16.5: Car Cultures and Faster Food
The rise of the car culture led to the development of new architectural styles that were designed to accommodate the needs of motorists. For example, drive-thru restaurants and car washes became popular, as did malls and buildings with large parking lots. Additionally, the use of glass and steel in architecture became more common, as these materials were seen as being more modern and car friendly.
California Car Culture
California has long been a car-centric state. The state's mild climate, vast open spaces, and love of the open road have all contributed to the rise of car culture in California. In fact, the automobile also influenced the advent of the Freeway. The Arroyo Seco Parkway, also known as the Pasadena Freeway, is one of the oldest freeways built in the United States, (the 110 Freeway would replace the California Cycleway, which was a 9-mile elevated bicycle tollway built in 1900). It connects Los Angeles with Pasadena alongside the Arroyo Seco seasonal river. It is notable not only for being an early freeway, mostly opened in 1940, but for representing the transitional phase between early parkways and modern freeways. It conformed to modern standards when it was built, but is now regarded as a narrow, outdated roadway. A 1953 extension brought the south end to the Four Level Interchange in downtown Los Angeles and a connection with the rest of the freeway system.
In 1903, the first Model A Ford was produced in Detroit, Michigan. The Model A was an affordable and reliable car that was well-suited for the roads of California. As a result, the Model A quickly became popular in California.
In the 1920s, the popularity of car culture in California exploded. The state's population was growing rapidly, and new roads were being built all over the state. This made it easier for people to get around by car, and car ownership became increasingly common.
The 1930s saw the rise of the "California Dream." This was the idea that California was a place where people could escape the hustle and bustle of the East Coast and start a new life. The California Dream was often associated with cars, as cars were seen as a symbol of freedom and mobility.
The 1940s and 1950s saw the golden age of car culture in California. The state's economy was booming, and new car models were being introduced all the time. Cars were becoming more affordable, and more people were able to own them. This led to a boom in car-related businesses, such as drive-in restaurants, motels, and gas stations.
The 1960s saw the rise of the counterculture movement in California. This movement challenged the traditional values of American society, and cars were often seen as a symbol of the status quo. In fact, in 1960 it was estimated that in California alone, 31% households had two or more cars, while places like New York only had 12% of the households owning two or more cars. As a result, some members of the counterculture movement turned to motorcycles and vans as an alternative to cars.
The 1970s saw the oil crisis, which led to a decline in car culture in California in a few different ways. As gas prices rose, people started to drive less, and new car ownership began to decline as Californians were pushed to purchase fuel economic vehicles, rather than based on social class and design. However, car culture in California never fully disappeared. In fact, it has made a resurgence in recent years, as people have become more interested in classic cars and hot rods.
U.S. Route 66 - The Will Rogers Memorial Highway
Route 66, also known as the "Mother Road," was one of the original highways in the United States Numbered Highway System. It was established on November 11 th , 1926 and was officially designated as U.S. Route 66 in 1936. The highway was named after its designation as the 66th numbered highway in the system and was originally called the Will Rogers Highway, after the famous American actor and humorist. The highway ran for 2,448 miles (3,940 km) from Chicago, Illinois, to Santa Monica, California, passing through eight states: Illinois, Missouri, Kansas, Oklahoma, Texas, New Mexico, and Arizona.
Route 66 was featured in several films and television shows, and movies, including "The Grapes of Wrath," "Easy Rider," and "Cars." In the 1950s, the Interstate Highway System began to be built, and many of the original sections of Route 66 were bypassed by the new highways. As a result, Route 66 began to decline in popularity. However, the highway has since been designated as a National Historic Trail, and it is now a popular destination for Roadtrippers and history buffs.
The beginning of the decline of US Route 66 can be traced to 1956, when President Dwight D. Eisenhower signed the Interstate Highway Act. Many Disney historians believe that Walt Disney's vision of a futuristic highway system, as seen in the Autopia attraction at Disneyland, may have influenced the development of the Interstate Highway System. The similarities between the two projects suggest that Disney's ideas about how to move people around quickly and efficiently may have been shared by some of the architects of the Interstate Highway System. This is especially likely given that the Autopia attraction opened in 1955, just one year before the act was signed.
Drive Thru and Fast-Food
The car culture also led to the development of fast food. Fast food restaurants were designed to be convenient for motorists, with drive-thru windows and parking lots that were easy to access. Additionally, fast-food restaurants often featured bright colors and simple designs that were meant to catch the attention of passing motorists.
Want to learn more? Either scan the QR code or visit this link to learn about the influence California had on perfecting Fast-Food and the influence these establishments had on the landscape. (Video length: 10:34min).
California may not have invented fast food, but it certainly perfected it. The history of fast food can be traced back 1916, when the first White Castle opened in Wichita, Kansas. White Castle was a hamburger restaurant that served food quickly and cheaply. The novelty of getting food fast quickly caught on, and other fast-food restaurants began to open all over the country, including in California.
In the 1940s, two of the most iconic fast-food chains in the world were founded in California: McDonald's and In-N-Out Burger. McDonald's was founded in San Bernardino, California, in 1940 by brothers Maurice and Richard McDonald. The McDonald brothers' restaurant was a simple drive-thru operation that served hamburgers, fries, and shakes. In-N-Out Burger was founded in Baldwin Park, California, in 1948 by Harry and Esther Snyder. The restaurant was the first drive-thru hamburger stand in California, allowing drivers to place orders via a two-way speaker system. This was a new and unique idea, since in post-World War II California, carhops were used to take orders and serve food.
The 1950s saw the golden age of fast food in California. The state's population was growing rapidly, and new roads and freeways were being built all over the state. This made it easier for people to get around by car, and fast-food restaurants became increasingly popular.
By the 1960s, fast food continued to grow in popularity in California especially with the creation of chains, such as Taco Bell, Der Weinerschnitzel, Del Taco, and the International House of Pancakes (IHOP).
Today, fast food is still a major part of the California food landscape, culturally and architecturally, and it is often seen as a symbol of the state's car culture and diversity.


7.3: Partitions
Theorems 7.42 and 7.43 imply that if \(\sim\) is an equivalence relation on a set \(A\) , then \(\sim\) breaks \(A\) up into pairwise disjoint “chunks", where each chunk is some \([a]\) for \(a\in A\) . As you have probably already noticed, equivalence relations are intimately related to the following concept.
Definition 7.51. A collection \(\Omega\) of subsets of a set \(A\) is said to be a partition of \(A\) if the elements of \(\Omega\) satisfy:
- Each \(X\in \Omega\) is nonempty,
- For all \(X,Y\in\Omega\) , \(X\cap Y=\emptyset\) when \(X\neq Y\) , and
- \(\displaystyle \bigcup_{X\in\Omega}X=A\) .
That is, the elements of \(\Omega\) are pairwise disjoint nonempty sets and their union is all of \(A\) . Each \(X\in \Omega\) is called a block of the partition.
Example 7.52. Consider the equivalence relation \(\sim\) on the set \(P\) described in Example 7.45. Recall that the equivalence classes correspond to collections of individuals with the same last name. Since each equivalence class is nonempty and each resident of the town belongs to exactly one equivalence class, the collection of equivalence classes forms a partition of \(P\) . That is, \(P/\mathord\sim\) is a partition of \(P\) , where the blocks of the partition correspond to sets of residents with the same last name.
Example 7.53. Each of the following is an example of a partition of the set given in parentheses.
- Democrat, Republican, Independent, Green Party, Libertarian, etc. (set of registered voters)
- Freshman, sophomore, junior, senior (set of high school students)
- Evens, odds (set of integers)
- Rationals, irrationals (set of real numbers)
Example 7.54. Let \(A=\{a,b,c,d,e,f\}\) and \(\Omega=\{\{a\}, \{b,c,d\}, \{e,f\}\}\) . Since the elements of \(\Omega\) are pairwise disjoint nonempty subsets of \(A\) such that their union is all of \(A\) , \(\Omega\) is a partition of \(A\) consisting of three blocks.
Problem 7.55. Consider the set \(A\) from Example 7.54.
- Find a partition of \(A\) consisting of four blocks.
- Find a collection of subsets of \(A\) that does not form a partition. See how many ways you can prevent your collection from being a partition.
Problem 7.56. For each of the following, find a partition of \(\mathbb{Z}\) with the given properties.
- A partition of \(\mathbb{Z}\) that consists of finitely many blocks, where each of the blocks is infinite.
- A partition of \(\mathbb{Z}\) that consists of infinitely many blocks, where each of the blocks is finite.
- A partition of \(\mathbb{Z}\) that consists of infinitely many blocks, where each of the blocks is infinite.
Problem 7.57. For each relation in Problem 7.34, determine whether the corresponding collection of the sets of relatives forms a partition of the given set.
Problem 7.58. Can we partition the empty set? If so, describe a partition. If not, explain why.
The next theorem spells out half of the close connection between partitions and equivalence relations. Theorem 7.73 yields the other half.
Problem 7.59. If \(\sim\) is an equivalence relation on a nonempty set \(A\) , then \(A/\mathord\sim\) forms a partition of \(A\) .
Problem 7.60. In the previous theorem, why did we require \(A\) to be nonempty?
Problem 7.61. Consider the equivalence relation \[\sim\ =\{(1,1),(1,2),(2,1), (2,2),(3,3),(4,4),(4,5),(5,4),(5,5),(6,6),(5,6),(6,5),(4,6),(6,4)\}\] on the set \(A=\{1,2,3,4,5,6\}\) . Find the partition determined by \(Rel(\sim)\) .
It turns out that we can reverse the situation, as well. That is, given a partition, we can form an equivalence relation such that the equivalence classes correspond to the blocks of the partition. Before proving this, we need a definition.
Definition 7.62. Let \(A\) be a set and \(\Omega\) any collection of subsets of \(A\) (not necessarily a partition). Define the relation \(R_{\Omega}\) on \(A\) via \(aR_{\Omega}b\) if there exists \(X\in \Omega\) that contains both \(a\) and \(b\) . This relation is called the relation on \(A\) associated to \(\Omega\) .
In other words, two elements are related exactly when they are in the same subset.
Problem 7.63. Let \(A=\{a,b,c,d,e,f\}\) and let \(\Omega=\{\{a,c\},\{b,c\},\{d,f\}\}\) . List the ordered pairs in \(R_{\Omega}\) and draw the corresponding digraph.
Problem 7.64. Let \(A\) and \(\Omega\) be as in Example 7.54. List the ordered pairs in \(R_{\Omega}\) and draw the corresponding digraph.
Problem 7.65. Consider Problem 7.24. Find the relation on \(A\) associated to \(Rel(\sim)\) and compare with what you obtained for \(R\) in Problem 7.24.
