In IPSUR: Introduction to Probability and Statistics Using R

Continuous Distributions {#cha-continuous-distributions}

#    IPSUR: Introduction to Probability and Statistics Using R
#    Copyright (C) 2018  G. Jay Kerns
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#    Chapter: Continuous Distributions
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# This chapter's package dependencies
library(actuar)
library(distrEx)
distroptions("WarningSim" = FALSE)   
         # switches off warnings as to (In)accuracy due to simulations
distroptions("WarningArith" = FALSE) 
         # switches off warnings as to arithmetics

The focus of the last chapter was on random variables whose support can be written down in a list of values (finite or countably infinite), such as the number of successes in a sequence of Bernoulli trials. Now we move to random variables whose support is a whole range of values, say, an interval ((a,b)). It is shown in later classes that it is impossible to write all of the numbers down in a list; there are simply too many of them.

This chapter begins with continuous random variables and the associated PDFs and CDFs The continuous uniform distribution is highlighted, along with the Gaussian, or normal, distribution. Some mathematical details pave the way for a catalogue of models.

The interested reader who would like to learn more about any of the assorted discrete distributions mentioned below should take a look at Continuous Univariate Distributions, Volumes 1 and 2 by Johnson et al [@Johnson1994], [@Johnson1995].

What do I want them to know?

• how to choose a reasonable continuous model under a variety of physical circumstances
• basic correspondence between continuous versus discrete random variables
• the general tools of the trade for manipulation of continuous random variables, integration, etc.
• some details on a couple of continuous models, and exposure to a bunch of other ones
• how to make new continuous random variables from old ones

Continuous Random Variables {#sec-continuous-random-variables}

Probability Density Functions {#sub-probability-density-functions}

Continuous random variables have supports that look like \begin{equation} S_{X}=[a,b]\mbox{ or }(a,b), \end{equation} or unions of intervals of the above form. Examples of random variables that are often taken to be continuous are:

• the height or weight of an individual,
• other physical measurements such as the length or size of an object, and
• durations of time (usually).

Every continuous random variable (X) has a probability density function (PDF) denoted (f_{X}) associated with it[^contdist-1] that satisfies three basic properties: 1. (f_{X}(x)>0) for (x\in S_{X}), 2. (\int_{x\in S_{X}}f_{X}(x)\,\mathrm{d} x=1), and 3. \label{enu-contrvcond3} (\mathbb{P}(X\in A)=\int_{x\in A}f_{X}(x)\:\mathrm{d} x), for an event (A\subset S_{X}).

\bigskip

```{block, type="remark"} We can say the following about continuous random variables:

• Usually, the set (A) in \@ref(enu-contrvcond3) takes the form of an interval, for example, (A=[c,d]), in which case \begin{equation} \mathbb{P}(X\in A)=\int_{c}^{d}f_{X}(x)\:\mathrm{d} x. \end{equation}
• It follows that the probability that (X) falls in a given interval is simply the area under the curve of (f_{X}) over the interval.
• Since the area of a line (x=c) in the plane is zero, (\mathbb{P}(X=c)=0) for any value (c). In other words, the chance that (X) equals a particular value (c) is zero, and this is true for any number (c). Moreover, when (a<b) all of the following probabilities are the same: \begin{equation} \mathbb{P}(a\leq X\leq b)=\mathbb{P}(a<X\leq b)=\mathbb{P}(a\leq X<b)=\mathbb{P}(a<X<b). \end{equation}
• The PDF (f_{X}) can sometimes be greater than 1. This is in contrast to the discrete case; every nonzero value of a PMF is a probability which is restricted to lie in the interval ([0,1]).

[^contdist-1]: Not true. There are pathological random variables with
no density function. (This is one of the crazy things that can happen
in the world of measure theory). But in this book we will not get even
close to these anomalous beasts, and regardless it can be proved that
the CDF always exists.

We met the cumulative distribution function, \(F_{X}\), in Chapter
\@ref(cha-discrete-distributions). Recall that it is defined by
\(F_{X}(t)=\mathbb{P}(X\leq t)\), for \(-\infty<t<\infty\). While in
the discrete case the CDF is unwieldy, in the continuous case the CDF
has a relatively convenient form:
\begin{equation}
F_{X}(t)=\mathbb{P}(X\leq t)=\int_{-\infty}^{t}f_{X}(x)\:\mathrm{d} x,\quad -\infty < t < \infty.
\end{equation}

\bigskip

```{block, type="remark"}
For any continuous CDF \(F_{X}\) the following are true.

* \(F_{X}\) is nondecreasing , that is, \(t_{1}\leq t_{2}\) implies
  \(F_{X}(t_{1})\leq F_{X}(t_{2})\).
* \(F_{X}\) is continuous (see Appendix
  \@ref(sec-differential-and-integral). Note the distinction from the
  discrete case: CDFs of discrete random variables are not continuous,
  they are only right continuous.
* \(\lim_{t\to-\infty}F_{X}(t)=0\) and
  \(\lim_{t\to\infty}F_{X}(t)=1\).

There is a handy relationship between the CDF and PDF in the continuous case. Consider the derivative of (F_{X}): \begin{equation} F'{X}(t)=\frac{\mathrm{d}}{\mathrm{d} t}F{X}(t)=\frac{\mathrm{d}}{\mathrm{d} t}\,\int_{-\infty}^{t}f_{X}(x)\,\mathrm{d} x=f_{X}(t), \end{equation} the last equality being true by the Fundamental Theorem of Calculus, part (2) (see Appendix \@ref(sec-differential-and-integral)). In short, ((F_{X})'=f_{X}) in the continuous case[^contdist-2].

[^contdist-2]: In the discrete case, (f_{X}(x)=F_{X}(x)-\lim_{t\to x^{-}}F_{X}(t)).

Expectation of Continuous Random Variables {#sub-expectation-of-continuous}

For a continuous random variable (X) the expected value of (g(X)) is \begin{equation} \mathbb{E} g(X)=\int_{x\in S}g(x)f_{X}(x)\:\mathrm{d} x, \end{equation} provided the (potentially improper) integral (\int_{S}|g(x)|\, f(x)\mathrm{d} x) is convergent. One important example is the mean (\mu), also known as (\mathbb{E} X): \begin{equation} \mu=\mathbb{E} X=\int_{x\in S}xf_{X}(x)\:\mathrm{d} x, \end{equation} provided (\int_{S}|x|f(x)\mathrm{d} x) is finite. Also there is the variance \begin{equation} \sigma^{2}=\mathbb{E}(X-\mu)^{2}=\int_{x\in S}(x-\mu)^{2}f_{X}(x)\,\mathrm{d} x, \end{equation} which can be computed with the alternate formula (\sigma^{2}=\mathbb{E} X^{2}-(\mathbb{E} X)^{2}). In addition, there is the standard deviation (\sigma=\sqrt{\sigma^{2}}). The moment generating function is given by \begin{equation} M_{X}(t)=\mathbb{E}\:\mathrm{e}^{tX}=\int_{-\infty}^{\infty}\mathrm{e}^{tx}f_{X}(x)\:\mathrm{d} x, \end{equation} provided the integral exists (is finite) for all (t) in a neighborhood of (t=0).

