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Ideas in Mathematics and Probability: The Uniform Distribution (2007) – Article by G. Stolyarov II

Ideas in Mathematics and Probability: The Uniform Distribution (2007) – Article by G. Stolyarov II

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The New Renaissance Hat
G. Stolyarov II
July 17, 2014
******************************
Note from the Author: This article was originally published on Associated Content (subsequently, Yahoo! Voices) in 2007.  The article earned over 4,800 page views on Associated Content/Yahoo! Voices, and I seek to preserve it as a valuable resource for readers, subsequent to the imminent closure of Yahoo! Voices. Therefore, this essay is being published directly on The Rational Argumentator for the first time.  ***
***
~ G. Stolyarov II, July 17, 2014
***

The uniform distribution is alternately known as the de Moivre distribution, in honor of the French mathematician Abraham de Moivre (1667-1754) who introduced it to probability theory. The fundamental assumption behind the uniform distribution is that none of the possible outcomes is more or less likely than any other. The uniform distribution applies to continuous random variables, i.e., variables that can assume any values within a specified range.***

Let us say that a given random variable X is uniformly distributed over the interval from a to b. That is, the smallest value X can assume is a and the largest value it can assume is b. To determine the probability density function (pdf) of such a random variable, we need only remember that the total area under the graph of the pdf must equal 1. Since the pdf is constant throughout the interval on which X can assume values, the area underneath its graph is that of a rectangle — which can be determined by multiplying its base by its height. But we know the base of the rectangle to be (b-a), the width of the interval over which the random variable is distributed, and its area to be 1. Thus, the height of the rectangle must be 1/(b-a), which is also the probability density function of a uniform random variable over the region from a to b.

What is the mean of a uniformly distributed random variable? It is, conveniently, the halfway point of the interval from a to b, since half of the entire area under the graph of the pdf will be to the right of such a midway point, and half will be to the left. So the mean or mathematical expectation of a uniformly distributed random variable is (b-a)/2.

It is also possible to arrive at a convenient formula for the variance of such a uniform variable. Let us consider the following equation used for determining variance:

Var(X) = E(X2) – E(X)2 , where X is our uniformly distributed random variable.

We already know that E(X) = (b-a)/2, so E(X)2 must equal (b-a)2/4. To find E(X2), we can use the definition of such an expectation as the definite integral of x2*f(x) evaluated from b to a, where f(x) is the pdf of our random variable. We already know that f(x) = 1/(b-a); so E(X2) is equal to the integral of x2/(b-a), or x3/3(b-a), evaluated from b to a, which becomes (b-a)3/3(b-a), or (b-a)2/3.

Thus, Var(X) = E(X2) – E(X)2 = (b-a)2/3 – (b-a)2/4 = (b-a)2/12, which is the variance for any uniformly distributed random variable.

Ideas in Mathematics and Probability: Covariance of Random Variables (2007) – Article by G. Stolyarov II

Ideas in Mathematics and Probability: Covariance of Random Variables (2007) – Article by G. Stolyarov II

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The New Renaissance Hat
G. Stolyarov II
July 17, 2014
******************************
Note from the Author: This article was originally published on Associated Content (subsequently, Yahoo! Voices) in 2007.  The article earned over 5,200 page views on Associated Content/Yahoo! Voices, and I seek to preserve it as a valuable resource for readers, subsequent to the imminent closure of Yahoo! Voices. Therefore, this essay is being published directly on The Rational Argumentator for the first time.  ***
***
~ G. Stolyarov II, July 17, 2014
***
Analyzing the variances of dependent variables and the sums of those variances is an essential aspect of statistics and actuarial science. The concept of covariance is an indispensable tool for such analysis.
***

Let us assume that there are two random variables, X and Y. We can call the mathematical expectations of each of these variables E(X) and E(Y) respectively, and their variances Var(X) and Var(Y) respectively. What do we do when we want to find the variance of the sum of the random variables, X+Y? If X and Y are independent variables, this is easy to determine; in that case, simple addition accomplishes the task: Var(X+Y) = Var(X) + Var(Y).

But what if X and Y are dependent? Then the variance of the sum most often does not simply equal sum of the variances. Instead, the idea of covariance must be applied to the analysis. We shall denote the covariance of X and Y as Cov(X, Y).

Two crucial formulas are needed in order to deal effectively with the covariance concept:

Var(X+Y) = Var(X) + Var(Y) + 2Cov(X, Y)

Cov(X, Y) = E(XY) – E(X)E(Y)

We note that these formulas work for both independent and dependent variables. For independent variables, Var(X+Y) = Var(X) + Var(Y), so Cov(X, Y) = 0. Similarly, for independent variables, E(XY) = E(X)E(Y), so Cov(X, Y) = 0.

