois(0:10,2), col=3, type="h", lwd=c(3))
legend(6.5, .6, c("lambda=0.5", "lambda=1", "lambda=2"), col = c(2, 1, 3), lty = c(1, 1, 1), lwd=c(3,3,3))
```

Let us compute the expectation of the given Poisson distribution:

```{r}
X.val <- 0:10
P.val <- dpois(X.val,2)
sum(X.val*P.val)
```

Observe that the outcome is almost, but not quite, equal to $2.00$,
which is the actual value of the expectation. The reason for the
inaccuracy is the fact that we have based the computation in `R` on the
first 11 values of the distribution only, instead of the infinite
sequence of values. A more accurate result may be obtained by the
consideration of the first 101 values:

```{r}
X.val <- 0:100
P.val <- dpois(X.val,2)
EX <- sum(X.val*P.val)
EX
sum((X.val-EX)^2*P.val)
```

In the last expression we have computed the variance of the Poisson
distribution and obtained that it is equal to the expectation. This
results can be validated mathematically. For the Poisson distribution it
is always the case that the variance is equal to the expectation, namely
to $\lambda$: $$\Expec(X) = \Var(X) = \lambda\;.$$

In Figure \@ref(fig:RVar5) you may find the probabilities of the
Poisson distribution for $\lambda = 0.5$, $\lambda = 1$ and
$\lambda = 2$. Notice once more that the sample space is the same for
all the Poisson distributions. What varies when we change the value of
$\lambda$ are the probabilities. Observe that as $\lambda$ increases
then probability of larger values increases as well.

```{example, exrvar3}
A radio active element decays by the release of
subatomic particles and energy. The decay activity is measured in terms
of the number of decays per second in a unit mass. A typical model for
the distribution of the number of decays is the Poisson distribution.
Observe that the number of decays in a second is a integer and, in
principle, it may obtain any integer value larger or equal to zero. The
event of a radio active decay of an atom is a relatively rare event.
Therefore, the Poisson model is likely to fit this phenomena[^5_3].
```

```{example, exrvar4}
Consider an overhead power line suspended between
two utility poles. During rain, drops of water may hit the power line.
The total number of drops that hit the line in a one minute period may
be modeled by a Poisson random variable.
```

## Continuous Random Variable {#RandomVar_5}

Many types of measurements, such as height, weight, angle, temperature,
etc., may in principle have a continuum of possible values. Continuous
random variables are used to model uncertainty regarding future values
of such measurements.

The main difference between discrete random variables, which is the type
we examined thus far, and continuous random variable, that are added now
to the list, is in the sample space, i.e., the collection of possible
outcomes. The former type is used when the possible outcomes are
separated from each other as the integers are. The latter type is used
when the possible outcomes are the entire line of real numbers or when
they form an interval (possibly an open ended one) of real numbers.

The difference between the two types of sample spaces implies
differences in the way the distribution of the random variables is being
described. For discrete random variables one may list the probability
associated with each value in the sample space using a table, a formula,
or a bar plot. For continuous random variables, on the other hand,
probabilities are assigned to intervals of values, and not to specific
values. Thence, densities are used in order to display the distribution.

Densities are similar to histograms, with areas under the plot
corresponding to probabilities. We will provide a more detailed
description of densities as we discuss the different examples of
continuous random variables.

In continuous random variables integration replaces summation and the
density replaces the probability in the computation of quantities such
as the probability of an event, the expectation, and the variance.

Hence, if the expectation of a discrete random variable is given in the
formula $\Expec(X) = \sum_x \big(x \times \Prob(x)\big)$, which involves
the summation over all values of the product between the value and the
probability of the value, then for continuous random variable the
definition becomes: 

$$\Expec(X) = \int \big(x \times f(x)\big)dx\;,$$
where $f(x)$ is the density of $X$ at the value $x$. Therefore, in the
expectation of a continuous random variable one multiplies the value by
the density at the value. This product is then integrated over the
sample space.

Likewise, the formula
$\Var(X) = \sum_x\big( (x-\Expec(X))^2 \times \Prob(x)\big)$ for the
variance is replaced by:

$$\Var(X) =\int\big((x-\Expec(X))^2 \times f(x) \big) dx\;.$$
Nonetheless, the intuitive interpretation of the expectation as the
central value of the distribution that identifies the location and the
interpretation of the standard deviation (the square root of the
variance) as the summary of the total spread of the distribution is
still valid.

In this section we will describe two types of continuous random
variables: Uniform and Exponential. In the next chapter another example
– the Normal distribution – will be introduced.

### The Uniform Random Variable

The Uniform distribution is used in order to model measurements that may
have values in a given interval, with all values in this interval
equally likely to occur.

For example, consider a random variable $X$ with the Uniform
distribution over the interval $[3,7]$, denoted by
“$X \sim \mathrm{Uniform}(3,7)$". The density function at given values
may be computed with the aid of the function “`dunif`". For instance let
us compute the density of the $\mathrm{Uniform}(3,7)$ distribution over
the integers $\{0, 1, \ldots, 10\}$:

```{r}
dunif(0:10,3,7)
```

Notice that for the values 0, 1, and 2, and the values 8, 9 and 10 that
are outside of the interval the density is equal to zero, indicating
that such values cannot occur in the given distribution. The values of
the density at integers inside the interval are positive and equal to
each other. The density is not restricted to integer values. For
example, at the point $4.73$ we get that the density is positive and of
the same height:

```{r}
dunif(4.73,3,7)
```

```{r RVar6, fig.cap='The Uniform(3,7) Distribution', echo=FALSE, message=FALSE,warning=FALSE, out.width = '60%', fig.align = "center"}

x <- seq(0,10,length=1000)
den <- dunif(x,3,7)
plot(x,den, type="l", ylab="Density",xlab="x", lwd=2)
polygon(c(3,3,4.73,4.73),c(0,.25,.25,0), density=10)
text(1,.20,"P(X <= 4.73)")
text(6,.06,"x = 4.73")
library(shape)
Arrows(4, .08,1.2,.18)
Arrows(4.73, 0,5.6,.04)

```

A plot of the $\mathrm{Uniform}(3,7)$ density is given in
Figure \@ref(fig:RVar6) in the form of a solid line. Observe that
the density is positive over the interval $[3,7]$ where its height is
1/4. Area under the curve in the density corresponds to probability.
Indeed, the fact that the total probability is one is reflected in the
total area under the curve being equal to 1. Over the interval $[3,7]$
the density forms a rectangle. The base of the rectangle is the length
of the interval $7-3=4$. The height of the rectangle is thus equal to
1/4 in order to produce a total area of $4 \times (1/4) = 1$.

The function “`punif`" computes the cumulative probability of the
uniform distribution. The probability $\Prob(X \leq 4.73)$, for
$X \sim \mathrm{Uniform}(3,7)$, is given by:

```{r}
punif(4.73,3,7)
```

This probability corresponds to the marked area to the left of the point
$x = 4.73$ in Figure \@ref(fig:RVar6). This area of the marked
rectangle is equal to the length of the base 4.73 - 3 = 1.73, times the
height of the rectangle 1/(7-3) = 1/4. Indeed:

```{r}
(4.73-3)/(7-3)
```

is the area of the marked rectangle and is equal to the probability.

Let us use `R` in order to plot the density and the cumulative
probability functions of the Uniform distribution. We produce first a
large number of points in the region we want to plot. The points are
produced with aid of the function “`seq`". The output of this function
is a sequence with equally spaced values. The starting value of the
sequence is the first argument in the input of the function and the last
value is the second argument in the input. The argument “`length=1000`"
sets the length of the sequence, 1,000 values in this case:

```{r, out.width = '60%', fig.align = "center"}
x <- seq(0,10,length=1000)
den <- dunif(x,3,7)
plot(x,den)
```

The object “`den`" is a sequence of length 1,000 that contains the
density of the $\mathrm{Uniform}(3,7)$ evaluated over the values of
“`x`". When we apply the function “`plot`" to the two sequences we get a
scatter plot of the 1,000 points.

A scatter plot is a plot of points. Each point in the scatter plot is
identify by its horizontal location on the plot (its “$x$" value) and by
its vertical location on the plot (its $y$ value). The horizontal value
of each point in the plot is determined by the first argument to the
function “`plot`" and the vertical value is determined by the second
argument. For example, the first value in the sequence “`x`" is 0. The
value of the Uniform density at this point is 0. Hence, the first value
of the sequence “`den`" is also 0. A point that corresponds to these
values is produced in the plot. The horizontal value of the point is 0
and the vertical value is 0. In a similar way the other 999 points are
plotted. The last point to be plotted has a horizontal value of 10 and a
vertical value of 0.

The number of points that are plotted is large and they overlap each
other in the graph and thus produce an impression of a continuum. In
order to obtain nicer looking plots we may choose to connect the points
to each other with segments and use smaller points. This may be achieved
by the addition of the argument “`type=l`", with the letter `l` for
line, to the plotting function:

```{r, out.width = '60%', fig.align = "center"}
plot(x,den,type="l")
```

The cumulative probability of the $\mathrm{Uniform}(3,7)$ is produced by the code:

```{r, out.width = '60%', fig.align = "center"}
cdf <- punif(x,3,7)
plot(x,cdf,type="l")
```

One can think of the density of the Uniform as an histogram[^5_4]. The
expectation of a Uniform random variable is the middle point of it’s
histogram. Hence, if $X \sim \mathrm{Uniform}(a,b)$ then:

$$\Expec(X) = \frac{a+b}{2}\;.$$ For the $X \sim \mathrm{Uniform}(3,7)$
distribution the expectation is $\Expec(X)= (3+7)/2 = 5$. Observe that 5
is the center of the Uniform density in Plot \@ref(fig:RVar6).

It can be shown that the variance of the $\mathrm{Uniform}(a,b)$ is
equal to 

$$\Var(X) = \frac{(b-a)^2}{12}\;,$$ with the standard deviation
being the square root of this value. Specifically, for
$X \sim \mathrm{Uniform}(3,7)$ we get that
$\Var(X) = (7-3)^2/12 = 1.333333$. The standard deviation is equal to
$\sqrt{1.333333} = 1.154701$.

```{example, exrvar5}
In Example \@ref(exm:exrvar4) we considered
rain drops that hit an overhead power line suspended between two utility
poles. The [**number**]{} of drops that hit the line can be modeled
using the Poisson distribution. The [**position**]{} between the two
poles where a rain drop hits the line can be modeled by the Uniform
distribution. The rain drop can hit any position between the two utility
poles. Hitting one position along the line is as likely as hitting any
other position.
```

```{example, exrvar6}
Meiosis is the process in which a diploid cell
that contains two copies of the genetic material produces an haploid
cell with only one copy (sperms or eggs, depending on the sex). The
resulting molecule of genetic material is linear molecule (chromosome)
that is composed of consecutive segments: a segment that originated from
one of the two copies followed by a segment from the other copy and vice
versa. The border points between segments are called points of
crossover. The Haldane model for crossovers states that the position of
a crossover between two given loci on the chromosome corresponds to the
Uniform distribution and the total number of crossovers between these
two loci corresponds to the Poisson distribution.
```

### The Exponential Random Variable

The Exponential distribution is frequently used to model times between
events. For example, times between incoming phone calls, the time until
a component becomes malfunction, etc. We denote the Exponential
distribution via “$X \sim \mathrm{Exponential}(\lambda)$", where
$\lambda$ is a parameter that characterizes the distribution and is
called the *rate* of the distribution. The overlap between the parameter
used to characterize the Exponential distribution and the one used for
the Poisson distribution is deliberate. The two distributions are
tightly interconnected. As a matter of fact, it can be shown that if the
distribution between occurrences of a phenomena has the Exponential
distribution with rate $\lambda$ then the total number of the
occurrences of the phenomena within a unit interval of time has a
$\mathrm{Poisson}(\lambda)$ distribution.

The sample space of an Exponential random variable contains all
non-negative numbers. Consider, for example,
$X \sim \mathrm{Exponential}(0.5)$. The density of the distribution in
the range between 0 and 10 is presented in Figure \@ref(fig:RVar7).
Observe that in the Exponential distribution smaller values are more
likely to occur in comparison to larger values. This is indicated by the
density being larger at the vicinity of 0. The density of the
exponential distribution given in the plot is positive, but hardly so,
for values larger than 10.