Problem 7.66. Give an example of a set \(A\) and a collection \(\Omega\) from \(\mathcal{P}(A)\) such that the relation \(R_{\Omega}\) is not reflexive.
Problem 7.67. Let \(A=\{1,2,3,4,5,6\}\) and \(\Omega=\{\{1,3,4\},\{2,4\},\{3,4\},\{6\}\}\) .
- Is \(\Omega\) a partition of \(A\) ?
- Find \(R_{\Omega}\) by listing ordered pairs or drawing a digraph.
- Is \(R_{\Omega}\) an equivalence relation?
- Find \(Rel(R_\Omega)\) (i.e., the collection of subsets of \(A\) determined by \(R_{\Omega}\) ). How are \(\Omega\) and \(Rel(R_\Omega)\) related?
Theorem 7.68. If \(\Omega\) is a collection of subsets of a nonempty set \(A\) (not necessarily a partition) such that \[\bigcup_{X\in\Omega}X=A,\] then \(R_{\Omega}\) is reflexive.
Problem 7.69. Is it necessary to require \(A\) to be nonempty in Theorem 7.68?
Theorem 7.70. If \(\Omega\) is a collection of subsets of a set \(A\) (not necessarily a partition), then \(R_{\Omega}\) is symmetric.
Theorem 7.71. If \(\Omega\) is a collection of subsets of a set \(A\) (not necessarily a partition) such that the elements of \(\Omega\) are pairwise disjoint, then \(R_{\Omega}\) is transitive.
Problem 7.72. Why didn’t we require \(A\) to be nonempty in Theorems 7.70 and 7.71?
Recall that Theorem 7.59 says that the equivalence classes for a relation on a nonempty set \(A\) determines a partition of \(A\) . The following theorem tells us that every partition of a set yields an equivalence relation where the equivalence classes correspond to the blocks of the partition. This result is a consequence of Theorems 7.68, 7.70, and 7.71.
Theorem 7.73. If \(\Omega\) is a partition of a set \(A\) , then \(R_{\Omega}\) is an equivalence relation.
Together, Theorems 7.59 and 7.73 tell us that equivalence relations and partitions are two different ways of viewing the same thing.
Corollary 7.74. If \(R\) is a relation on a nonempty set \(A\) such that the collection of the set of relatives with respect to \(R\) is a partition of \(A\) , then \(R\) is an equivalence relation.
Problem 7.75. Let \(A=\{\circ, \triangle, \blacktriangle, \square, \blacksquare, \bigstar \}\) . Make up a partition \(\Omega\) on \(A\) and then draw the digraph corresponding to \(R_{\Omega}\) .


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5.1: Separation of group III cations
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After insoluble chlorides and acid-insoluble sulfides of group I and group II has been removed, the medium is made alkaline at pH ~9 where group III cations precipitate out as hydroxides, i.e., iron(II) hydroxide, iron (III) hydroxide, chromium(III) hydroxide, and nickel(II) hydroxides. These ions also precipitate out as insoluble sulfides in the alkaline medium.
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5.2: Separation and confirmation if individual ions in group III precipitates
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Nickel sulfide is separated based on its insolubility in HCl, then dissolved in aqua regia and confirmed by making a red precipitate of its complex. Iron(II) ions are oxidized to iron(III), dissolved in HCl, and confirmed by making a red color complex. Chromate ion is reduced to chromium(III) ion by acidic hydrogen peroxide, then confirmed by precipitating it upon making the medium alkaline.
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5.3: Procedure, flowchart, and datasheets for separation and confirmation of group III cations
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The procedure, flowchart, datasheets and filling instructions for the known sample that contains all of the group III cations and for the unknown sample that may contain some of the group III cations.


1.14.16: Energy and First Law of Thermodynamics
A central axiom of chemical thermodynamics is that a given system has a property called energy. In fact the First Law of Thermodynamics centres on the concept of energy. In its broadest terms, the law requires that the energy of the universe is constant [1]. This is a rather overwhelming statement. A more attractive statement is that the thermodynamic energy \(\mathrm{U}\) of a typical chemistry laboratory is constant.
\[\mathrm{U} = \text{ constant} \label{a} \]
The latter is the principle of conservation of energy; energy can be neither created nor destroyed. A chemist ‘watches’ energy “move” between system and surroundings. As a consequence of Equation \ref{a} we state that,
\[\Delta \mathrm{U}(\text { system })=-\Delta \mathrm{U}(\text { surroundings }) \label{b} \]
We cannot know the actual energy \(\mathrm{U}\) of a closed system although we agree that it is an extensive property of a system. In describing the energy changes we need a convention. We use the acquisitive convention, describing all changes in terms of how the system is affected. Thus \(\Delta \mathrm{U} < 0\), means that the energy of the system falls whereas \(\Delta \mathrm{U} > 0\) means that the energy increases [2]. In the context of chemistry, chemists agree that the energy of a given closed system can be increased in two ways:
- heat \(\mathrm{q}\) passing from the surroundings into the system, and
- work \(\mathrm{w}\) done by the surroundings on the system. In a wider context the concept of energy is linked with the First Law of Thermodynamics which is based on the following axiom.
\[\Delta \mathrm{U}=\mathrm{q}+\mathrm{w} \label{c} \]
As it stands the symbols \(\mathrm{U}\), \(\mathrm{q}\) and \(\mathrm{w}\) seem rather uninformative. It is the task of chemists to flesh out the meaning of these terms. If only ‘\(\mathrm{p}-\mathrm{V}\)’ work is involved,
\[\mathrm{w}=-\mathrm{p} \, \mathrm{dV} \label{d} \]
The point of Equation \ref{c} is to separate the work term from the heat term. The significance for chemists is that \(\mathrm{q}\) links to the Second Law of Thermodynamics. Thus chemists know that heat flows spontaneously from high to low temperatures. This concept of ‘spontaneous change’ is picked up with enormous impact in the second law.
Footnotes
[1] Peter Atkins (Galileo’s Finger, Oxford University press, 2003, page 107) speculates that the total energy of the universe ‘may be exactly zero’.
[2] In principle it is possible to calculate the total energy of a given system using a scale in conjunction with Einstein’s famous equation, \(\mathrm{E}=\mathrm{m} \, \mathrm{c}^{2}\). However the mass corresponding to \(1 \mathrm{~kJ}\) is only about \(10^{-14} \mathrm{~kg}\).


4.2: FORMAL OR CRITICAL ANALYSIS
While restricting our attention only to a description of the formal elements of an artwork may at first seem limited or even tedious, a careful and methodical examination of the physical components of an artwork is an important first step in “decoding” its meaning. It is useful, therefore, to begin at the beginning. There are four aspects of a formal analysis: description, analysis, interpretation, and evaluation . In addition to defining these terms, we will look at examples.
4.2.1 Description
What can we notice at first glance about a work of art? Is it two-dimensional or three-dimen- sional? What is the medium? What kinds of actions were required in its production? How big is the work? What are the elements of design used within it?
Starting with line: is it soft or hard, jagged or straight, expressive or mechanical? How is line being used to describe space?
Considering shape: are the shapes large or small, hard-edged or soft? What is the relationship between shapes? Do they compete with one another for prominence? What shapes are in front? Which ones fade into the background?
Indicating mass and volume: if two-dimensional, what means if any are used to give the illu- sion that the presented forms have weight and occupy space? If three-dimensional, what space is occupied or filled by the work? What is the mass of the work?
Organizing space: does the artist use perspective? If so, what kind? If the work uses linear perspective, where are the horizon line and vanishing point(s) located?
On texture: how is texture being used? Is it actual or implied texture?
In terms of color: what kinds of colors are used? Is there a color scheme? Is the image overall light, medium, or dark?
4.2.2 Analysis
Once the elements of the artwork have been identified, next come questions of how these ele- ments are related. How are the elements arranged? In other words, how have principles of design been employed?
What elements in the work were used to create unity and provide variety? How have the ele- ments been used to do so?
What is the scale of the work? Is it larger or smaller than what it represents (if it does depict someone or something)? Are the elements within the work in proportion to one another?
Is the work symmetrically or asymmetrically balanced?
What is used within the artwork to create emphasis? Where are the areas of emphasis? How has movement been conveyed in the work, for example, through line or placement of figures?
Are there any elements within the work that create rhythm? Are any shapes or colors repeated?
4.2.3 Interpretation
Interpretation comes as much from the individual viewer as it does from the artwork. It de- rives from the intersection of what an object symbolizes to the artist and what it means to the viewer. It also often records how the meaning of objects has been changed by time and culture. Interpretation, then, is a process of unfolding. A work that may seem to mean one thing on first inspection may come to mean something more when studied further. Just as when re-reading a favorite book or re-watching a favorite movie, we often notice things not seen on the first viewing; interpretations of art objects can also reveal themselves slowly. Claims about meaning can be made but are better when they are backed up with supporting evidence. Interpretations can also change and some interpretations are better than others.
4.2.4 Evaluation
All this work of description, analysis, and interpretation, is done with one goal in mind: to make an evaluation about a work of art. Just as interpretations vary, so do evaluations. Your evaluation includes what you have discovered about the work during your examination as well as what you have learned, about the work, yourself, and others in the process. Your reaction to the artwork is an important component of your evaluation: what do you feel when you look at it? And, do you like the work? How and why do you find it visually pleasing, in some way disturbing, emotionally engaging?
Evaluating and judging contemporary works of art is more difficult than works that are hundreds or thousands of years old because the verdict of history has not yet been passed on them. Museums are full of paintings by contemporary artists who were considered the next Michelange- lo but who have since faded from the cultural forefront.
The best art of a culture and period is that work which exemplifies the thought of the age from which it derives. What we think about our own culture is probably not what will be thought of it a century from now. The art that we believe best embodies our time may or may not last. As time moves on, our evaluations and judgments of our own time may not prove to be the most accurate ones. We live in a world full of art, and it is almost impossible to avoid making evaluations—pos- sibly mistaken—about its value. Nonetheless, informed evaluations are still possible and useful even in the short term.