This leads us to the general insight that the covariance of independent variables is equal to zero. Indeed, this makes conceptual sense as well. The covariance of two variables is a tool that tells us how much of an effect the variation in one of the variables has on the other variable. If two variables are independent, what happens to one has no effect on the other, so the variables’ covariance must be zero.

Covariances can be positive or negative, and the sign of the covariance can give useful information about the kind of relationship that exists between the random variables in question. If the covariance is positive, then there exists a direct relationship between two random variables; an increase in the values of one tends to also increase the values of the other. If the covariance is negative, then there exists an inverse relationship between two random variables; an increase in the values of one tends to decrease the values of the other, and vice versa.

In some problems involving covariance, it is possible to work from even the most basic information to determine the solution. When given random variables X and Y, if one can compute E(X), E(Y), E(X2), E(Y2), and E(XY), one will have all the data necessary to solve for Cov(X, Y) and Var(X+Y). From the way each random variable is defined, one can derive the mathematical expectations above and use them to arrive at the covariance and the variance of the sums for the two variables.

Concepts in Probability Theory: Mathematical Expectation (2007) – Article by G. Stolyarov II

Concepts in Probability Theory: Mathematical Expectation (2007) – Article by G. Stolyarov II

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The New Renaissance Hat
G. Stolyarov II
July 17, 2014
******************************
Note from the Author: This article was originally published on Associated Content (subsequently, Yahoo! Voices) in 2007.  The article earned over 10,000 page views on Associated Content/Yahoo! Voices, and I seek to preserve it as a valuable resource for readers, subsequent to the imminent closure of Yahoo! Voices. Therefore, this essay is being published directly on The Rational Argumentator for the first time.  ***
***
~ G. Stolyarov II, July 17, 2014
***

The idea of expectation is crucial to probability theory and its applications. As one who has successfully passed actuarial Exam P on Probability, I would like to educate the general public about this interesting and useful mathematical concept.

The idea of expectation relies on some set of possible outcomes, each of which has a known probability and a known, quantifiable payoff — which can be positive or negative. Let us presume that we are playing a game called X with possible outcomes A, B, and C on a given turn. Each of these outcomes has a known probability P(A), P(B), and P(C) respectively. Each of the outcomes is associated with set payoffs a, b, and c, respectively. How much can one expect to win on an average turn of playing this game?

This is where the concept of expectation comes in. There is a P(A) probability of getting payoff a, a P(B) probability of getting payoff b, and a P(C) probability of getting payoff c. The expectation for a given turn of game X, E(X) is equal to the sum of the products of the probabilities for each given event and the payoffs for that event. So, in this case,

E(X) = a*P(A) + b*P(B) + c*P(C).

Now let us substitute some numbers to see how this concept could be applied. Let us say that event A has a probability of 0.45 of occurring, and if A occurs, you win $50. B has probability of 0.15 of occurring, and if B occurs, you lose $5. C has a probability of 0.4 of occurring, and if C occurs, you lose $60. Should you play this game? Let us find out.

E(X) = a*P(A) + b*P(B) + c*P(C). Substituting the values given above, we find that E(X) = 50*0.45 + (-5)(0.15) + (-60)(0.40) = -2.25. So, on an average turn of the game, you can be expected to lose about $2.25.

Note that this corresponds to neither of the three possible outcomes A, B, and C. But it does inform you of the kinds of results that you will approach if you play this game for a large number of turns. The Law of Large Numbers implies that the more times you play such a game, the more likely your average payoff per turn is to approach the expected value E(X). So if you play the game for 5 turns, you can be expected to lose 5*2.25 = $11.25, but you will likely experience some deviation from this in the real world. Yet if you play the game for 100 turns, you can be expected to lose 100*2.25 = $225, and your real-world outcome will most likely be quite close to this expected value.

In its more general form for some random variable X, the expectation of X or E(X) can be phrased as the sum of the products of all the possible outcomes x and their probabilities p(x). In mathematical notation, E(X) = sigma(x*p(x)) for all values of x. You can apply this formula to any discrete random variable, i. e., a random variable which assumes only a finite set of particular values.

For a continuous random variable Y, the mathematical expectation is equal to the integral of y*f(y) over the region on which the variable is defined. The function f(y) is called the probability density function of Y; its height over a given domain on a graph can be an indication the likelihood of the random variable assuming values over that domain.