\bigskip

```{example, label="cont-pdf3x2"} Let the continuous random variable (X) have PDF [ f_{X}(x)=3x^{2},\quad 0\leq x\leq 1. ] We will see later that (f_{X}) belongs to the Beta family of distributions. It is easy to see that (\int_{-\infty}^{\infty}f(x)\mathrm{d} x=1). \begin{align} \int_{-\infty}^{\infty}f_{X}(x)\mathrm{d} x & =\int_{0}^{1}3x^{2}\:\mathrm{d} x\ & =\left.x^{3}\right|_{x=0}^{1}\ & =1^{3}-0^{3}\ & =1. \end{align} This being said, we may find (\mathbb{P}(0.14\leq X<0.71)). \begin{align} \mathbb{P}(0.14\leq X<0.71) & =\int_{0.14}^{0.71}3x^{2}\mathrm{d} x,\ & =\left.x^{3}\right|_{x=0.14}^{0.71}\ & =0.71^{3}-0.14^{3}\ & \approx0.355167. \end{align} We can find the mean and variance in an identical manner. \begin{align} \mu=\int_{-\infty}^{\infty}xf_{X}(x)\mathrm{d} x & =\int_{0}^{1}x\cdot3x^{2}\:\mathrm{d} x,\ & =\frac{3}{4}x^{4}|_{x=0}^{1},\ & =\frac{3}{4}. \end{align} It would perhaps be best to calculate the variance with the shortcut formula (\sigma^{2}=\mathbb{E} X^{2}-\mu^{2}): \begin{align} \mathbb{E} X^{2}=\int_{-\infty}^{\infty}x^{2}f_{X}(x)\mathrm{d} x & =\int_{0}^{1}x^{2}\cdot3x^{2}\:\mathrm{d} x\ & =\left.\frac{3}{5}x^{5}\right|_{x=0}^{1}\ & =3/5. \end{align} which gives (\sigma^{2}=3/5-(3/4)^{2}=3/80).

\bigskip

```{example, label="cont-pdf-3x4"}
We will try one with unbounded support to brush
up on improper integration. Let the random variable \(X\) have PDF \[
f_{X}(x)=\frac{3}{x^{4}},\quad x>1.  \] We can show that
\(\int_{-\infty}^{\infty}f(x)\mathrm{d} x=1\):
\begin{align*}
\int_{-\infty}^{\infty}f_{X}(x)\mathrm{d} x & =\int_{1}^{\infty}\frac{3}{x^{4}}\:\mathrm{d} x,\\
 & =\lim_{t\to\infty}\int_{1}^{t}\frac{3}{x^{4}}\:\mathrm{d} x,\\
 & =\lim_{t\to\infty}\ \left.3\,\frac{1}{-3}x^{-3}\right|_{x=1}^{t},\\
 & =-\left(\lim_{t\to\infty}\frac{1}{t^{3}}-1\right),\\
 & =1.
\end{align*}
We calculate \(\mathbb{P}(3.4\leq X<7.1)\):
\begin{align*}
\mathbb{P}(3.4\leq X<7.1) & =\int_{3.4}^{7.1}3x^{-4}\mathrm{d} x,\\
 & =\left.3\,\frac{1}{-3}x^{-3}\right|_{x=3.4}^{7.1},\\
 & =-1(7.1^{-3}-3.4^{-3}),\\
 & \approx0.0226487123.
\end{align*}
We locate the mean and variance just like before.
\begin{align*}
\mu=\int_{-\infty}^{\infty}xf_{X}(x)\mathrm{d} x & =\int_{1}^{\infty}x\cdot\frac{3}{x^{4}}\:\mathrm{d} x,\\
 & =\left.3\,\frac{1}{-2}x^{-2}\right|_{x=1}^{\infty},\\
 & =-\frac{3}{2}\left(\lim_{t\to\infty}\frac{1}{t^{2}}-1\right),\\
 & =\frac{3}{2}.
\end{align*}
Again we use the shortcut \(\sigma^{2}=\mathbb{E} X^{2}-\mu^{2}\):
\begin{align*}
\mathbb{E} X^{2}=\int_{-\infty}^{\infty}x^{2}f_{X}(x)\mathrm{d} x & =\int_{1}^{\infty}x^{2}\cdot\frac{3}{x^{4}}\:\mathrm{d} x,\\
 & =\left.3\:\frac{1}{-1}x^{-1}\right|_{x=1}^{\infty},\\
 & =-3\left(\lim_{t\to\infty}\frac{1}{t^{2}}-1\right),\\
 & =3,
\end{align*}
which closes the example with \(\sigma^{2}=3-(3/2)^{2}=3/4\).

How to do it with R

There exist utilities to calculate probabilities and expectations for general continuous random variables, but it is better to find a built-in model, if possible. Sometimes it is not possible. We show how to do it the long way, and the distr \index{R packages@\textsf{R} packages!distr@\texttt{distr}} package [@distr] way.

\bigskip

Let \(X\) have PDF \(f(x)=3x^{2}\), \(0<x<1\) and find
\(\mathbb{P}(0.14\leq X\leq0.71)\). (We will ignore that \(X\) is a
beta random variable for the sake of argument.)

f <- function(x) 3*x^2
integrate(f, lower = 0.14, upper = 0.71)

Compare this to the answer we found in Example \@ref(ex:cont-pdf3x2). We could integrate the function (x \cdot f(x)= 3x^3) from zero to one to get the mean, and use the shortcut (\sigma^{2}=\mathbb{E} X^{2}-\left(\mathbb{E} X\right)^{2}) for the variance.

\bigskip

Let \(X\) have PDF \(f(x)=3/x^{4}\), \(x>1\). We may integrate the
function \(g(x) = x \cdot f(x)= 3/x^3\) from zero to infinity to get
the mean of \(X\).

g <- function(x) 3/x^3
integrate(g, lower = 1, upper = Inf)

Compare this to the answer we got in Example \@ref(ex:cont-pdf-3x4). Use -Inf for (-\infty).

\bigskip

Let us redo Example \@ref(ex:cont-pdf3x2) with the `distr` package. 

The method is similar to that encountered in Section \@ref(sub-disc-rv-how-r) in Chapter \@ref(cha-discrete-distributions). We define an absolutely continuous random variable:

f <- function(x) 3*x^2
X <- AbscontDistribution(d = f, low1 = 0, up1 = 1)
p(X)(0.71) - p(X)(0.14)

Compare this to the answer we found earlier. Now let us try expectation with the distrEx package [@distrEx]:

E(X); var(X); 3/80

Compare these answers to the ones we found in Example \@ref(ex:cont-pdf3x2). Why are they different? Because the distrEx package resorts to numerical methods when it encounters a model it does not recognize. This means that the answers we get for calculations may not exactly match the theoretical values. Be careful.