4.2.5 Examples of Formal Analysis
Snow Storm—Steam-Boat off a Harbour’s Mouth by J. M. W. Turner
Snow Storm—Steam-Boat off a Harbour’s Mouth by Joseph Mallord William Turner (1775- 1851, England) is a chaotic, atmospheric oil on canvas painting. (Figure 4.1) First, on the level of description, the dark structure of the foundering steamboat is hinted at in the center of the work, while heavy smoke from the vessel, pitching waves, and swirling snow surround it. The brown and gray curving lines are created with long strokes of heavily applied paint that expand to the edges of the composition. Second, on the level of analysis, we note that the paint application, heavy, with long strokes, adds dramatic movement to the image. We see that the design principle of scale and proportion is being used in the small size of the steamboat in relation to the overall canvas. Now let us interpret these elements and their relation: The artist has emphasized the maelstrom of sea, snow, and wind. A glimpse of blue sky through the smoke and snow above the vessel is the only indication of space beyond this gripping scene of danger, and provides the only place for the viewer’s eyes to rest from the tumult. This scene is of humanity’s struggle for survival against powerful forces of nature. And finally, we are ready to evaluate this work. Is it powerfully effective in reminding us of the transitory nature of our own limited existence, a memento morii, perhaps? Or is it a wise caution of the limits of our human power to control our destiny? Does the work have sufficient power and value to be accepted by us as significant? The verdict of history tells us it is. J.M.W. Turner is considered a significant artist of his time, and this work is one that is thought to support that verdict. In the end, however, each of us can accept or reject this historical verdict for our own reasons. We may fear the sea. We may reject the use of technology as valiantly heroic. We may see the British colonial period as one of oppression and tyranny and this work as an illustration of the hubris of that time. Whatever we conclude, this work of art stands as a catalyst for this important dialogue.
Another example of formal analysis. Consider Lady at the Tea Table by Mary Cassatt Mary Cassatt (1844-1925, USA, lived France) is best known for her paintings, drawings, and prints of mothers and children. In those works, she focused on the bond between them as well as the strength and dignity of women within the predominantly domestic and maternal roles they played in the nineteenth century.
Lady at the Tea Table is a depiction of a woman in a later period of her life, and captures the sense of calm power a matriarch held within the home. (Figure 4.2) First, a description of the elements being used in this work: The white of the wall behind the woman and the tablecloth be- fore her provide a strong contrast to the black of her clothing and the blue of the tea set. The gold frame of the artwork on the wall, the gold rings on her fingers, and the gold bands on the china link those three main elements of the painting. Analysis shows the organizing principle of variety is employed in the rectangles behind the woman’s head and the multiple circles and arcs of the individual pieces of the tea set. The composition is a stable triangle formed by the woman’s head and body, and extending to the pieces of china that span the foreground from one edge of the composition to the other. Let us interpret these observations. There is little evidence of movement in the work other than the suggestion that the woman’s hand, resting on the handle of the teapot, may soon move. Her gaze, directed away from the viewer and out of the picture frame, implies she is in the midst of pouring tea, but her stillness suggests she is lost in thought. How to evaluate this work? The artist expresses a restrained but powerful strength of character in her treatment of this subject. Is the lack of obvious movement in the work a comment on the emergence of women’s roles in society, a hope or a demand for change? Or is it a monument to the quiet dignity of the domestic life of Victorian era Paris? The gold of the frame, the rings, and the china dishes appear to unify three disparate objects into one statement of value. Do they symbolize art, fidelity, and service? Is this a comment on the restrictions of French domestic society, or a claim to its strength? One indication of the quality of a work of art is its power to evoke multiple interpretations. This open and poetic richness is one reason why the work of Mary Cassatt is considered to be important. The above examples are only one of many ways in which we can interpret and evalu- ate works of art. We will examine a few more approaches to analysis and critique. The point of this exercise is to equip the interested student with tools to become more fully aware of the dynamics and content of works of art, not only in museums and textbooks, but in the world of images that continually surrounds us today.


14.2: The Exponential Distribution
Basic Theory
The Memoryless Property
Recall that in the basic model of the Poisson process, we have
points
that occur randomly in time. The sequence of inter-arrival times is \(\bs{X} = (X_1, X_2, \ldots)\). The strong renewal assumption states that at each arrival time and at each fixed time, the process must probabilistically restart, independent of the past. The first part of that assumption implies that \(\bs{X}\) is a sequence of independent, identically distributed variables. The second part of the assumption implies that if the first arrival has not occurred by time \(s\), then the time remaining until the arrival occurs must have the same distribution as the first arrival time itself. This is known as the
memoryless property
and can be stated in terms of a general random variable as follows:
Suppose that \( X \) takes values in \( [0, \infty) \). Then \( X \) has the memoryless property if the conditional distribution of \(X - s\) given \(X \gt s\) is the same as the distribution of \(X\) for every \( s \in [0, \infty) \). Equivalently, \[ \P(X \gt t + s \mid X \gt s) = \P(X \gt t), \quad s, \; t \in [0, \infty) \]
The memoryless property determines the distribution of \(X\) up to a positive parameter, as we will see now.
Distribution functions
Suppose that \(X\) takes values in \( [0, \infty) \) and satisfies the memoryless property.
\(X\) has a continuous distribution and there exists \(r \in (0, \infty)\) such that the distribution function \(F\) of \(X\) is \[ F(t) = 1 - e^{-r\,t}, \quad t \in [0, \infty) \]
Proof
Let \(F^c = 1 - F\) denote the denote the right-tail distribution function of \(X\) (also known as the reliability function ), so that \(F^c(t) = \P(X \gt t)\) for \(t \ge 0\). From the definition of conditional probability, the memoryless property is equivalent to the law of exponents : \[ F^c(t + s) = F^c(s) F^c(t), \quad s, \; t \in [0, \infty) \] Let \(a = F^c(1)\). Implicit in the memoryless property is \(\P(X \gt t) \gt 0\) for \(t \in [0, \infty)\), so \(a \gt 0\). If \(n \in \N_+\) then \[ F^c(n) = F^c\left(\sum_{i=1}^n 1\right) = \prod_{i=1}^n F^c(1) = \left[F^c(1)\right]^n = a^n \] Next, if \(n \in \N_+\) then \[ a = F^c(1) = F^c\left(\frac{n}{n}\right) = F^c\left(\sum_{i=1}^n \frac{1}{n}\right) = \prod_{i=1}^n F^c\left(\frac{1}{n}\right) = \left[F^c\left(\frac{1}{n}\right)\right]^n \] so \(F^c\left(\frac{1}{n}\right) = a^{1/n}\). Now suppose that \(m \in \N\) and \(n \in \N_+\). Then \[ F^c\left(\frac{m}{n}\right) = F^c\left(\sum_{i=1}^m \frac{1}{n}\right) = \prod_{i=1}^m F^c\left(\frac{1}{n}\right) = \left[F^c\left(\frac{1}{n}\right)\right]^m = a^{m/n} \] Thus we have \(F^c(q) = a^q\) for rational \(q \in [0, \infty)\). For \(t \in [0, \infty)\), there exists a sequence of rational numbers \((q_1, q_2, \ldots)\) with \(q_n \downarrow t\) as \(n \uparrow \infty\). We have \(F^c(q_n) = a^{q_n}\) for each \(n \in \N_+\). But \(F^c\) is continuous from the right, so taking limits gives \(a^t = F^c(t) \). Now let \(r = -\ln(a)\). Then \(F^c(t) = e^{-r\,t}\) for \(t \in [0, \infty)\).
The probability density function of \(X\) is \[ f(t) = r \, e^{-r\,t}, \quad t \in [0, \infty) \]
- \( f \) is decreasing on \( [0, \infty) \).
- \( f \) is concave upward on \( [0, \infty) \).
- \( f(t) \to 0 \) as \( t \to \infty \).
Proof
This follows since \( f = F^\prime \). The properties in parts (a)–(c) are simple.
A random variable with the distribution function above or equivalently the probability density function in the last theorem is said to have the exponential distribution with rate parameter \(r\). The reciprocal \(\frac{1}{r}\) is known as the scale parameter (as will be justified below ). Note that the mode of the distribution is 0, regardless of the parameter \( r \), not very helpful as a measure of center.
In the gamma experiment, set \(n = 1\) so that the simulated random variable has an exponential distribution. Vary \(r\) with the scroll bar and watch how the shape of the probability density function changes. For selected values of \(r\), run the experiment 1000 times and compare the empirical density function to the probability density function.
The quantile function of \(X\) is \[ F^{-1}(p) = \frac{-\ln(1 - p)}{r}, \quad p \in [0, 1) \]
- The median of \(X\) is \(\frac{1}{r} \ln(2) \approx 0.6931 \frac{1}{r}\)
- The first quartile of \(X\) is \(\frac{1}{r}[\ln(4) - \ln(3)] \approx 0.2877 \frac{1}{r}\)
- The third quartile \(X\) is \(\frac{1}{r} \ln(4) \approx 1.3863 \frac{1}{r}\)
- The interquartile range is \(\frac{1}{r} \ln(3) \approx 1.0986 \frac{1}{r}\)
Proof
The formula for \( F^{-1} \) follows easily from solving \( p = F^{-1}(t) \) for \( t \) in terms of \( p \).
In the special distribution calculator, select the exponential distribution. Vary the scale parameter (which is \( 1/r \)) and note the shape of the distribution/quantile function. For selected values of the parameter, compute a few values of the distribution function and the quantile function.
Returning to the Poisson model, we have our first formal definition:
A process of random points in time is a Poisson process with rate \( r \in (0, \infty) \) if and only the interarrvial times are independent, and each has the exponential distribution with rate \( r \).
Constant Failure Rate
Suppose now that \(X\) has a continuous distribution on \([0, \infty)\) and is interpreted as the lifetime of a device. If \(F\) denotes the distribution function of \(X\), then \(F^c = 1 - F\) is the reliability function of \(X\). If \(f\) denotes the probability density function of \(X\) then the failure rate function \( h \) is given by \[ h(t) = \frac{f(t)}{F^c(t)}, \quad t \in [0, \infty) \] If \(X\) has the exponential distribution with rate \(r \gt 0\), then from the results above, the reliability function is \(F^c(t) = e^{-r t}\) and the probability density function is \(f(t) = r e^{-r t}\), so trivially \(X\) has constant rate \(r\). The converse is also true.
If \(X\) has constant failure rate \(r \gt 0\) then \(X\) has the exponential distribution with parameter \(r\).
Proof
Recall that in general, the distribution of a lifetime variable \(X\) is determined by the failure rate function \(h\). Specifically, if \(F^c = 1 - F\) denotes the reliability function, then \((F^c)^\prime = -f\), so \(-h = (F^c)^\prime / F^c\). Integrating and then taking exponentials gives \[ F^c(t) = \exp\left(-\int_0^t h(s) \, ds\right), \quad t \in [0, \infty) \] In particular, if \(h(t) = r\) for \(t \in [0, \infty)\), then \(F^c(t) = e^{-r t}\) for \(t \in [0, \infty)\).
The memoryless and constant failure rate properties are the most famous characterizations of the exponential distribution, but are by no means the only ones. Indeed, entire books have been written on characterizations of this distribution.
Moments
Suppose again that \(X\) has the exponential distribution with rate parameter \(r \gt 0\). Naturaly, we want to know the the mean, variance, and various other moments of \(X\).