```{r RVar7, fig.cap='The Exponential(0.5) Distribution', echo=FALSE, message=FALSE,warning=FALSE, out.width = '60%', fig.align = "center"}

x.c = seq(0,10,.1)
plot(c(0,x.c),c(0,dexp(x.c, .5)), type="l", ylab="Density",xlab="x")
x.c = seq(2,6,.1)
polygon(c(2,x.c,6),c(0,dexp(x.c, .5),0), density=10)
text(5.2,.22,"P(2 < x < 6)")
library(shape)
Arrows(4,.04,4.8, .18)

```

The density of the Exponential distribution can be computed with the aid
of the function “`dexp`"[^5_5]. The cumulative probability can be computed
with the function “`pexp`". For illustration, assume
$X \sim \mathrm{Exponential}(0.5)$. Say one is interested in the
computation of the probability $\Prob(2 < X \leq 6)$ that the random
variable obtains a value that belongs to the interval $(2,6]$. The
required probability is indicated as the marked area in
Figure \@ref(fig:RVar7). This area can be computed as the difference
between the probability $\Prob(X \leq 6)$, the area to the left of 6,
and the probability $\Prob(X \leq 2)$, the area to the left of 2:

```{r}
pexp(6,0.5) - pexp(2,0.5)
```

The difference is the probability of belonging to the interval, namely
the area marked in the plot.

The expectation of $X$, when $X \sim \mathrm{Exponential}(\lambda)$, is
given by the equation: 

$$\Expec(X) = 1/\lambda\;,$$ and the variance is
given by: 

$$\Var(X) =1/\lambda^2\;.$$ The standard deviation is the
square root of the variance, namely $1/\lambda$. Observe that the larger
is the rate the smaller are the expectation and the standard deviation.

```{r RVar8, fig.cap='The Exponential Distribution for Various Values of $\\lambda$', echo=FALSE, message=FALSE,warning=FALSE, out.width = '60%', fig.align = "center"}
x.coord = seq(0.01,10,.1)
plot(c(0,x.coord), c(0,dexp(x.coord,1)), ylab ="Probability", lty=1, xlab="x",ylim=c(0, 2), type="l")
lines(c(0,x.coord)-.03, c(0,dexp(x.coord,.5)), col=2)
lines(c(0,x.coord)+.03, c(0,dexp(x.coord,2)), col=3)
legend(7.4, 2, c("lambda=0.5", "lambda=1", "lambda=2"), col = c(2, 1, 3), lty = c(1, 1, 1))
```

In Figure \@ref(fig:RVar8) the densities of the Exponential
distribution are plotted for $\lambda = 0.5$, $\lambda = 1$, and
$\lambda = 2$. Notice that with the increase in the value of the
parameter then the values of the random variable tends to become
smaller. This inverse relation makes sense in connection to the Poisson
distribution. Recall that the Poisson distribution corresponds to the
total number of occurrences in a unit interval of time when the time
between occurrences has an Exponential distribution. A larger
expectation $\lambda$ of the Poisson corresponds to a larger number of
occurrences that are likely to take place during the unit interval of
time. The larger is the number of occurrences the smaller are the time
intervals between occurrences.

```{example, exrvar7}
Consider Examples \@ref(exm:exrvar4)
and \@ref(exm:exrvar5) that deal with rain dropping on a power line.
The times between consecutive hits of the line may be modeled by the
Exponential distribution. Hence, the time to the first hit has an
Exponential distribution. The time between the first and the second hit
is also Exponentially distributed, and so on.
```

```{example, exrvar8}
Return to Example \@ref(exm:exrvar3) that
deals with the radio activity of some element. The total count of decays
per second is model by the Poisson distribution. The times between radio
active decays is modeled according to the Exponential distribution. The
rate $\lambda$ of that Exponential distribution is equal to the
expectation of the total count of decays in one second, i.e. the
expectation of the Poisson distribution.
```

## Exercises {#sec:RVarExercises}

```{exercise} 
A particular measles vaccine produces a reaction (a
fever higher that 102 Fahrenheit) in each vaccinee with probability of
0.09. A clinic vaccinates 500 people each day.