The Continuous Uniform Distribution {#sec-the-continuous-uniform}

A random variable (X) with the continuous uniform distribution on the interval ((a,b)) has PDF \begin{equation} f_{X}(x)=\frac{1}{b-a}, \quad a < x < b. \end{equation} The associated R function is (\mathsf{dunif}(\mathtt{min}=a,\,\mathtt{max}=b)). We write (X\sim\mathsf{unif}(\mathtt{min}=a,\,\mathtt{max}=b)). Due to the particularly simple form of this PDF we can also write down explicitly a formula for the CDF (F_{X}): \begin{equation} \label{eq-unif-cdf} F_{X}(t) = \begin{cases} 0, & t < 0,\ \frac{t-a}{b-a}, & a\leq t < b,\ 1, & t \geq b. \end{cases} \end{equation}

The continuous uniform distribution is the continuous analogue of the discrete uniform distribution; it is used to model experiments whose outcome is an interval of numbers that are "equally likely" in the sense that any two intervals of equal length in the support have the same probability associated with them.

\bigskip

Choose a number in \([0,1]\) at random, and let \(X\) be the number
chosen. Then \(X\sim\mathsf{unif}(\mathtt{min}=0,\,\mathtt{max}=1)\).
The mean of \(X\sim\mathsf{unif}(\mathtt{min}=a,\,\mathtt{max}=b)\) is
relatively simple to calculate:
\begin{align*}
\mu=\mathbb{E} X & =\int_{-\infty}^{\infty}x\, f_{X}(x)\,\mathrm{d} x,\\
 & =\int_{a}^{b}x\ \frac{1}{b-a}\ \mathrm{d} x,\\
 & =\left.\frac{1}{b-a}\ \frac{x^{2}}{2}\ \right|_{x=a}^{b},\\
 & =\frac{1}{b-a}\ \frac{b^{2}-a^{2}}{2},\\
 & =\frac{b+a}{2},
\end{align*}
using the popular formula for the difference of squares. The variance
is left to Exercise \@ref(xca-variance-dunif).

The Normal Distribution {#sec-the-normal-distribution}

We say that (X) has a normal distribution if it has PDF \begin{equation} f_{X}(x)=\frac{1}{\sigma\sqrt{2\pi}}\exp \left{ \frac{-(x-\mu)^{2}}{2\sigma^{2}} \right},\quad -\infty < x < \infty. \end{equation} We write (X\sim\mathsf{norm}(\mathtt{mean}=\mu,\,\mathtt{sd}=\sigma)), and the associated R function is dnorm(x, mean = 0, sd = 1).

The familiar bell-shaped curve, the normal distribution is also known as the Gaussian distribution because the German mathematician C. F. Gauss largely contributed to its mathematical development. This distribution is by far the most important distribution, continuous or discrete. The normal model appears in the theory of all sorts of natural phenomena, from to the way particles of smoke dissipate in a closed room, to the journey of a bottle floating in the ocean, to the white noise of cosmic background radiation.

When (\mu=0) and (\sigma=1) we say that the random variable has a standard normal distribution and we typically write (Z\sim\mathsf{norm}(\mathtt{mean}=0,\,\mathtt{sd}=1)). The lowercase Greek letter phi ((\phi)) is used to denote the standard normal PDF and the capital Greek letter phi (\Phi) is used to denote the standard normal CDF: for (-\infty<z<\infty), \begin{equation} \phi(z)=\frac{1}{\sqrt{2\pi}}\,\mathrm{e}^{-z^{2}/2}\mbox{ and }\Phi(t)=\int_{-\infty}^{t}\phi(z)\,\mathrm{d} z. \end{equation}

\bigskip

If \(X\sim\mathsf{norm}(\mathtt{mean}=\mu,\,\mathtt{sd}=\sigma)\) then
\begin{equation}
Z=\frac{X-\mu}{\sigma}\sim\mathsf{norm}(\mathtt{mean}=0,\,\mathtt{sd}=1).
\end{equation}

The MGF of (Z\sim\mathsf{norm}(\mathtt{mean}=0,\,\mathtt{sd}=1)) is relatively easy to derive: \begin{eqnarray} M_{Z}(t) & = & \int_{-\infty}^{\infty}\mathrm{e}^{tz}\frac{1}{\sqrt{2\pi}}\mathrm{e}^{-z^{2}/2}\mathrm{d} z,\ & = & \int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}\exp { -\frac{1}{2}\left(z^{2}+2tz+t^{2}\right)+\frac{t^{2}}{2} } \mathrm{d} z,\ & = & \mathrm{e}^{t^{2}/2}\left(\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}\mathrm{e}^{-[z-(-t)]^{2}/2}\mathrm{d} z\right), \end{eqnarray} and the quantity in the parentheses is the total area under a (\mathsf{norm}(\mathtt{mean}=-t,\,\mathtt{sd}=1)) density, which is one. Therefore, \begin{equation} M_{Z}(t)=\mathrm{e}^{t^{2}/2},\quad -\infty < t < \infty. \end{equation}

\bigskip

The MGF of
\(X\sim\mathsf{norm}(\mathtt{mean}=\mu,\,\mathtt{sd}=\sigma)\) is then
not difficult either because \[ Z=\frac{X-\mu}{\sigma},\mbox{ or
rewriting, }X=\sigma Z+\mu.  \] Therefore \[
M_{X}(t)=\mathbb{E}\mathrm{e}^{tX}=\mathbb{E}\mathrm{e}^{t(\sigma
Z+\mu)}=\mathbb{E}\mathrm{e}^{\sigma
tZ}\mathrm{e}^{t\mu}=\mathrm{e}^{t\mu}M_{Z}(\sigma t), \] and we know
that \(M_{Z}(t)=\mathrm{e}^{t^{2}/2}\), thus substituting we get \[
M_{X}(t)=\mathrm{e}^{t\mu}\mathrm{e}^{(\sigma t)^{2}/2}=\exp\left\{
\mu t+\sigma^{2}t^{2}/2\right\} , \] for \(-\infty<t<\infty\).

\bigskip

```{block, type="fact"} The same argument above shows that if (X) has MGF (M_{X}(t)) then the MGF of (Y=a+bX) is \begin{equation} M_{Y}(t)=\mathrm{e}^{ta}M_{X}(bt). \end{equation}

\bigskip

```{example, name="The 68-95-99.7 Rule"}
We saw in Section \@ref(sub-measures-of-spread)
that when an empirical distribution is approximately bell shaped there
are specific proportions of the observations which fall at varying
distances from the (sample) mean. We can see where these come from --
and obtain more precise proportions -- with the following:

pnorm(1:3) - pnorm(-(1:3))

\bigskip

```{example, label="iq-model"} Let the random experiment consist of a person taking an IQ test, and let (X) be the score on the test. The scores on such a test are typically standardized to have a mean of 100 and a standard deviation of 15, and IQ tests have (approximately and notoriously) a bell-shaped distribution. What is (\mathbb{P}(85\leq X\leq115))?

**Solution:** this one is easy because the limits 85 and 115 fall
exactly one standard deviation (below and above, respectively) from
the mean of 100. The answer is therefore approximately 68%.



### Normal Quantiles and the Quantile Function {#sub-normal-quantiles-qf}


Until now we have been given two values and our task has been to find
the area under the PDF between those values. In this section, we go in
reverse: we are given an area, and we would like to find the value(s)
that correspond to that area.

\bigskip