If \(n \in \N\) then \(\E\left(X^n\right) = n! \big/ r^n\).
Proof
By the change of variables theorem for expected value, \[ \E\left(X^n\right) = \int_0^\infty t^n r e^{-r\,t} \, dt\] Integrating by parts gives \(\E\left(X^n\right) = \frac{n}{r} \E\left(X^{n-1}\right)\) for \(n \in \N+\). Of course \(\E\left(X^0\right) = 1\) so the result now follows by induction.
More generally, \(\E\left(X^a\right) = \Gamma(a + 1) \big/ r^a\) for every \(a \in [0, \infty)\), where \(\Gamma\) is the gamma function.
In particular.
- \(\E(X) = \frac{1}{r}\)
- \(\var(X) = \frac{1}{r^2}\)
- \(\skw(X) = 2\)
- \(\kur(X) = 9\)
In the context of the Poisson process, the parameter \(r\) is known as the rate of the process. On average, there are \(1 / r\) time units between arrivals, so the arrivals come at an average rate of \(r\) per unit time. The Poisson process is completely determined by the sequence of inter-arrival times, and hence is completely determined by the rate \( r \).
Note also that the mean and standard deviation are equal for an exponential distribution, and that the median is always smaller than the mean. Recall also that skewness and kurtosis are standardized measures, and so do not depend on the parameter \(r\) (which is the reciprocal of the scale parameter).
The moment generating function of \(X\) is \[ M(s) = \E\left(e^{s X}\right) = \frac{r}{r - s}, \quad s \in (-\infty, r) \]
Proof
By the change of variables theorem \[ M(s) = \int_0^\infty e^{s t} r e^{-r t} \, dt = \int_0^\infty r e^{(s - r)t} \, dt \] The integral evaluates to \( \frac{r}{r - s} \) if \( s \lt r \) and to \( \infty \) if \( s \ge r \).
In the gamma experiment, set \(n = 1\) so that the simulated random variable has an exponential distribution. Vary \(r\) with the scroll bar and watch how the mean\( \pm \)standard deviation bar changes. For various values of \(r\), run the experiment 1000 times and compare the empirical mean and standard deviation to the distribution mean and standard deviation, respectively.
Additional Properties
The exponential distribution has a number of interesting and important mathematical properties. First, and not surprisingly, it's a member of the general exponential family.
Suppose that \( X \) has the exponential distribution with rate parameter \( r \in (0, \infty) \). Then \( X \) has a one parameter general exponential distribution, with natural parameter \( -r \) and natural statistic \( X \).
Proof
This follows directly from the form of the PDF, \( f(x) = r e^{-r x} \) for \( x \in [0, \infty) \), and the definition of the general exponential family.
The Scaling Property
As suggested earlier, the exponential distribution is a scale family, and \(1/r\) is the scale parameter.
Suppose that \(X\) has the exponential distribution with rate parameter \(r \gt 0\) and that \(c \gt 0\). Then \(c X\) has the exponential distribution with rate parameter \(r / c\).
Proof
For \(t \ge 0\), \(\P(c\,X \gt t) = \P(X \gt t / c) = e^{-r (t / c)} = e^{-(r / c) t}\).
Recall that multiplying a random variable by a positive constant frequently corresponds to a change of units (minutes into hours for a lifetime variable, for example). Thus, the exponential distribution is preserved under such changes of units. In the context of the Poisson process, this has to be the case, since the memoryless property, which led to the exponential distribution in the first place, clearly does not depend on the time units.
In fact, the exponential distribution with rate parameter 1 is referred to as the standard exponential distribution . From the previous result, if \( Z \) has the standard exponential distribution and \( r \gt 0 \), then \( X = \frac{1}{r} Z \) has the exponential distribution with rate parameter \( r \). Conversely, if \( X \) has the exponential distribution with rate \( r \gt 0 \) then \( Z = r X \) has the standard exponential distribution.
Similarly, the Poisson process with rate parameter 1 is referred to as the standard Poisson process . If \( Z_i \) is the \( i \)th inter-arrival time for the standard Poisson process for \( i \in \N_+ \), then letting \( X_i = \frac{1}{r} Z_i \) for \( i \in \N_+ \) gives the inter-arrival times for the Poisson process with rate \( r \). Conversely if \( X_i \) is the \( i \)th inter-arrival time of the Poisson process with rate \( r \gt 0 \) for \( i \in \N_+ \), then \( Z_i = r X_i \) for \( i \in \N_+ \) gives the inter-arrival times for the standard Poisson process.
Relation to the Geometric Distribution
In many respects, the geometric distribution is a discrete version of the exponential distribution. In particular, recall that the geometric distribution on \( \N_+ \) is the only distribution on \(\N_+\) with the memoryless and constant rate properties. So it is not surprising that the two distributions are also connected through various transformations and limits.
Suppose that \(X\) has the exponential distribution with rate parameter \(r \gt 0\). Then
- \(\lfloor X \rfloor\) has the geometric distributions on \(\N\) with success parameter \(1 - e^{-r}\).
- \(\lceil X \rceil\) has the geometric distributions on \(\N_+\) with success parameter \(1 - e^{-r}\).
Proof
- For \(n \in \N\) note that \(\P(\lfloor X \rfloor = n) = \P(n \le X \lt n + 1) = F(n + 1) - F(n)\). Substituting into the distribution function and simplifying gives \(\P(\lfloor X \rfloor = n) = (e^{-r})^n (1 - e^{-r})\).
- For \(n \in \N_+\) note that \(\P(\lceil X \rceil = n) = \P(n - 1 \lt X \le n) = F(n) - F(n - 1)\). Substituting into the distribution function and simplifying gives \(\P(\lceil X \rceil = n) = (e^{-r})^{n - 1} (1 - e^{-r})\).
The following connection between the two distributions is interesting by itself, but will also be very important in the section on splitting Poisson processes. In words, a random, geometrically distributed sum of independent, identically distributed exponential variables is itself exponential.
Suppose that \(\bs{X} = (X_1, X_2, \ldots)\) is a sequence of independent variables, each with the exponential distribution with rate \(r\). Suppose that \(U\) has the geometric distribution on \(\N_+\) with success parameter \(p\) and is independent of \(\bs{X}\). Then \(Y = \sum_{i=1}^U X_i\) has the exponential distribution with rate \(r p\).
Proof
Recall that the moment generating function of \(Y\) is \(P \circ M\) where \(M\) is the common moment generating function of the terms in the sum, and \(P\) is the probability generating function of the number of terms \(U\). But \(M(s) = r \big/ (r - s)\) for \(s \lt r\) and \(P(s) = p s \big/ \left[1 - (1 - p)s\right]\) for \(s \lt 1 \big/ (1 - p)\). Thus, \[ (P \circ M)(s) = \frac{p r \big/ (r - s)}{1 - (1 - p) r \big/ (r - s)} = \frac{pr}{pr - s}, \quad s \lt pr \] It follows that \(Y\) has the exponential distribution with parameter \(p r\)
The next result explores the connection between the Bernoulli trials process and the Poisson process that was begun in the Introduction.
For \( n \in \N_+ \), suppose that \( U_n \) has the geometric distribution on \( \N_+ \) with success parameter \( p_n \), where \( n p_n \to r \gt 0 \) as \( n \to \infty \). Then the distribution of \( U_n / n \) converges to the exponential distribution with parameter \( r \) as \( n \to \infty \).
Proof
Let \( F_n \) denote the CDF of \( U_n / n \). Then for \( x \in [0, \infty) \) \[ F_n(x) = \P\left(\frac{U_n}{n} \le x\right) = \P(U_n \le n x) = \P\left(U_n \le \lfloor n x \rfloor\right) = 1 - \left(1 - p_n\right)^{\lfloor n x \rfloor} \] But by a famous limit from calculus, \( \left(1 - p_n\right)^n = \left(1 - \frac{n p_n}{n}\right)^n \to e^{-r} \) as \( n \to \infty \), and hence \( \left(1 - p_n\right)^{n x} \to e^{-r x} \) as \( n \to \infty \). But by definition, \( \lfloor n x \rfloor \le n x \lt \lfloor n x \rfloor + 1\) or equivalently, \( n x - 1 \lt \lfloor n x \rfloor \le n x \) so it follows that \( \left(1 - p_n \right)^{\lfloor n x \rfloor} \to e^{- r x} \) as \( n \to \infty \). Hence \( F_n(x) \to 1 - e^{-r x} \) as \( n \to \infty \), which is the CDF of the exponential distribution.
To understand this result more clearly, suppose that we have a sequence of Bernoulli trials processes. In process \( n \), we run the trials at a rate of \( n \) per unit time, with probability of success \( p_n \). Thus, the actual time of the first success in process \( n \) is \( U_n / n \). The last result shows that if \( n p_n \to r \gt 0 \) as \( n \to \infty \), then the sequence of Bernoulli trials processes converges to the Poisson process with rate parameter \( r \) as \( n \to \infty \). We will return to this point in subsequent sections.
Orderings and Order Statistics
Suppose that \(X\) and \(Y\) have exponential distributions with parameters \(a\) and \(b\), respectively, and are independent. Then \[ \P(X \lt Y) = \frac{a}{a + b} \]
Proof
This result can be proved in a straightforward way by integrating the joint PDF of \((X, Y)\) over \(\{(x, y): 0 \lt x \lt y \lt \infty\}\). A more elegant proof uses conditioning and the moment generating function above: \[ \P(Y \gt X) = \E\left[\P(Y \gt X \mid X)\right] = \E\left(e^{-b X}\right) = \frac{a}{a + b}\]
The following theorem gives an important random version of the memoryless property.
Suppose that \(X\) and \(Y\) are independent variables taking values in \([0, \infty)\) and that \(Y\) has the exponential distribution with rate parameter \(r \gt 0\). Then \(X\) and \(Y - X\) are conditionally independent given \(X \lt Y\), and the conditional distribution of \(Y - X\) is also exponential with parameter \(r\).