1.  What is the expected number of people that will develop a reaction
    each day?

2.  What is the standard deviation of the number of people that will
    develop a reaction each day?

3.  In a given day, what is the probability that more than 40 people
    will develop a reaction?

4.  In a given day, what is the probability that the number of people
    that will develop a reaction is between 50 and 45 (inclusive)?

```


```{r RVar10, fig.cap='Bar Plots of the Negative-Binomial Distribution', echo=FALSE, message=FALSE,warning=FALSE, out.width = '60%', fig.align = "center"}
par(mfrow=c(3,1), oma=c(0,0,0,0), mar=c(2,4,1,1),cex=.71)

plot(0:10, dnbinom(0:10, 4,.5), type="h", lwd=3, ylim=c(0,.3), ylab="", xlab="",yaxt="n")
axis(2,las=2)
plot(0:10, dnbinom(0:10, 8,.8), type="h", lwd=3, ylim=c(0,.3), ylab="", xlab="",yaxt="n")
axis(2,las=2)
title(ylab="Probability", line=3)
plot(0:10, dnbinom(0:10, 2,.5), type="h", lwd=3, ylim=c(0,.3), ylab="", xlab="",yaxt="n")
axis(2,las=2)
title(xlab="x", line=2)

```


```{exercise}
The Negative-Binomial distribution is yet another
example of a discrete, integer valued, random variable. The sample space
of the distribution are all non-negative integers $\{0, 1, 2, \ldots\}$.
The fact that a random variable $X$ has this distribution is marked by
“$X \sim \mbox{Negative-Binomial}(r,p)$", where $r$ and $p$ are
parameters that specify the distribution.

Consider 3 random variables from the Negative-Binomial distribution:

-   $X_1 \sim \mbox{Negative-Binomial}(2,0.5)$

-   $X_2 \sim \mbox{Negative-Binomial}(4,0.5)$

-   $X_3 \sim \mbox{Negative-Binomial}(8,0.8)$

The bar plots of these random variables are presented in
Figure \@ref(fig:RVar10), re-organizer in a random order.

1.  Produce bar plots of the distributions of the random variables
    $X_1$, $X_2$, $X_3$ in the range of integers between 0 and 15 and
    thereby identify the pair of parameters that produced each one of
    the plots in Figure \@ref(fig:RVar10). Notice that the bar plots
    can be produced with the aid of the function “`plot`" and the
    function “`dnbinom(x,r,p)`", where “`x`" is a sequence of integers
    and “`r`" and “`p`" are the parameters of the distribution. Pay
    attention to the fact that you should use the argument “`type = h`"
    in the function “`plot`" in order to produce the horizontal bars.

2.  Below is a list of pairs that includes an expectation and a
    variance. Each of the pairs is associated with one of the random
    variables $X_1$, $X_2$, and $X_3$:

    1.  $\Expec(X) = 4$, $\Var(X) = 8$.

    2.  $\Expec(X) = 2$, $\Var(X) =  4$.

    3.  $\Expec(X) = 2$, $\Var(X) =  2.5$.