```{example, label="iq-quantile-state-problem"}
Assuming the IQ model of Example
\@ref(ex:iq-model), what is the lowest possible IQ score that a person can
have and still be in the top 1% of all IQ scores?  *Solution*: If a
person is in the top 1%, then that means that 99% of the people have
lower IQ scores. So, in other words, we are looking for a value \(x\)
such that \(F(x)=\mathbb{P}(X\leq x)\) satisfies \(F(x)=0.99\), or yet
another way to say it is that we would like to solve the equation
\(F(x)-0.99=0\). For the sake of argument, let us see how to do this
the long way. We define the function \(g(x)=F(x)-0.99\), and then look
for the root of \(g\) with the `uniroot` function. It uses numerical
procedures to find the root so we need to give it an interval of \(x\)
values in which to search for the root. We can get an educated guess
from the Empirical Rule \@ref(fac-empirical-rule); the root should be
somewhere between two and three standard deviations (15 each) above
the mean (which is 100).

g <- function(x) pnorm(x, mean = 100, sd = 15) - 0.99
uniroot(g, interval = c(130, 145))

temp <- round(uniroot(g, interval = c(130, 145))$root, 4)

The answer is shown in $root which is approximately r temp, that is, a person with this IQ score or higher falls in the top 1% of all IQ scores.

The discussion in Example \@ref(ex:iq-quantile-state-problem) was centered on the search for a value (x) that solved an equation (F(x)=p), for some given probability (p), or in mathematical parlance, the search for (F^{-1}), the inverse of the CDF of (X), evaluated at (p). This is so important that it merits a definition all its own.

\bigskip

The *quantile function*[^contdist-3] of a random variable \(X\) is
the inverse of its cumulative distribution function:
\begin{equation}
Q_{X}(p)=\min\left\{ x:\ F_{X}(x)\geq p\right\} ,\quad 0 < p <1.
\end{equation}

[^contdist-3]: The precise definition of the quantile function is (Q_{X}(p)=\inf { x:\ F_{X}(x)\geq p }), so at least it is well defined (though perhaps infinite) for the values (p=0) and (p=1).

\bigskip

```{block, type="remark"} Here are some properties of quantile functions:

1. The quantile function is defined and finite for all (0<p<1).
2. (Q_{X}) is left-continuous (see Appendix \@ref(sec-differential-and-integral)). For discrete random variables it is a step function, and for continuous random variables it is a continuous function.
3. In the continuous case the graph of (Q_{X}) may be obtained by reflecting the graph of (F_{X}) about the line (y=x). In the discrete case, before reflecting one should: 1) connect the dots to get rid of the jumps -- this will make the graph look like a set of stairs, 2) erase the horizontal lines so that only vertical lines remain, and finally 3) swap the open circles with the solid dots. Please see Figure \@ref(fig:binom-plot-distr) for a comparison.
4. The two limits [ \lim_{p\to0^{+}}Q_{X}(p)\quad \mbox{and}\quad \lim_{p\to1^{-}}Q_{X}(p) ] always exist, but may be infinite (that is, sometimes (\lim_{p\to0}Q(p)=-\infty) and/or (\lim_{p\to1}Q(p)=\infty)).

As the reader might expect, the standard normal distribution is a very
special case and has its own special notation.

\bigskip

```{definition}
For \(0<\alpha<1\), the symbol \(z_{\alpha}\) denotes the unique
solution of the equation \(\mathbb{P} ( Z > z_{\alpha}) = \alpha\),
where \(Z \sim \mathsf{norm}(\mathtt{mean} = 0,\,\mathtt{sd} =
1)\). It can be calculated in one of two equivalent ways:
\(\mathtt{qnorm(} 1 - \alpha \mathtt{)}\) and \(\mathtt{qnorm(}
\alpha \mathtt{, lower.tail = FALSE)}\).

There are a few other very important special cases which we will encounter in later chapters.

How to do it with R

Quantile functions are defined for all of the base distributions with the q prefix to the distribution name, except for the ECDF whose quantile function is exactly the (Q_{x}(p) = \mathsf{quantile}(x, \mathtt{probs} = p, \mathtt{type} = 1)) function.

\bigskip

Back to Example \@ref(ex:iq-quantile-state-problem), we are looking
for \(Q_{X}(0.99)\), where
\(X\sim\mathsf{norm}(\mathtt{mean}=100,\,\mathtt{sd}=15)\). It could
not be easier to do with R.

qnorm(0.99, mean = 100, sd = 15)

Compare this answer to the one obtained earlier with uniroot.

\bigskip

Find the values \(z_{0.025}\), \(z_{0.01}\), and \(z_{0.005}\) (these
will play an important role from Chapter \@ref(cha-estimation) onward).

qnorm(c(0.025, 0.01, 0.005), lower.tail = FALSE)

Note the lower.tail argument. We would get the same answer with qnorm(c(0.975, 0.99, 0.995)).

Functions of Continuous Random Variables {#sec-functions-of-continuous}

The goal of this section is to determine the distribution of (U=g(X)) based on the distribution of (X). In the discrete case all we needed to do was back substitute for (x=g^{-1}(u)) in the PMF of (X) (sometimes accumulating probability mass along the way). In the continuous case, however, we need more sophisticated tools. Now would be a good time to review Appendix \@ref(sec-differential-and-integral).

The PDF Method

```{proposition, label="func-cont-rvs-pdf-formula"} Let (X) have PDF (f_{X}) and let (g) be a function which is one-to-one with a differentiable inverse (g^{-1}). Then the PDF of (U=g(X)) is given by \begin{equation} \label{eq-univ-trans-pdf-long} f_{U}(u)=f_{X}\left[g^{-1}(u)\right]\ \left|\frac{\mathrm{d}}{\mathrm{d} u}g^{-1}(u)\right|. \end{equation}

\bigskip

```{block, type="remark"}
The formula in Equation \eqref{eq-univ-trans-pdf-long} is nice, but does not
really make any sense. It is better to write in the intuitive form
\begin{equation}
\label{eq-univ-trans-pdf-short}
f_{U}(u)=f_{X}(x)\left|\frac{\mathrm{d} x}{\mathrm{d} u}\right|.
\end{equation}

\bigskip

```{example, label="lnorm-transformation"} Let (X\sim\mathsf{norm}(\mathtt{mean}=\mu,\,\mathtt{sd}=\sigma)), and let (Y=\mathrm{e}^{X}). What is the PDF of (Y)? Solution: Notice first that (\mathrm{e}^{x}>0) for any (x), so the support of (Y) is ((0,\infty)). Since the transformation is monotone, we can solve (y=\mathrm{e}^{x}) for (x) to get (x=\ln\, y), giving (\mathrm{d} x/\mathrm{d} y=1/y). Therefore, for any (y>0), [ f_{Y}(y)=f_{X}(\ln y)\cdot\left|\frac{1}{y}\right|=\frac{1}{\sigma\sqrt{2\pi}}\exp\left{ \frac{(\ln y-\mu)^{2}}{2\sigma^{2}}\right} \cdot\frac{1}{y}, ] where we have dropped the absolute value bars since (y>0). The random variable (Y) is said to have a lognormal distribution; see Section \@ref(sec-other-continuous-distributions).

\bigskip

```{example, label="lin-trans-norm"}
Suppose
\(X\sim\mathsf{norm}(\mathtt{mean}=0,\,\mathtt{sd}=1)\) and let
\(Y=4-3X\). What is the PDF of \(Y\)?