Proof
Suppose that \(A \subseteq [0, \infty)\) (measurable of course) and \(t \ge 0\). Then \[ \P(X \in A, Y - X \ge t \mid X \lt Y) = \frac{\P(X \in A, Y - X \ge t)}{\P(X \lt Y)} \] But conditioning on \(X\) we can write the numerator as \[ \P(X \in A, Y - X \gt t) = \E\left[\P(X \in A, Y - X \gt t \mid X)\right] = \E\left[\P(Y \gt X + t \mid X), X \in A\right] = \E\left[e^{-r(t + X)}, X \in A\right] = e^{-rt} \E\left(e^{-r\,X}, X \in A\right) \] Similarly, conditioning on \(X\) gives \(\P(X \lt Y) = \E\left(e^{-r\,X}\right)\). Thus \[ \P(X \in A, Y - X \gt t \mid X \lt Y) = e^{-r\,t} \frac{\E\left(e^{-r\,X}, X \in A\right)}{\E\left(e^{-rX}\right)} \] Letting \(A = [0, \infty)\) we have \(\P(Y \gt t) = e^{-r\,t}\) so given \(X \lt Y\), the variable \(Y - X\) has the exponential distribution with parameter \(r\). Letting \(t = 0\), we see that given \(X \lt Y\), variable \(X\) has the distribution \[ A \mapsto \frac{\E\left(e^{-r\,X}, X \in A\right)}{\E\left(e^{-r\,X}\right)} \] Finally, because of the factoring, \(X\) and \(Y - X\) are conditionally independent given \(X \lt Y\).
For our next discussion, suppose that \(\bs{X} = (X_1, X_2, \ldots, X_n)\) is a sequence of independent random variables, and that \(X_i\) has the exponential distribution with rate parameter \(r_i \gt 0\) for each \(i \in \{1, 2, \ldots, n\}\).
Let \(U = \min\{X_1, X_2, \ldots, X_n\}\). Then \(U\) has the exponential distribution with parameter \(\sum_{i=1}^n r_i\).
Proof
Recall that in general, \(\{U \gt t\} = \{X_1 \gt t, X_2 \gt t, \ldots, X_n \gt t\}\) and therefore by independence, \(F^c(t) = F^c_1(t) F^c_2(t) \cdots F^c_n(t)\) for \(t \ge 0\), where \(F^c\) is the reliability function of \(U\) and \(F^c_i\) is the reliability function of \(X_i\) for each \(i\). When \(X_i\) has the exponential distribution with rate \(r_i\) for each \(i\), we have \(F^c(t) = \exp\left[-\left(\sum_{i=1}^n r_i\right) t\right]\) for \(t \ge 0\).
In the context of reliability, if a series system has independent components, each with an exponentially distributed lifetime, then the lifetime of the system is also exponentially distributed, and the failure rate of the system is the sum of the component failure rates. In the context of random processes, if we have \(n\) independent Poisson process, then the new process obtained by combining the random points in time is also Poisson, and the rate of the new process is the sum of the rates of the individual processes (we will return to this point latter).
Let \(V = \max\{X_1, X_2, \ldots, X_n\}\). Then \(V\) has distribution function \( F \) given by \[ F(t) = \prod_{i=1}^n \left(1 - e^{-r_i t}\right), \quad t \in [0, \infty) \]
Proof
Recall that in general, \(\{V \le t\} = \{X_1 \le t, X_2 \le t, \ldots, X_n \le t\}\) and therefore by independence, \(F(t) = F_1(t) F_2(t) \cdots F_n(t)\) for \(t \ge 0\), where \(F\) is the distribution function of \(V\) and \(F_i\) is the distribution function of \(X_i\) for each \(i\).
Consider the special case where \( r_i = r \in (0, \infty) \) for each \( i \in \N_+ \). In statistical terms, \(\bs{X}\) is a random sample of size \( n \) from the exponential distribution with parameter \( r \). From the last couple of theorems, the minimum \(U\) has the exponential distribution with rate \(n r\) while the maximum \(V\) has distribution function \(F(t) = \left(1 - e^{-r t}\right)^n\) for \(t \in [0, \infty)\). Recall that \(U\) and \(V\) are the first and last order statistics, respectively.
In the order statistic experiment, select the exponential distribution.
- Set \(k = 1\) (this gives the minimum \(U\)). Vary \(n\) with the scroll bar and note the shape of the probability density function. For selected values of \(n\), run the simulation 1000 times and compare the empirical density function to the true probability density function.
- Vary \(n\) with the scroll bar, set \(k = n\) each time (this gives the maximum \(V\)), and note the shape of the probability density function. For selected values of \(n\), run the simulation 1000 times and compare the empirical density function to the true probability density function.
Curiously, the distribution of the maximum of independent, identically distributed exponential variables is also the distribution of the sum of independent exponential variables, with rates that grow linearly with the index.
Suppose that \( r_i = i r \) for each \( i \in \{1, 2, \ldots, n\} \) where \( r \in (0, \infty) \). Then \( Y = \sum_{i=1}^n X_i \) has distribution function \( F \) given by \[ F(t) = (1 - e^{-r t})^n, \quad t \in [0, \infty) \]
Proof
By assumption, \( X_k \) has PDF \( f_k \) given by \( f_k(t) = k r e^{-k r t} \) for \( t \in [0, \infty) \). We want to show that \( Y_n = \sum_{i=1}^n X_i\) has PDF \( g_n \) given by \[ g_n(t) = n r e^{-r t} (1 - e^{-r t})^{n-1}, \quad t \in [0, \infty) \] The PDF of a sum of independent variables is the convolution of the individual PDFs, so we want to show that \[ f_1 * f_2 * \cdots * f_n = g_n, \quad n \in \N_+ \] The proof is by induction on \( n \). Trivially \( f_1 = g_1 \), so suppose the result holds for a given \( n \in \N_+ \). Then \begin{align*} g_n * f_{n+1}(t) & = \int_0^t g_n(s) f_{n+1}(t - s) ds = \int_0^t n r e^{-r s}(1 - e^{-r s})^{n-1} (n + 1) r e^{-r (n + 1) (t - s)} ds \\ & = r (n + 1) e^{-r(n + 1)t} \int_0^t n(1 - e^{-rs})^{n-1} r e^{r n s} ds \end{align*} Now substitute \( u = e^{r s} \) so that \( du = r e^{r s} ds \) or equivalently \(r ds = du / u\). After some algebra, \begin{align*} g_n * f_{n+1}(t) & = r (n + 1) e^{-r (n + 1)t} \int_1^{e^{rt}} n (u - 1)^{n-1} du \\ & = r(n + 1) e^{-r(n + 1) t}(e^{rt} - 1)^n = r(n + 1)e^{-rt}(1 - e^{-rt})^n = g_{n+1}(t) \end{align*}
This result has an application to the Yule process , named for George Yule. The Yule process, which has some parallels with the Poisson process, is studied in the chapter on Markov processes. We can now generalize the order probability above:
For \(i \in \{1, 2, \ldots, n\}\), \[ \P\left(X_i \lt X_j \text{ for all } j \ne i\right) = \frac{r_i}{\sum_{j=1}^n r_j} \]
Proof
First, note that \(X_i \lt X_j\) for all \(i \ne j\) if and only if \(X_i \lt \min\{X_j: j \ne i\}\). But the minimum on the right is independent of \(X_i\) and, by result on minimums above, has the exponential distribution with parameter \(\sum_{j \ne i} r_j\). The result now follows from order probability for two events above.
Suppose that for each \(i\), \(X_i\) is the time until an event of interest occurs (the arrival of a customer, the failure of a device, etc.) and that these times are independent and exponentially distributed. Then the first time \(U\) that one of the events occurs is also exponentially distributed, and the probability that the first event to occur is event \(i\) is proportional to the rate \(r_i\).
The probability of a total ordering is \[ \P(X_1 \lt X_2 \lt \cdots \lt X_n) = \prod_{i=1}^n \frac{r_i}{\sum_{j=i}^n r_j} \]
Proof
Let \( A = \left\{X_1 \lt X_j \text{ for all } j \in \{2, 3, \ldots, n\}\right\} \). then \[ \P(X_1 \lt X_2 \lt \cdots \lt X_n) = \P(A, X_2 \lt X_3 \lt \cdots \lt X_n) = \P(A) \P(X_2 \lt X_3 \lt \cdots \lt X_n \mid A) \] But \( \P(A) = \frac{r_1}{\sum_{i=1}^n r_i} \) from the previous result, and \( \{X_2 \lt X_3 \lt \cdots \lt X_n\} \) is independent of \( A \). Thus we have \[ \P(X_1 \lt X_2 \lt \cdots \lt X_n) = \frac{r_1}{\sum_{i=1}^n r_i} \P(X_2 \lt X_3 \lt \cdots \lt X_n) \] so the result follows by induction.
Of course, the probabilities of other orderings can be computed by permuting the parameters appropriately in the formula on the right.
The result on minimums and the order probability result above are very important in the theory of continuous-time Markov chains. But for that application and others, it's convenient to extend the exponential distribution to two degenerate cases: point mass at 0 and point mass at \( \infty \) (so the first is the distribution of a random variable that takes the value 0 with probability 1, and the second the distribution of a random variable that takes the value \( \infty \) with probability 1). In terms of the rate parameter \( r \) and the distribution function \( F \), point mass at 0 corresponds to \( r = \infty \) so that \( F(t) = 1 \) for \( 0 \lt t \lt \infty \). Point mass at \( \infty \) corresponds to \( r = 0 \) so that \( F(t) = 0 \) for \( 0 \lt t \lt \infty \). The memoryless property, as expressed in terms of the reliability function \( F^c \), still holds for these degenerate cases on \( (0, \infty) \): \[ F^c(s) F^c(t) = F^c(s + t), \quad s, \, t \in (0, \infty) \] We also need to extend some of results above for a finite number of variables to a countably infinite number of variables. So for the remainder of this discussion, suppose that \( \{X_i: i \in I\} \) is a countable collection of independent random variables, and that \( X_i \) has the exponential distribution with parameter \( r_i \in (0, \infty) \) for each \( i \in I \).
Let \( U = \inf\{X_i: i \in I\} \). Then \( U \) has the exponential distribution with parameter \( \sum_{i \in I} r_i \)
Proof
The proof is almost the same as the one above for a finite collection. Note that \( \{U \ge t\} = \{X_i \ge t \text{ for all } i \in I\} \) and so \[ \P(U \ge t) = \prod_{i \in I} \P(X_i \ge t) = \prod_{i \in I} e^{-r_i t} = \exp\left[-\left(\sum_{i \in I} r_i\right)t \right] \] If \( \sum_{i \in I} r_i \lt \infty \) then \( U \) has a proper exponential distribution with the sum as the parameter. If \( \sum_{i \in I} r_i = \infty \) then \( P(U \ge t) = 0 \) for all \( t \in (0, \infty) \) so \( P(U = 0) = 1 \).