    Use Figure \@ref(fig:RVar10) in order to match the random
    variable with its associated pair. Do not use numerical computations
    or formulae for the expectation and the variance in the
    Negative-Binomial distribution in order to carry out the
    matching[^5_6]. Use, instead, the structure of the bar-plots.

```


## Summary

### Glossary {#glossary .unnumbered}

Binomial Random Variable:

:   The number of successes among $n$ repeats of independent trials with
    a probability $p$ of success in each trial. The distribution is
    marked as $\mathrm{Binomial}(n,p)$.

Poisson Random Variable:

:   An approximation to the number of occurrences of a rare event, when
    the expected number of events is $\lambda$. The distribution is
    marked as $\mathrm{Poisson}(\lambda)$.

Density:

:   Histogram that describes the distribution of a continuous random
    variable. The area under the curve corresponds to probability.

Uniform Random Variable:

:   A model for a measurement with equally likely outcomes over an
    interval $[a,b]$. The distribution is marked as
    $\mathrm{Uniform}(a,b)$.

Exponential Random Variable:

:   A model for times between events. The distribution is marked as
    $\mathrm{Exponential}(\lambda)$.

### Discuss in the Forum {#discuss-in-the-forum .unnumbered}

This unit deals with two types of discrete random variables, the
Binomial and the Poisson, and two types of continuous random variables,
the Uniform and the Exponential. Depending on the context, these types
of random variables may serve as theoretical models of the uncertainty
associated with the outcome of a measurement.

In your opinion, is it or is it not useful to have a theoretical model
for a situation that occurs in real life?

When forming your answer to this question you may give an example of a
situation from you own field of interest for which a random variable,
possibly from one of the types that are presented in this unit, can
serve as a model. Discuss the importance (or lack thereof) of having a
theoretical model for the situation.

For example, the Exponential distribution may serve as a model for the
time until an atom of a radio active element decays by the release of
subatomic particles and energy. The decay activity is measured in terms
of the number of decays per second. This number is modeled as having a
Poisson distribution. Its expectation is the rate of the Exponential
distribution. For the radioactive element Carbon-14
(${}^{\mbox{\tiny 14}}\mathrm{C}$) the decay rate is
$3.8394 \times 10^{-12}$ particles per second. Computations that are
based on the Exponential model may be used in order to date ancient
specimens.

### Summary of Formulas {#summary-of-formulas .unnumbered}

Discrete Random Variable:

:   

$$\begin{aligned}
        \Expec(X) &= \sum_x \big(x \times \Prob(x)\big) \\
        \Var(X) &= \sum_x\big( (x-\Expec(X))^2 \times \Prob(x)\big) \end{aligned}$$

Continuous Random Variable:

:   

$$\begin{aligned}
        \Expec(X) &= \int \big(x \times f(x)\big)dx \\
        \Var(X) &= \int\big((x-\Expec(X))^2 \times f(x) \big) dx \end{aligned}$$

Binomial:

:   $\Expec(X) = n p \;, \quad \Var(X) = n p(1-p)$

Poisson:

:   $\Expec(X) = \lambda\;, \quad \Var(X) = \lambda$

Uniform:

:   $\Expec(X) = (a+b)/2\;, \quad \Var(X)= (b-a)^2/12$

Exponential:

:   $\Expec(X) = 1/\lambda\;, \quad \Var(X)= 1/\lambda^2$

[^5_1]: If $X\sim \mathrm{Binomial}(n,p)$ then
    $\Prob(X = x) = {n \choose x} p^x (1-p)^{n-x}$, for
    $x = 0, 1, \ldots, n$.

[^5_2]: If $X \sim \mathrm{Poisson}(\lambda)$ then
    $\Prob(X=x) = e^{-\lambda}\lambda^x/x!$, for $x=0,1,2,\ldots$.

[^5_3]: The number of decays may also be considered in the
    $\mathrm{Binomial}(n,p)$ setting. The number $n$ is the total number
    of atoms in the unit mass and $p$ is the probability that an atom
    decays within the given second. However, since $n$ is very large and
    $p$ is very small we get that the Poisson distribution is an
    appropriate model for the count.

[^5_4]: If $X \sim \mathrm{Uniform}(a,b)$ then the density is
    $f(x) = 1/(b-a)$, for $a \leq x \leq b$, and it is equal to 0 for
    other values of $x$.

[^5_5]: If $X \sim \mathrm{Exponential}(\lambda)$ then the density 
    is $f(x) =\lambda e^{-\lambda x}$, for $0 \leq x$, and it is equal 
    to 0 for $x < 0$.

[^5_6]: It can be shown, or else found on the web, that if
    $X\sim \mbox{Negative-Binomial}(r,p)$ then $\Expec(X) = r(1-p)/p$
    and $\Var(X) = r(1-p)/p^2$.