The support of (X) is ((-\infty,\infty)), and as (x) goes from (-\infty) to (\infty), the quantity (y=4-3x) also traverses ((-\infty,\infty)). Solving for (x) in the equation (y=4-3x) yields (x=-(y-4)/3) giving (\mathrm{d} x/\mathrm{d} y=-1/3). And since [ f_{X}(x)=\frac{1}{\sqrt{2\pi}}\mathrm{e}^{-x^{2}/2}, \quad -\infty < x < \infty , ] we have \begin{eqnarray} f_{Y}(y) & = & f_{X}\left(\frac{y-4}{3}\right)\cdot\left|-\frac{1}{3}\right|,\quad -\infty < y < \infty,\ & = & \frac{1}{3\sqrt{2\pi}}\mathrm{e}^{-(y-4)^{2}/2\cdot3^{2}},\quad -\infty < y < \infty. \end{eqnarray} We recognize the PDF of (Y) to be that of a (\mathsf{norm}(\mathtt{mean}=4,\,\mathtt{sd}=3)) distribution. Indeed, we may use an identical argument as the above to prove the following fact:

\bigskip

```{block, type="fact", label="fac-lin-trans-norm-is-norm"} If (X\sim\mathsf{norm}(\mathtt{mean}=\mu,\,\mathtt{sd}=\sigma)) and if (Y=a+bX) for constants (a) and (b), with (b\neq0), then (Y\sim\mathsf{norm}(\mathtt{mean}=a+b\mu,\,\mathtt{sd}=|b|\sigma)).

Note that it is sometimes easier to *postpone* solving for the inverse
transformation \(x=x(u)\). Instead, leave the transformation in the
form \(u=u(x)\) and calculate the derivative of the *original*
transformation
\begin{equation}
\mathrm{d} u/\mathrm{d} x=g'(x).
\end{equation}
Once this is known, we can get the PDF of \(U\) with
\begin{equation}
f_{U}(u)=f_{X}(x)\left|\frac{1}{\mathrm{d} u/\mathrm{d} x}\right|.
\end{equation}
In many cases there are cancellations and the work is shorter. Of course, it is not always true that
\begin{equation}
\label{eq-univ-jacob-recip}
\frac{\mathrm{d} x}{\mathrm{d} u}=\frac{1}{\mathrm{d} u/\mathrm{d} x},
\end{equation}
but for the well-behaved examples in this book the trick works just fine.

\bigskip

```{block, type="remark"}
In the case that \(g\) is not monotone we cannot apply Proposition
\@ref(pro:func-cont-rvs-pdf-formula) directly. However, hope is not
lost. Rather, we break the support of \(X\) into pieces such that
\(g\) is monotone on each one. We apply Proposition
\@ref(pro:func-cont-rvs-pdf-formula) on each piece, and finish up by
adding the results together.

The CDF method

We know from Section \@ref(sec-continuous-random-variables) that (f_{X}=F_{X}') in the continuous case. Starting from the equation (F_{Y}(y)=\mathbb{P}(Y\leq y)), we may substitute (g(X)) for (Y), then solve for (X) to obtain (\mathbb{P}[X\leq g^{-1}(y)]), which is just another way to write (F_{X}[g^{-1}(y)]). Differentiating this last quantity with respect to (y) will yield the PDF of (Y).

\bigskip

Suppose \(X\sim\mathsf{unif}(\mathtt{min}=0,\,\mathtt{max}=1)\) and
suppose that we let \(Y=-\ln\, X\). What is the PDF of \(Y\)?

The support set of (X) is ((0,1),) and (y) traverses ((0,\infty)) as (x) ranges from (0) to (1), so the support set of (Y) is (S_{Y}=(0,\infty)). For any (y>0), we consider [ F_{Y}(y)=\mathbb{P}(Y\leq y)=\mathbb{P}(-\ln\, X\leq y)=\mathbb{P}(X\geq\mathrm{e}^{-y})=1-\mathbb{P}(X<\mathrm{e}^{-y}), ] where the next to last equality follows because the exponential function is monotone (this point will be revisited later). Now since (X) is continuous the two probabilities (\mathbb{P}(X<\mathrm{e}^{-y})) and (\mathbb{P}(X\leq\mathrm{e}^{-y})) are equal; thus [ 1-\mathbb{P}(X < \mathrm{e}^{-y})=1-\mathbb{P}(X\leq\mathrm{e}^{-y})=1-F_{X}(\mathrm{e}^{-y}). ] Now recalling that the CDF of a (\mathsf{unif}(\mathtt{min}=0,\,\mathtt{max}=1)) random variable satisfies (F(u)=u) (see Equation \eqref{eq-unif-cdf}), we can say [ F_{Y}(y)=1-F_{X}(\mathrm{e}^{-y})=1-\mathrm{e}^{-y},\quad \mbox{for }y>0. ] We have consequently found the formula for the CDF of (Y); to obtain the PDF (f_{Y}) we need only differentiate (F_{Y}): [ f_{Y}(y)=\frac{\mathrm{d}}{\mathrm{d} y}\left(1-\mathrm{e}^{-y}\right)=0-\mathrm{e}^{-y}(-1), ] or (f_{Y}(y)=\mathrm{e}^{-y}) for (y>0). This turns out to be a member of the exponential family of distributions, see Section \@ref(sec-other-continuous-distributions).

\bigskip

```{proposition, name="The Probability Integral Transform"} Given a continuous random variable (X) with strictly increasing CDF (F_{X}), let the random variable (Y) be defined by (Y=F_{X}(X)). Then the distribution of (Y) is (\mathsf{unif}(\mathtt{min}=0,\,\mathtt{max}=1)).

\bigskip

```{proof}
We employ the CDF method. First note that the support of \(Y\) is
\((0,1)\). Then for any \(0<y<1\), \[ F_{Y}(y)=\mathbb{P}(Y\leq
y)=\mathbb{P}(F_{X}(X)\leq y).  \] Now since \(F_{X}\) is strictly
increasing, it has a well defined inverse function
\(F_{X}^{-1}\). Therefore, \[ \mathbb{P}(F_{X}(X)\leq
y)=\mathbb{P}(X\leq F_{X}^{-1}(y))=F_{X}[F_{X}^{-1}(y)]=y.  \]
Summarizing, we have seen that \(F_{Y}(y)=y\), \(0<y<1\). But this is
exactly the CDF of a
\(\mathsf{unif}(\mathtt{min}=0,\,\mathtt{max}=1)\) random variable.