For \(i \in \N_+\), \[ \P\left(X_i \lt X_j \text{ for all } j \in I - \{i\}\right) = \frac{r_i}{\sum_{j \in I} r_j} \]
Proof
First note that since the variables have continuous distributions and \( I \) is countable, \[ \P\left(X_i \lt X_j \text{ for all } j \in I - \{i\} \right) = \P\left(X_i \le X_j \text{ for all } j \in I - \{i\}\right)\] Next note that \(X_i \le X_j\) for all \(j \in I - \{i\}\) if and only if \(X_i \le U_i \) where \(U_i = \inf\left\{X_j: j \in I - \{i\}\right\}\). But \( U_i \) is independent of \(X_i\) and, by previous result , has the exponential distribution with parameter \(s_i = \sum_{j \in I - \{i\}} r_j\). If \( s_i = \infty \), then \( U_i \) is 0 with probability 1, and so \( P(X_i \le U_i) = 0 = r_i / s_i \). If \( s_i \lt \infty \), then \( X_i \) and \( U_i \) have proper exponential distributions, and so the result now follows from order probability for two variables above.
We need one last result in this setting: a condition that ensures that the sum of an infinite collection of exponential variables is finite with probability one.
Let \( Y = \sum_{i \in I} X_i \) and \( \mu = \sum_{i \in I} 1 / r_i \). Then \( \mu = \E(Y) \) and \( \P(Y \lt \infty) = 1 \) if and only if \( \mu \lt \infty \).
Proof
The result is trivial if \( I \) is finite, so assume that \( I = \N_+ \). Recall that \( \E(X_i) = 1 / r_i \) and hence \( \mu = \E(Y) \). Trivially if \( \mu \lt \infty \) then \( \P(Y \lt \infty) = 1 \). Conversely, suppose that \( \P(Y \lt \infty) = 1 \). Then \( \P(e^{-Y} \gt 0) = 1 \) and hence \( \E(e^{-Y}) \gt 0 \). Using independence and the moment generating function above, \[ \E(e^{-Y}) = \E\left(\prod_{i=1}^\infty e^{-X_i}\right) = \prod_{i=1}^\infty \E(e^{-X_i}) = \prod_{i=1}^\infty \frac{r_i}{r_i + 1} \gt 0\] Next recall that if \( p_i \in (0, 1) \) for \( i \in \N_+ \) then \[ \prod_{i=1}^\infty p_i \gt 0 \text{ if and only if } \sum_{i=1}^\infty (1 - p_i) \lt \infty \] Hence it follows that \[ \sum_{i=1}^\infty \left(1 - \frac{r_i}{r_i + 1}\right) = \sum_{i=1}^\infty \frac{1}{r_i + 1} \lt \infty \] In particular, this means that \( 1/(r_i + 1) \to 0 \) as \( i \to \infty \) and hence \( r_i \to \infty \) as \( i \to \infty \). But then \[ \frac{1/(r_i + 1)}{1/r_i} = \frac{r_i}{r_i + 1} \to 1 \text{ as } i \to \infty \] By the comparison test for infinite series, it follows that \[ \mu = \sum_{i=1}^\infty \frac{1}{r_i} \lt \infty \]
Computational Exercises
Show directly that the exponential probability density function is a valid probability density function.
Solution
Clearly \( f(t) = r e^{-r t} \gt 0 \) for \( t \in [0, \infty) \). Simple integration that \[ \int_0^\infty r e^{-r t} \, dt = 1 \]
Suppose that the length of a telephone call (in minutes) is exponentially distributed with rate parameter \(r = 0.2\). Find each of the following:
- The probability that the call lasts between 2 and 7 minutes.
- The median, the first and third quartiles, and the interquartile range of the call length.
Answer
Let \(X\) denote the call length.
- \(\P(2 \lt X \lt 7) = 0.4237\)
- \(q_1 = 1.4384\), \(q_2 = 3.4657\), \(q_3 = 6.9315\), \(q_3 - q_1 = 5.4931\)
Suppose that the lifetime of a certain electronic component (in hours) is exponentially distributed with rate parameter \(r = 0.001\). Find each of the following:
- The probability that the component lasts at least 2000 hours.
- The median, the first and third quartiles, and the interquartile range of the lifetime.
Answer
Let \(T\) denote the lifetime
- \(\P(T \ge 2000) = 0.1353\)
- \(q_1 = 287.682\), \(q_2 = 693.147\), \(q_3 = 1386.294\), \(q_3 - q_1 = 1098.612\)
Suppose that the time between requests to a web server (in seconds) is exponentially distributed with rate parameter \(r = 2\). Find each of the following:
- The mean and standard deviation of the time between requests.
- The probability that the time between requests is less that 0.5 seconds.
- The median, the first and third quartiles, and the interquartile range of the time between requests.
Answer
Let \(T\) denote the time between requests.
- \(\E(T) = 0.5\), \(\sd(T) = 0.5\)
- \(\P(T \lt 0.5) = 0.6321\)
- \(q_1 = 0.1438\), \(q_2 = 0.3466\), \(q_3 = 0.6931\), \(q_3 - q_1 = 0.5493\)
Suppose that the lifetime \(X\) of a fuse (in 100 hour units) is exponentially distributed with \(\P(X \gt 10) = 0.8\). Find each of the following:
- The rate parameter.
- The mean and standard deviation.
- The median, the first and third quartiles, and the interquartile range of the lifetime.
Answer
Let \(X\) denote the lifetime.
- \(r = 0.02231\)
- \(\E(X) = 44.814\), \(\sd(X) = 44.814\)
- \(q_1 = 12.8922\), \(q_2 = 31.0628\), \(q_3 = 62.1257\), \(q_3 - q_1 = 49.2334\)
The position \(X\) of the first defect on a digital tape (in cm) has the exponential distribution with mean 100. Find each of the following:
- The rate parameter.
- The probability that \(X \lt 200\) given \(X \gt 150\).
- The standard deviation.
- The median, the first and third quartiles, and the interquartile range of the position.
Answer
Let \(X\) denote the position of the first defect.
- \(r = 0.01\)
- \(\P(X \lt 200 \mid X \gt 150) = 0.3935\)
- \(\sd(X) = 100\)
- \(q_1 = 28.7682\), \(q_2 = 69.3147\), \(q_3 = 138.6294\), \(q_3 - q_1 = 109.6812\)
Suppose that \( X, \, Y, \, Z \) are independent, exponentially distributed random variables with respective parameters \( a, \, b, \, c \in (0, \infty) \). Find the probability of each of the 6 orderings of the variables.
Proof
- \( \P(X \lt Y \lt Z) = \frac{a}{a + b + c} \frac{b}{b + c} \)
- \( \P(X \lt Z \lt Y) = \frac{a}{a + b + c} \frac{c}{b + c} \)
- \( \P(Y \lt X \lt Z) = \frac{b}{a + b + c} \frac{a}{a + c} \)
- \( \P(Y \lt Z \lt X) = \frac{b}{a + b + c} \frac{c}{a + c} \)
- \( \P(Z \lt X \lt Y) = \frac{c}{a + b + c} \frac{a}{a + b} \)
- \( \P(Z \lt Y \lt X) = \frac{c}{a + b + c} \frac{b}{a + b} \)


 The key signature reminds the performer which sharps or flats are in the scale (or key) of the piece and prevents the composer or arranger from writing every sharp or flat from the scale every time it occurs.
There are 15 major key signatures. The key of C major has no sharps or flats in the key signature. The other key signatures can have between 1 to 7 sharps and 1 to 7 flats, giving us the other 14 key signatures.
It is important to memorize the order of sharps and flats, since you will be writing key signatures regularly. The order of sharps is FF–CC–GG–DD–AA–EE–BB, often remembered by a mnemonic. One common mnemonic for the order of sharps is “Fast Cars Go Dangerously Around Every Bend.” The order of flats is BB–EE–AA–DD–GG–CC–FF. It is the reverse of the order of sharps. It is easy to remember since the first four letters make the word BEAD, and GCF is something most students learn as “Greatest Common Factor” when studying math in elementary school. A mnemonic that works forward and backward is “Father Charles Goes Down And Ends Battle,” which reversed is “Battle Ends And Down Goes Charles’ Father.” A helpful learning device to remember the order of keys in relation to the order of sharps and flats is the circle of fifths . As you ascend in fifths (clockwise), key signatures get one degree “sharper.” (CC to GG is a fifth because CC=1, DD=2, EE=3, FF=4, and GG=5.) As you descend in fifths (counterclockwise), key signatures get one degree “flatter.”
Note the overlapping keys at the bottom of the circle. BB major is enharmonically the same as CC♭ major, FF♯ major is enharmonically the same as GG♭ major, and CC♯major is enharmonically the same as DD♭ major.
2.3.1 Identifying Key Signatures
While it is preferable to memorize key signatures, use the following method to determine major key signatures based on the sharps or flats in the key signature.
- For key signatures withs sharps : Go up a half step from the last sharp to find the key.
- For key signatures with flats : The second-to-last flat is the key.


5.4: Sonata Form - Thematic Modules
The four thematic modules that comprise a typical two-part sonata exposition (P, TR, S, and C) create a thematic rotation that is found again in the recapitulation and often referenced in the development.
Primary theme module (P)
The P theme has several functions: (1) establish the home key, (2) present the primary melodic material to begins the thematic cycle, and (3) begin the motion toward the MC and the EEC.
Most often a P theme is harmonically closed, ending with a I:PAC or, less typically, with a I:HC. (In smaller sonatas, P might be a single phrase (antecedent or presentation) that does not end with a cadence.) When a P theme ends with a PAC and is followed by another theme also ending with a PAC, the P module contains a “primary theme group.” The themes in a primary theme group are differentiated with subscripts: P1 and P2.
A P theme may exhibit any standard theme type (sentence, period, hybrid, compound), though sentences are more common. Typically, a P theme is relatively “tight-knit” as compared to other thematic modules, containing a more straightforward presentation of tonic and less phrase deviations.
Transition module (TR)
The TR module’s principal functional role is to drive toward the MC that marks its end. This is both a harmonic motion and a rhetorical motion, characterized by energy gain . An analysis of a TR module should center around the MC and how the composer approaches it.
Harmonic motion
In a sonata, a transition may be “modulating” or “non-modulating.” A non-modulating TR will lead to a I:HC MC, while a modulating TR leads to a V (or III): HC MC.
Melodic/motivic characteristics
We will follow Hepokoski & Darcy’s practice of locating the beginning of TR at the start of a phrase. In general, once you hear TR function clearly projected, track back to the beginning of that phrase and label it the beginning of TR. Unlike primary themes, transitions are much “looser” thematically. This module of the sonata is often associated with phrase expansions and compression. More generally, anything that can be associated with continuation function fits transition function, as well: fragmentation, liquidation, acceleration of melodic or harmonic rhythm, etc.
Melodically, we classify transitions as (1) independent , or (2) dissolving .
An independent TR begins with new thematic material. In other words, it is not P-based.
Dissolving TR modules all take some part of the P theme and dissolve, degenerate, or liquidate as the module gains energy and moves toward the MC.
- In the case of a dissolving restatement, P ends and then seems to begin again. This restatement of P dissolves into TR function, and we can subsequently reinterpret the whole theme as a dissolving-restatement type of TR.
- In a dissolving consequent, continuation, or hybrid, the P module will consist of the opening phrase of a theme, and the closing phrase of the theme will begin as usual but degenerate into TR.
- A dissolving P-codetta will introduce post-cadential material to reinforce the cadence at the end of P, and that post-cadential material will dissolve into TR rhetoric.
In the following example from Beethoven’s Piano Sonata in G minor, Op. 49, No. 1, a III: HC MC is clearly found in m. 15 and coincides with the structural HC. Backtracking, notice the continuational characteristics at m. 12 which indicate that the transition must begin at m. 9. (Remember, we always begin a transition at the start of a phrase.) Both the primary theme and transition begin with the same presentation, and therefore, this is a modulating , dissolving transition. As a listener, we will likely not realize that the transition is a transition until we reach m. 12, where the continuation begins to project characteristics associated with TR.
Beethoven, Piano Sonata in G minor, Op. 49, No. 1, i
Subordinate theme module (S)
The chief function of S , which is in the subordinate key of the sonata, is to lead to a PAC in that key — the essential expositional closure (EEC) . Because of its role in relation to this central harmonic event (and its corresponding cadence in the recapitulation, the ESC), the S module is of immense importance and interest in a sonata-form movement. Hepokoski and Darcy go so far as to say that “what happens in S makes a sonata a sonata” (p. 117). Because the EEC is of such structural importance, its arrival is often delayed and dramatized to a great degree.
Thematically, the S theme is much looser than the P theme. It is typical to find extended continuations and expanded cadential passages, all as a means towards dramatizing the eventual arrival at the EEC.
It is important to note that new melodic material is not a requirement of S. In fact, many sonatas – especially those composed by Haydn – have S themes that resemble the P theme. (These are called mono-thematic sonatas.)
Closing module (C)
The definitive characteristics of C are that it follows the EEC, and that it is not S. C modules can present wholly new thematic material, or they can borrow from P or TR. They cannot, by definition, be S-based, since that would be a continuation of S. (Keep in mind that the EEC must go on to new material, otherwise the S module continues and the EEC has not been reached yet.)
The C module will always be in the secondary key. It is post-cadential, and the harmonic goal of the exposition has already been reached. If C goes somewhere else, it is not C.
Retransition (RT)
A retransition is like a turnaround in pop/rock or blues music. It is a dominant chord or arrival in the home key that prepares the return to the home key at the beginning of the repeat of the exposition. The difference between an RT and a turnaround is that an RT follows a modulation. When the RT follows a secondary key of V, it turns I in the key of V into V into the key of I by repetition, melodic figuration, or the adding of a chordal seventh.


20.4: Capillary Exchange
By the end of this section, you will be able to:
- Identify the primary mechanisms of capillary exchange
- Distinguish between capillary hydrostatic pressure and blood colloid osmotic pressure, explaining the contribution of each to net filtration pressure
- Compare filtration and reabsorption
- Explain the fate of fluid that is not reabsorbed from the tissues into the vascular capillaries
The primary purpose of the cardiovascular system is to circulate gases, nutrients, wastes, and other substances to and from the cells of the body. Small molecules, such as gases, lipids, and lipid-soluble molecules, can diffuse directly through the membranes of the endothelial cells of the capillary wall. Glucose, amino acids, and ions—including sodium, potassium, calcium, and chloride—use transporters to move through specific channels in the membrane by facilitated diffusion. Glucose, ions, and larger molecules may also leave the blood through intercellular clefts. Larger molecules can pass through the pores of fenestrated capillaries, and even large plasma proteins can pass through the great gaps in the sinusoids. Some large proteins in blood plasma can move into and out of the endothelial cells packaged within vesicles by endocytosis and exocytosis. Water moves by osmosis.
Bulk Flow
The mass movement of fluids into and out of capillary beds requires a transport mechanism far more efficient than mere diffusion. This movement, often referred to as bulk flow, involves two pressure-driven mechanisms: Volumes of fluid move from an area of higher pressure in a capillary bed to an area of lower pressure in the tissues via filtration . In contrast, the movement of fluid from an area of higher pressure in the tissues into an area of lower pressure in the capillaries is reabsorption . Two types of pressure interact to drive each of these movements: hydrostatic pressure and osmotic pressure.
Hydrostatic Pressure
The primary force driving fluid transport between the capillaries and tissues is hydrostatic pressure, which can be defined as the pressure of any fluid enclosed in a space. Blood hydrostatic pressure is the force exerted by the blood confined within blood vessels or heart chambers. Even more specifically, the pressure exerted by blood against the wall of a capillary is called capillary hydrostatic pressure (CHP) , and is the same as capillary blood pressure. CHP is the force that drives fluid out of capillaries and into the tissues.
As fluid exits a capillary and moves into tissues, the hydrostatic pressure in the interstitial fluid correspondingly rises. This opposing hydrostatic pressure is called the interstitial fluid hydrostatic pressure (IFHP) . Generally, the CHP originating from the arterial pathways is considerably higher than the IFHP, because lymphatic vessels are continually absorbing excess fluid from the tissues. Thus, fluid generally moves out of the capillary and into the interstitial fluid. This process is called filtration.
Osmotic Pressure
The net pressure that drives reabsorption—the movement of fluid from the interstitial fluid back into the capillaries—is called osmotic pressure (sometimes referred to as oncotic pressure). Whereas hydrostatic pressure forces fluid out of the capillary, osmotic pressure draws fluid back in. Osmotic pressure is determined by osmotic concentration gradients, that is, the difference in the solute-to-water concentrations in the blood and tissue fluid. A region higher in solute concentration (and lower in water concentration) draws water across a semipermeable membrane from a region higher in water concentration (and lower in solute concentration).
As we discuss osmotic pressure in blood and tissue fluid, it is important to recognize that the formed elements of blood do not contribute to osmotic concentration gradients. Rather, it is the plasma proteins that play the key role. Solutes also move across the capillary wall according to their concentration gradient, but overall, the concentrations should be similar and not have a significant impact on osmosis. Because of their large size and chemical structure, plasma proteins are not truly solutes, that is, they do not dissolve but are dispersed or suspended in their fluid medium, forming a colloid rather than a solution.
The pressure created by the concentration of colloidal proteins in the blood is called the blood colloidal osmotic pressure (BCOP) . Its effect on capillary exchange accounts for the reabsorption of water. The plasma proteins suspended in blood cannot move across the semipermeable capillary cell membrane, and so they remain in the plasma. As a result, blood has a higher colloidal concentration and lower water concentration than tissue fluid. It therefore attracts water. We can also say that the BCOP is higher than the interstitial fluid colloidal osmotic pressure (IFCOP) , which is always very low because interstitial fluid contains few proteins. Thus, water is drawn from the tissue fluid back into the capillary, carrying dissolved molecules with it. This difference in colloidal osmotic pressure accounts for reabsorption.
Interaction of Hydrostatic and Osmotic Pressures
The normal unit used to express pressures within the cardiovascular system is millimeters of mercury (mm Hg). When blood leaving an arteriole first enters a capillary bed, the CHP is quite high—about 35 mm Hg. Gradually, this initial CHP declines as the blood moves through the capillary so that by the time the blood has reached the venous end, the CHP has dropped to approximately 18 mm Hg. In comparison, the plasma proteins remain suspended in the blood, so the BCOP remains fairly constant at about 25 mm Hg throughout the length of the capillary and considerably above the osmotic pressure in the interstitial fluid.
The net filtration pressure (NFP) represents the interaction of the hydrostatic and osmotic pressures, driving fluid out of the capillary. It is equal to the difference between the CHP and the BCOP. Since filtration is, by definition, the movement of fluid out of the capillary, when reabsorption is occurring, the NFP is a negative number.
NFP changes at different points in a capillary bed (Figure \(\PageIndex{1}\)). Close to the arterial end of the capillary, it is approximately 10 mm Hg, because the CHP of 35 mm Hg minus the BCOP of 25 mm Hg equals 10 mm Hg. Recall that the hydrostatic and osmotic pressures of the interstitial fluid are essentially negligible. Thus, the NFP of 10 mm Hg drives a net movement of fluid out of the capillary at the arterial end. At approximately the middle of the capillary, the CHP is about the same as the BCOP of 25 mm Hg, so the NFP drops to zero. At this point, there is no net change of volume: Fluid moves out of the capillary at the same rate as it moves into the capillary. Near the venous end of the capillary, the CHP has dwindled to about 18 mm Hg due to loss of fluid. Because the BCOP remains steady at 25 mm Hg, water is drawn into the capillary, that is, reabsorption occurs. Another way of expressing this is to say that at the venous end of the capillary, there is an NFP of −7 mm Hg.
The Role of Lymphatic Capillaries
Since overall CHP is higher than BCOP, it is inevitable that more net fluid will exit the capillary through filtration at the arterial end than enters through reabsorption at the venous end. Considering all capillaries over the course of a day, this can be quite a substantial amount of fluid: Approximately 24 liters per day are filtered, whereas 20.4 liters are reabsorbed. This excess fluid is picked up by capillaries of the lymphatic system. These extremely thin-walled vessels have copious numbers of valves that ensure unidirectional flow through ever-larger lymphatic vessels that eventually drain into the subclavian veins in the neck. An important function of the lymphatic system is to return the fluid (lymph) to the blood. Lymph may be thought of as recycled blood plasma . (Seek additional content for more detail on the lymphatic system.)
Interactive Link
Watch this video to explore capillaries and how they function in the body. Capillaries are never more than 100 micrometers away. What is the main component of interstitial fluid?


12.8: Using Symmetry to Solve Secular Determinants
As we continue with this course, we will discover that there are many times when we would like to know whether a particular integral is necessarily zero, or whether there is a chance that it may be non-zero. We can often use group theory to differentiate these two cases.
You will have already used symmetry properties of functions to determine whether or not a one-dimensional integral is zero. For example, cos(x) is an ‘even’ function (symmetric with respect to reflection through the origin), and it follows from this that
\[\int^{\infty}_{-\infty} \cos(x) dx = 0 \nonumber \]
In general integral between these limits for any other even function will be also be zero.
In the general case we may have an integral of more than one dimension. The key to determining whether a general integral is necessarily zero lies in the fact that because an integral is just a number, it must be invariant to any symmetry operation. For example, bonding in a diatomic (see next section) depends on the presence of a non-zero overlap between atomic orbitals on adjacent atoms, which may be quantified by an overlap integral . You would not expect the bonding in a molecule to change if you rotated the molecule through some angle \(\theta\) , so the integral must be invariant to rotation, and indeed to any other symmetry operation.