\bigskip

```{block, type="fact"} The Probability Integral Transform is true for all continuous random variables with continuous CDFs, not just for those with strictly increasing CDFs (but the proof is more complicated). The transform is not true for discrete random variables, or for continuous random variables having a discrete component (that is, with jumps in their CDF).

\bigskip

```{example, label="distn-of-z-squared"}
Let
\(Z\sim\mathsf{norm}(\mathtt{mean}=0,\,\mathtt{sd}=1)\) and let
\(U=Z^{2}\). What is the PDF of \(U\)?  Notice first that
\(Z^{2}\geq0\), and thus the support of \(U\) is \([0,\infty)\). And
for any \(u\geq0\), \[ F_{U}(u)=\mathbb{P}(U\leq
u)=\mathbb{P}(Z^{2}\leq u).  \] But \(Z^{2}\leq u\) occurs if and only
if \(-\sqrt{u}\leq Z\leq\sqrt{u}\). The last probability above is
simply the area under the standard normal PDF from \(-\sqrt{u}\) to
\(\sqrt{u}\), and since \(\phi\) is symmetric about 0, we have \[
\mathbb{P}(Z^{2}\leq u)=2\mathbb{P}(0\leq
Z\leq\sqrt{u})=2\left[F_{Z}(\sqrt{u})-F_{Z}(0)\right]=2\Phi(\sqrt{u})-1,
\] because \(\Phi(0)=1/2\). To find the PDF of \(U\) we differentiate
the CDF recalling that \(\Phi'= \phi\).  \[
f_{U}(u)=\left(2\Phi(\sqrt{u})-1\right)'=2\phi(\sqrt{u})\cdot\frac{1}{2\sqrt{u}}=u^{-1/2}\phi(\sqrt{u}).
\] Substituting, \[ f_{U}(u) =
u^{-1/2}\frac{1}{\sqrt{2\pi}}\,\mathrm{e}^{-(\sqrt{u})^{2}/2}=(2\pi
u)^{-1/2}\mathrm{e}^{-u/2},\quad u > 0.  \] This is what we will later
call a *chi-square distribution with 1 degree of freedom*. See Section
\@ref(sec-other-continuous-distributions).

How to do it with R

The distr package [@distr] has functionality to investigate transformations of univariate distributions. There are exact results for ordinary transformations of the standard distributions, and distr takes advantage of these in many cases. For instance, the distr package can handle the transformation in Example \@ref(ex:lin-trans-norm) quite nicely:

X <- Norm(mean = 0, sd = 1)
Y <- 4 - 3*X
Y

So distr "knows" that a linear transformation of a normal random variable is again normal, and it even knows what the correct mean and sd should be. But it is impossible for distr to know everything, and it is not long before we venture outside of the transformations that distr recognizes. Let us try Example \@ref(ex:lnorm-transformation):

Y <- exp(X)
Y

The result is an object of class AbscontDistribution, which is one of the classes that distr uses to denote general distributions that it does not recognize (it turns out that (Z) has a lognormal distribution; see Section \@ref(sec-other-continuous-distributions)). A simplified description of the process that distr undergoes when it encounters a transformation (Y=g(X)) that it does not recognize is

1. Randomly generate many, many copies (X_{1}), (X_{2}), ..., (X_{n}) from the distribution of (X),
2. Compute (Y_{1}=g(X_{1})), (Y_{2}=g(X_{2})), ..., (Y_{n}=g(X_{n})) and store them for use.
3. Calculate the PDF, CDF, quantiles, and random variates using the simulated values of (Y).

As long as the transformation is sufficiently nice, such as a linear transformation, the exponential, absolute value, etc., the d-p-q functions are calculated analytically based on the d-p-q functions associated with (X). But if we try a crazy transformation then we are greeted by a warning:

W <- sin(exp(X) + 27)
W

The warning (not shown here) confirms that the d-p-q functions are not calculated analytically, but are instead based on the randomly simulated values of (Y). We must be careful to remember this. The nature of random simulation means that we can get different answers to the same question: watch what happens when we compute (\mathbb{P}(W\leq0.5)) using the (W) above, then define (W) again, and compute the (supposedly) same (\mathbb{P}(W\leq0.5)) a few moments later.

p(W)(0.5)
W <- sin(exp(X) + 27)
p(W)(0.5)

The answers are not the same! Furthermore, if we were to repeat the process we would get yet another answer for (\mathbb{P}(W\leq0.5)).

The answers were close, though. And the underlying randomly generated (X)'s were not the same so it should hardly be a surprise that the calculated (W)'s were not the same, either. This serves as a warning (in concert with the one that distr provides) that we should be careful to remember that complicated transformations computed by R are only approximate and may fluctuate slightly due to the nature of the way the estimates are calculated.

Other Continuous Distributions {#sec-other-continuous-distributions}

Waiting Time Distributions {#sub-waiting-time-distributions}

In some experiments, the random variable being measured is the time until a certain event occurs. For example, a quality control specialist may be testing a manufactured product to see how long it takes until it fails. An efficiency expert may be recording the customer traffic at a retail store to streamline scheduling of staff.

The Exponential Distribution {#sub-the-exponential-distribution}

We say that (X) has an exponential distribution and write (X\sim\mathsf{exp}(\mathtt{rate}=\lambda)). \begin{equation} f_{X}(x)=\lambda\mathrm{e}^{-\lambda x},\quad x>0 \end{equation} The associated R functions are dexp(x, rate = 1), pexp, qexp, and rexp, which give the PDF, CDF, quantile function, and simulate random variates, respectively.

The parameter (\lambda) measures the rate of arrivals (to be described later) and must be positive. The CDF is given by the formula \begin{equation} F_{X}(t)=1-\mathrm{e}^{-\lambda t},\quad t>0. \end{equation} The mean is (\mu=1/\lambda) and the variance is (\sigma^{2}=1/\lambda^{2}).

The exponential distribution is closely related to the Poisson distribution. If customers arrive at a store according to a Poisson process with rate (\lambda) and if (Y) counts the number of customers that arrive in the time interval ([0,t)), then we saw in Section \@ref(sec-other-discrete-distributions) that (Y \sim \mathsf{pois}(\mathtt{lambda}=\lambda t).) Now consider a different question: let us start our clock at time 0 and stop the clock when the first customer arrives. Let (X) be the length of this random time interval. Then (X\sim\mathsf{exp}(\mathtt{rate}=\lambda)). Observe the following string of equalities: \begin{align} \mathbb{P}(X>t) & =\mathbb{P}(\mbox{first arrival after time \emph{t}}),\ & =\mathbb{P}(\mbox{no events in [0,\emph{t})}),\ & =\mathbb{P}(Y=0),\ & =\mathrm{e}^{-\lambda t}, \end{align} where the last line is the PMF of (Y) evaluated at (y=0). In other words, (\mathbb{P}(X\leq t)=1-\mathrm{e}^{-\lambda t}), which is exactly the CDF of an (\mathsf{exp}(\mathtt{rate}=\lambda)) distribution.

The exponential distribution is said to be memoryless because exponential random variables "forget" how old they are at every instant. That is, the probability that we must wait an additional five hours for a customer to arrive, given that we have already waited seven hours, is exactly the probability that we needed to wait five hours for a customer in the first place. In mathematical symbols, for any (s,\, t>0), \begin{equation} \mathbb{P}(X>s+t\,|\, X>t)=\mathbb{P}(X>s). \end{equation} See Exercise \@ref(xca-prove-the-memoryless).

The Gamma Distribution {#sub-the-gamma-distribution}

This is a generalization of the exponential distribution. We say that (X) has a gamma distribution and write (X\sim\mathsf{gamma}(\mathtt{shape}=\alpha,\,\mathtt{rate}=\lambda)). It has PDF \begin{equation} f_{X}(x)=\frac{\lambda^{\alpha}}{\Gamma(\alpha)}\: x^{\alpha-1}\mathrm{e}^{-\lambda x},\quad x>0. \end{equation}

The associated R functions are dgamma(x, shape, rate = 1), pgamma, qgamma, and rgamma, which give the PDF, CDF, quantile function, and simulate random variates, respectively. If (\alpha=1) then (X\sim\mathsf{exp}(\mathtt{rate}=\lambda)). The mean is (\mu=\alpha/\lambda) and the variance is (\sigma^{2}=\alpha/\lambda^{2}).

To motivate the gamma distribution recall that if (X) measures the length of time until the first event occurs in a Poisson process with rate (\lambda) then (X\sim\mathsf{exp}(\mathtt{rate}=\lambda)). If we let (Y) measure the length of time until the (\alpha^{\mathrm{th}}) event occurs then (Y\sim\mathsf{gamma}(\mathtt{shape}=\alpha,\,\mathtt{rate}=\lambda)). When (\alpha) is an integer this distribution is also known as the Erlang distribution.

\bigskip

At a car wash, two customers arrive per hour on the average. We decide
to measure how long it takes until the third customer arrives. If
\(Y\) denotes this random time then
\(Y\sim\mathsf{gamma}(\mathtt{shape}=3,\,\mathtt{rate}=2)\).

The Chi square, Student's t, and Snedecor's (F) Distributions {#sub-the-chi-Square-t-f}

The Chi square Distribution {#sub-the-chi-square}

A random variable (X) with PDF \begin{equation} f_{X}(x)=\frac{1}{\Gamma(p/2)2^{p/2}}x^{p/2-1}\mathrm{e}^{-x/2},\quad x>0, \end{equation} is said to have a chi-square distribution with (p) degrees of freedom. We write (X\sim\mathsf{chisq}(\mathtt{df}=p)). The associated R functions are dchisq(x, df), pchisq, qchisq, and rchisq, which give the PDF, CDF, quantile function, and simulate random variates, respectively. See Figure \@ref(fig:chisq-dist-vary-df). In an obvious notation we may define (\chi_{\alpha}^{2}(p)) as the number on the (x)-axis such that there is exactly (\alpha) area under the (\mathsf{chisq}(\mathtt{df}=p)) curve to its right.

The code to produce Figure \@ref(fig:chisq-dist-vary-df) is

curve(dchisq(x, df = 3), from = 0, to = 20, ylab = "y")
ind <- c(4, 5, 10, 15)
for (i in ind) curve(dchisq(x, df = i), 0, 20, add = TRUE)

(ref:cap-chisq-dist-vary-df) \small The chi square distribution for various degrees of freedom.

```{block, type="remark"} Here are some useful things to know about the chi-square distribution.

1. If (Z\sim\mathtt{norm}(\mathtt{mean}=0,\,\mathtt{sd}=1)), then (Z^{2}\sim\mathsf{chisq}(\mathtt{df}=1)). We saw this in Example \@ref(ex:distn-of-z-squared), and the fact is important when it comes time to find the distribution of the sample variance, (S^{2}). See Theorem \@ref(thm:xbar-ands) in Section \@ref(sub-samp-var-dist).
2. The chi-square distribution is supported on the positive (x)-axis, with a right-skewed distribution.
3. The (\mathsf{chisq}(\mathtt{df}=p)) distribution is the same as a (\mathsf{gamma}(\mathtt{shape}=p/2,\,\mathtt{rate}=1/2)) distribution.
4. The MGF of (X\sim\mathsf{chisq}(\mathtt{df}=p)) is \begin{equation} \label{eq-mgf-chisq} M_{X}(t)=\left(1-2t\right)^{-p},\quad t < 1/2. \end{equation}

#### Student's t distribution {#sub-students-t-distribution}

A random variable \(X\) with PDF
\begin{equation}
f_{X}(x) = \frac{\Gamma\left[ (r+1)/2\right] }{\sqrt{r\pi}\,\Gamma(r/2)}\left( 1 + \frac{x^{2}}{r} \right)^{-(r+1)/2},\quad -\infty < x < \infty
\end{equation}
is said to have *Student's* \(t\) distribution with \(r\) *degrees of
freedom*, and we write \(X\sim\mathsf{t}(\mathtt{df}=r)\). The
associated R functions are `dt`,`pt`, `qt`, and `rt`,
which give the PDF, CDF, quantile function, and simulate random
variates, respectively. See Section \@ref(sec-sampling-from-normal-dist).

#### Snedecor's F distribution {#sub-snedecor-F-distribution}


A random variable \(X\) with PDF
\begin{equation}
f_{X}(x)=\frac{\Gamma[(m+n)/2]}{\Gamma(m/2)\Gamma(n/2)}\left(\frac{m}{n}\right)^{m/2}x^{m/2-1}\left(1+\frac{m}{n}x\right)^{-(m+n)/2},\quad x>0.
\end{equation}
is said to have an \(F\) distribution with \((m,n)\) degrees of
freedom. We write
\(X\sim\mathsf{f}(\mathtt{df1}=m,\,\mathtt{df2}=n)\). The associated
R functions are `df(x, df1, df2)`, `pf`, `qf`, and `rf`,
which give the PDF, CDF, quantile function, and simulate random
variates, respectively. We define \(F_{\alpha}(m,n)\) as the number on
the \(x\)-axis such that there is exactly \(\alpha\) area under the
\(\mathsf{f}(\mathtt{df1}=m,\,\mathtt{df2}=n)\) curve to its right.

\bigskip

```{block, type="remark"}
Here are some notes about the \(F\) distribution.

1. If \(X\sim\mathsf{f}(\mathtt{df1}=m,\,\mathtt{df2}=n)\) and
   \(Y=1/X\), then
   \(Y\sim\mathsf{f}(\mathtt{df1}=n,\,\mathtt{df2}=m)\). Historically,
   this fact was especially convenient. In the old days, statisticians
   used printed tables for their statistical calculations. Since the
   \(F\) tables were symmetric in \(m\) and \(n\), it meant that
   publishers could cut the size of their printed tables in half. It
   plays less of a role today now that personal computers are
   widespread.
1. If \(X\sim\mathsf{t}(\mathtt{df}=r)\), then
   \(X^{2}\sim\mathsf{f}(\mathtt{df1}=1,\,\mathtt{df2}=r)\). We will
   see this again in Section \@ref(sub-slr-overall-f-statistic).