In group theoretical terms, for an integral to be non-zero , the integrand must transform as the totally symmetric irreducible representation in the appropriate point group . In practice, the integrand may not transform as a single irreducible representation, but it must include the totally symmetric irreducible representation. These ideas should become more clear in the next section.
It should be noted that even when the irreducible representations spanned by the integrand do include the totally symmetric irreducible representation, it is still possible for the integral to be zero. All group theory allows us to do is identify integrals that are necessarily zero based on the symmetry (or lack thereof) of the integrand.
Secular Equations
As we have seen already, any set of linear equations may be rewritten as a matrix equation \(A\textbf{x}\) = \(\textbf{b}\). Linear equations are classified as simultaneous linear equations or homogeneous linear equations , depending on whether the vector \(\textbf{b}\) is non-zero or zero. For a set of simultaneous linear equations (non-zero \(\textbf{b}\)) it is fairly apparent that if a unique solution exists, it can be found by multiplying both sides by the inverse matrix \(A^{-1}\) (since \(A^{-1}A\) on the left hand side is equal to the identity matrix, which has no effect on the vector \(\textbf{x}\))
\[\begin{array}{rcl} A\textbf{x} & = & \textbf{b} \\ A^{-1}A\textbf{x} & = & A^{-1}\textbf{b} \\ \textbf{x} & = & A^{-1}\textbf{b} \end{array} \label{21.1} \]
In practice, there are easier matrix methods for solving simultaneous equations than finding the inverse matrix, but these need not concern us here. We previously argued that in order for a matrix to have an inverse, it must have a non-zero determinant . Since \(A^{-1}\) must exist in order for a set of simultaneous linear equations to have a solution, this means that the determinant of the matrix \(A\) must be non-zero for the equations to be solvable.
For a matrix to have an inverse, it must have a non-zero determinant.
The reverse is true for homogeneous linear equations. In this case the set of equations only has a solution if the determinant of \(A\) is equal to zero. The secular equations we want to solve are homogeneous equations, and we will use this property of the determinant to determine the molecular orbital energies. An important property of homogeneous equations is that if a vector \(\textbf{x}\) is a solution, so is any multiple of \(\textbf{x}\), meaning that the solutions (the molecular orbitals) can be normalized without causing any problems.
Solving for the orbital energies and expansion coefficients
Recall the secular equations for the \(A_1\) orbitals of \(NH_3\) derived in the previous section
\[\begin{array}{rcl} c_1(H_{11} - ES_{11}) + c_2(H_{12} - ES_{12}) & = & 0 \\ c_1(H_{12} - ES_{12}) + c_2(H_{22} - ES_{22}) & = & 0 \end{array} \label{21.2} \]
where \(c_1\) and \(c_2\) are the coefficients in the linear combination of the SALCs \(\phi_1\) = \(s_N\) and \(\phi_2\) = \(\dfrac{1}{\sqrt{3}}(s_1 + s_2 + s_3)\) used to construct the molecular orbital. Writing this set of homogeneous linear equations in matrix form gives
\[\begin{pmatrix} H_{11} - ES_{11} & H_{12} - ES_{12} \\ H_{12} - ES_{12} & H_{22} - ES_{22} \end{pmatrix} \begin{pmatrix} c_1 \\ c_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \label{21.3} \]
In order for the equations to have a solution, the determinant of the matrix must be equal to zero. Writing out the determinant will give us a polynomial equation in \(E\) that we can solve to obtain the orbital energies in terms of the Hamiltonian matrix elements \(H_{ij}\) and overlap integrals \(S_{ij}\). The number of energies obtained by ‘solving the secular determinant’ in this way is equal to the order of the matrix, in this case two.
The secular determinant for Equation (21.3) is (noting that \(S_{11}\) = \(S_{22} = 1\) since the SALCs are normalized)
\[(H_{11} - E)(H_{22} - E) - (H_{12} - ES_{12})^2 = 0 \label{21.4} \]
Expanding and collecting terms in \(E\) gives
\[E^2(1-S_{12}^2) + E(2H_{12}S_{12} - H_{11} - H_{22}) + (H_{11}H_{22} - H_{12}^2) = 0 \label{21.5} \]
which can be solved using the quadratic formula to give the energies of the two molecular orbitals.
\[E_\pm = \dfrac{-(2H_{12}S_{12} - H_{11} - H_{22}) \pm \sqrt{(2H_{12}S_{12} - H_{11} - H_{22})^2 - 4(1-S_{12}^2)(H_{11}H_{22} - H_{12}^2)}}{2(1-S_{12}^2)} \label{21.6} \]
To obtain numerical values for the energies, we need to evaluate the integrals \(H_{11}\), \(H_{22}\), \(H_{12}\), and \(S_{12}\). This would be quite a challenge to do analytically, but luckily there are a number of computer programs that can be used to calculate the integrals. One such program gives the following values.
\[\begin{array}{rcl} H_{11} & = & -26.0000 \: eV \\ H_{22} & = & -22.2216 \: eV \\ H_{12} & = & -29.7670 \: eV \\ S_{12} & = & \: 0.8167 \: eV \end{array} \label{21.7} \]
When we substitute these into our equation for the energy levels, we get:
\[\begin{array}{rcl} E_+ & = & \: 29.8336 \: eV \\ E_- & = & -31.0063 \: eV \end{array} \label{21.8} \]
We now have the orbital energies and the next step is to find the orbital coefficients. The coefficients for an orbital of energy \(E\) are found by substituting the energy into the secular equations and solving for the coefficients \(c_i\). Since the two secular equations are not linearly independent (i.e. they are effectively only one equation), when we solve them to find the coefficients what we will end up with is the relative values of the coefficients. This is true in general: in a system with \(N\) coefficients, solving the secular equations will allow all \(N\) of the coefficients \(c_i\) to be obtained in terms of, say, \(c_1\). The absolute values of the coefficients are found by normalizing the wavefunction.
Since the secular equations for the orbitals of energy \(E_+\) and \(E_-\) are not linearly independent, we can choose to solve either one of them to find the orbital coefficients. We will choose the first.
\[(H_{11} - E_{\pm})c_1 + (H_{12} - E_{\pm}S_{12})c_2 = 0 \label{21.9} \]
For the orbital with energy \(E_-\) = -31.0063 eV, substituting numerical values into this equation gives
\[\begin{array}{rcl} 5.0063 c_1 - 4.4442 c_2 & = & 0 \\ c_2 & = & 1.1265 c_1 \end{array} \label{21.10} \]
The molecular orbital is therefore
\[\Psi = c_1(\phi_1 + 1.1265\phi_2) \label{21.11} \]
Normalizing to find the constant \(c_1\) (by requiring \(<\Psi|\Psi>\) = 1) gives
\[\begin{array}{rcll} \Psi_1 & = & 0.4933\phi_1 + 0.5557\phi_2 & \\ & = & 0.4933s_N + 0.3208(s_1 + s_2 + s_3) & (\text{substituting the SALCs for} \: \phi_1 \: \text{and} \: \phi_2) \end{array} \label{21.12} \]
For the second orbital, with energy \(E_+\) = 29.8336 eV, the secular equation is
\[\begin{array}{rcl} -55.8336c_1 - 54.1321c_2 & = & 0 \\ c_2 & = & -1.0314c_1 \end{array} \label{21.13} \]
giving
\[\begin{array}{rcll} \Psi_2 & = & c_1(\phi_1 - 1.0314\phi_2) & \\ & = & 1.6242\phi_1 - 1.6752\phi_2 & \text{(after normalization)} \\ & = & 1.6242s_N -0.9672(s_1 + s_2 + s_3) \end{array} \label{21.14} \]
These two \(A_1\) molecular orbitals \(\Psi_1\) and \(\Psi_2\) , one bonding and one antibonding, are shown below.
The remaining two SALCs arising from the \(s\) orbitals of \(NH_3\) (\(\phi_3 = \dfrac{1}{\sqrt{6}}\begin{pmatrix} 2s_1 - s_2 - s_3 \end{pmatrix}\) and \(\phi_4 = \dfrac{1}{\sqrt{2}}\begin{pmatrix} s_2 - s_3 \end{pmatrix}\) ), form an orthogonal pair of molecula r orbitals of \(E\) symmetry. We can show this by solving the secular determinant to find the orbital energies. The secular equations in this case are:
\[\begin{array}{rcl} c_1(H_{33} - ES_{33}) + c_2(H_{34} -ES_{34}) & = & 0 \\ c_1(H_{34} -ES_{34}) + c_2(H_{44} - ES_{44}) & = & 0 \end{array} \label{21.15} \]
Solving the secular determinant gives
\[E_\pm = \dfrac{-(2H_{34}S_{34} - H_{33} - H_{44}) \pm \sqrt{(2H_{34}S_{34} - H_{33} - H_{44})^2 - 4(1-S_{34}^2)(H_{33}H_{44} - H_{34}^2)}}{2(1-S_{34}^2)} \label{21.16} \]
The integrals required are
\[\begin{array}{rcl} H_{33} & = & -9.2892 \: eV \\ H_{44} & = & -9.2892 \: eV \\ H_{34} & = & 0 \\ S_{34} & = & 0 \end{array} \label{21.17} \]
Using the fact that \(H_{34}\) = \(S_{34} = 0\), the expression for the energies reduces to
\[E_\pm = \dfrac{(H_{33} + H_{44}) \pm (H_{33} - H_{44})}{2} \label{21.18} \]
giving \(E_+\) = \(H_{33}\) = -9.2892 eV and \(E_-\) = \(H_{44}\) = -9.2892 eV. Each SALC therefore forms a molecular orbital by itself, and the two orbitals have the same energy; the two SALCs form an orthogonal pair of degenerate orbitals. These two molecular orbitals of \(E\) symmetry are shown below.
- Choose a basis set of functions \(f_i\) consisting of the valence atomic orbitals on each atom in the system, or some chosen subset of these orbitals.
- With the help of the appropriate character table, determine which irreducible representations are spanned by the basis set using Equation (15.20) to determine the number of times \(a_k\) that the \(k^{th}\) irreducible representation appears in the representation. \[a_k = \dfrac{1}{h}\sum_C n_C \chi(g) \chi_k(g) \label{22.1} \]
- Construct the SALCs \(\phi_i\) that transform as each irreducible representation using Equation 16.1 \[\phi_i = \sum_g \chi_k(g) g f_i \label{22.2} \]
- Write down expressions for the molecular orbitals by taking linear combinations of all the irreducible representations of the same symmetry species.
- Write down the secular equations for the system.
- Solve the secular determinant to obtain the energies of the molecular orbitals.
- Substitute each energy in turn back into the secular equations and solve to obtain the coefficients appearing in your molecular orbital expressions in step 4.
- Normalize the orbitals.