Other Popular Distributions {#sub-other-popular-distributions}

The Cauchy Distribution {#sub-the-cauchy-distribution}

This is a special case of the Student's (t) distribution. It has PDF \begin{equation} f_{X}(x) = \frac{1}{\beta\pi} \left[ 1+\left( \frac{x-m}{\beta} \right)^{2} \right]^{-1},\quad -\infty < x < \infty. \end{equation} We write (X \sim \mathsf{cauchy}(\mathtt{location} = m,\,\mathtt{scale} = \beta)). The associated R function is dcauchy(x, location = 0, scale = 1).

It is easy to see that a (\mathsf{cauchy}(\mathtt{location} = 0,\,\mathtt{scale} = 1)) distribution is the same as a (\mathsf{t}(\mathtt{df}=1)) distribution. The (\mathsf{cauchy}) distribution looks like a (\mathsf{norm}) distribution but with very heavy tails. The mean (and variance) do not exist, that is, they are infinite. The median is represented by the (\mathtt{location}) parameter, and the (\mathtt{scale}) parameter influences the spread of the distribution about its median.

The Beta Distribution {#sub-the-beta-distribution}

This is a generalization of the continuous uniform distribution. \begin{equation} f_{X}(x)=\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}\, x^{\alpha-1}(1-x)^{\beta-1},\quad 0 < x < 1. \end{equation} We write (X\sim\mathsf{beta}(\mathtt{shape1}=\alpha,\,\mathtt{shape2}=\beta)). The associated R function is dbeta(x, shape1, shape2). The mean and variance are \begin{equation} \mu=\frac{\alpha}{\alpha+\beta}\mbox{ and }\sigma^{2}=\frac{\alpha\beta}{\left(\alpha+\beta\right)^{2}\left(\alpha+\beta+1\right)}. \end{equation} See Example \@ref(ex:cont-pdf3x2). This distribution comes up a lot in Bayesian statistics because it is a good model for one's prior beliefs about a population proportion (p), (0\leq p\leq1).

The Logistic Distribution {#sub-the-logistic-distribution}

\begin{equation} f_{X}(x)=\frac{1}{\sigma}\exp\left(-\frac{x-\mu}{\sigma}\right)\left[1+\exp\left(-\frac{x-\mu}{\sigma}\right)\right]^{-2},\quad -\infty < x < \infty. \end{equation} We write (X\sim\mathsf{logis}(\mathtt{location}=\mu,\,\mathtt{scale}=\sigma)). The associated R function is dlogis(x, location = 0, scale = 1). The logistic distribution comes up in differential equations as a model for population growth under certain assumptions. The mean is (\mu) and the variance is (\pi^{2}\sigma^{2}/3).

The Lognormal Distribution {#sub-the-lognormal-distribution}

This is a distribution derived from the normal distribution (hence the name). If (U\sim\mathtt{norm}(\mathtt{mean}=\mu,\,\mathtt{sd}=\sigma)), then (X = \mathrm{e}^{U}) has PDF \begin{equation} f_{X}(x)=\frac{1}{\sigma x\sqrt{2\pi}}\exp\left[\frac{-(\ln x-\mu)^{2}}{2\sigma^{2}}\right], \quad 0 < x < \infty. \end{equation} We write (X\sim\mathsf{lnorm}(\mathtt{meanlog}=\mu,\,\mathtt{sdlog}=\sigma)). The associated R function is dlnorm(x, meanlog = 0, sdlog = 1). Notice that the support is concentrated on the positive (x) axis; the distribution is right-skewed with a heavy tail. See Example \@ref(ex:lnorm-transformation).

The Weibull Distribution {#sub-the-weibull-distribution}

This has PDF \begin{equation} f_{X}(x)=\frac{\alpha}{\beta}\left(\frac{x}{\beta}\right)^{\alpha-1}\exp\left(\frac{x}{\beta}\right)^{\alpha},\quad x>0. \end{equation} We write (X\sim\mathsf{weibull}(\mathtt{shape}=\alpha,\,\mathtt{scale}=\beta)). The associated R function is dweibull(x, shape, scale = 1).

How to do it with R

There is some support of moments and moment generating functions for some continuous probability distributions included in the actuar package [@actuar]. The convention is m in front of the distribution name for raw moments, and mgf in front of the distribution name for the moment generating function. At the time of this writing, the following distributions are supported: gamma, inverse Gaussian, (non-central) chi-squared, exponential, and uniform.

\bigskip

Calculate the first four raw moments for
\(X\sim\mathsf{gamma}(\mathtt{shape}=13,\,\mathtt{rate}=1)\) and plot
the moment generating function.

We load the actuar package and use the functions mgamma and mgfgamma:

mgamma(1:4, shape = 13, rate = 1)

For the plot we can use the function in the following form:

plot(function(x){mgfgamma(x, shape = 13, rate = 1)}, 
     from=-0.1, to=0.1, ylab = "gamma mgf")

(ref:cap-gamma-mgf) \small A plot of the \textsf{gamma}(shape = 13, rate = 1) MGF.

Exercises

```{block, type="xca"} Find the constant (C) so that the given function is a valid PDF of a random variable (X). 1. (f(x) = Cx^{n},\quad 0 < x <1). 1. (f(x) = Cx\mathrm{e}^{-x},\quad 0 < x < \infty). 1. (f(x) = \mathrm{e}^{-(x - C)}, \quad 7 < x < \infty.) 1. (f(x) = Cx^{3}(1 - x)^{2},\quad 0 < x < 1.) 1. (f(x) = C(1 + x^{2}/4)^{-1}, \quad -\infty < x < \infty.)

\bigskip

```{block, type="xca"}
For the following random experiments, decide what the distribution of
\(X\) should be. In nearly every case, there are additional
assumptions that should be made for the distribution to apply;
identify those assumptions (which may or may not strictly hold in
practice).

1. We throw a dart at a dart board. Let \(X\) denote the squared
   linear distance from the bulls-eye to the where the dart landed.
1. We randomly choose a textbook from the shelf at the bookstore and
   let \(P\) denote the proportion of the total pages of the book
   devoted to exercises.
1. We measure the time it takes for the water to completely drain out
   of the kitchen sink.
1. We randomly sample strangers at the grocery store and ask them how
   long it will take them to drive home.

\bigskip

```{block, type="xca"} If (Z) is (\mathsf{norm}(\mathtt{mean} = 0,\,\mathtt{sd} = 1)), find

1. (\mathbb{P}(Z > 2.64))
2. (\mathbb{P}(0 \leq Z < 0.87))
3. (\mathbb{P}(|Z| > 1.39)) (Hint: draw a picture!)

\bigskip

```{block, type="xca", label="xca-variance-dunif"}
Calculate the variance of
\(X\sim\mathsf{unif}(\mathtt{min}=a,\,\mathtt{max}=b)\). *Hint:* First
calculate \(\mathbb{E} X^{2}\).

\bigskip

```{block, type="xca", label="xca-prove-the-memoryless"} Prove the memoryless property for exponential random variables. That is, for (X \sim \mathsf{exp}(\mathtt{rate} = \lambda)) show that for any (s,t > 0), [ \mathbb{P}(X > s + t\,|\, X > t) = \mathbb{P}(X > s). ]

```

IPSUR documentation built on May 2, 2019, 9:15 a.m.