Epsilon-N Proof of a Limit of a Sequence: Lim[2n/(3n+2)] = 2/3 – Article by G. Stolyarov II

Epsilon-N Proof of a Limit of a Sequence: Lim[2n/(3n+2)] = 2/3 – Article by G. Stolyarov II

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The New Renaissance Hat
G. Stolyarov II
July 12, 2014
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Note from the Author: This proof was originally published on Associated Content (subsequently, Yahoo! Voices) in 2007.  The article earned over 17,250 page views on Associated Content/Yahoo! Voices, and I seek to preserve it as a valuable resource for readers, subsequent to the imminent closure of Yahoo! Voices. Therefore, this essay is being published directly on The Rational Argumentator for the first time.  ***
***
~ G. Stolyarov II, July 12, 2014
***

This is the first in a series of formal mathematical proofs I intend to present in order to assist anybody who has ventured into the challenging but fascinating world of advanced calculus and real analysis. I start with a fairly basic proof: the limit of the nth term of a sequence as n becomes increasingly large. This is an epsilon-N proof, which uses the following definition: lim(n approaches ∞)xn= L iff for each real number ε>0, there exists a positive integer N(ε) such that if n ≥ N(ε), then │xn-L│< ε, i.e., L- ε < xn< L+ ε.

The epsilon-N proof has two component steps; first, we assume that our ε>0 is given and work backward to transform the inequality │xn-L│< ε in order to find an appropriate value of N(ε) that corresponds to a given value of ε. Then, using the value of N(ε) we found, we work forward to show that if n ≥ N(ε), then │xn-L│< ε.

The proof also uses the Archimedean Property, which states that the set of positive integers is not bounded above. There are actually four equivalent conditions that are known as the Archimedean Property:

1. If a,b are in R, a>0, b>0, then there is a positive integer n such that na>b

2. The set of positive integers is not bounded above.

3. For each real number x, there exists an integer n such that n ≤x < n+1

4. For each positive real number x, there exists a positive integer n such that 1/n ≤ x

Prove: lim(n approaches ∞)[2n/(3n+2)] = 2/3

Proof: Let ε>0 be given. Find N(ε)є Z+ such that if N(ε) < n, then │2n/(3n+2)-2/3│< ε.

Working backward to transform the inequality: │2n/(3n+2)-2/3│< ε

│6n/[3(3n+2)]-2(3n+2)/[3(3n+2)]│< ε

│[6n-2(3n+2)]/[3(3n+2)]│< ε

│[6n-6n-4]/[3(3n+2)]│< ε

│-4/[3(3n+2)]│< ε

Since ε>0, (3n+2)>0, the above inequality is the same as

4/[3(3n+2)] < ε

4/(3ε) < (3n+2)

4/(3ε)- 2 < 3n

4/(9ε)- 2/3 < n

Now I work forward to prove the original result:

Let N(ε)є Z+ э 4/(9ε)- 2/3 < N(ε).

Since 4/(9ε)- 2/3 is a real number, by the Archimedean Property it must be the case that some integer exists which is greater than that real number-since the set of positive integers is not bounded above. Here, we call that integer N(ε).

For all n>N, if 4/(9ε)- 2/3 < N < n, then:

4/(3ε)- 2 < 3n

4/(3ε) < (3n+2)

4/[3(3n+2)] < ε

│-4/[3(3n+2)]│< ε

│6n/[3(3n+2)]-2(3n+2)/[3(3n+2)]│< ε

│2n/(3n+2)-2/3│< ε

I have demonstrated the above inequality, which is sufficient to demonstrate that lim(n approaches ∞)[2n/(3n+2)] = 2/3.

I have hence proved what was desired. Another way to express that the desired objective has been obtained (which I shall use in future proofs of this sort) is the abbreviation

“Q. E. D.” of the Latin “Quod Erat Demonstraterum,” which means “that which was to be demonstrated.”

Conciseness on Actuarial Essay Exams (2010) – Article by G. Stolyarov II

Conciseness on Actuarial Essay Exams (2010) – Article by G. Stolyarov II

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The New Renaissance Hat
G. Stolyarov II
July 11, 2014
******************************
This essay, originally written and published on Associated Content/Yahoo! Voices in 2010, has helped many actuarial candidates to prepare for essay exams. I seek to preserve it as a valuable resource for readers, subsequent to the imminent closure of Yahoo! Voices. Therefore, this essay is being published directly on The Rational Argumentator for the first time. 

***
~ G. Stolyarov II, July 11, 2014
***

Upper-level actuarial exams are in essay format, requiring both conceptual discussions and extensive calculations to answer 30-50 questions within a 4-hour timeframe. Even for highly knowledgeable candidates, the sheer time constraints of the exam render it difficult to respond both thoroughly and within the allotted time. Thus, conciseness, without compromising the communication of understanding, becomes a priority.

The following ideas for condensing actuarial exam responses were derived from reviewing past sample answers released by the Casualty Actuarial Society. By understanding which answers received full credit while employing certain shortcuts of presentation, I was able to arrive at ideas that, when used in combination, may save candidates tens of minutes on the exam. This time can be devoted to reviewing one’s answers or to answering more questions than would otherwise be possible. While, as an outsider to the grading process, I can offer no guarantees, I plan to personally use these approaches to the extent they are relevant.

If other actuarial candidates have additional ideas to facilitate concise, effective exam answers, I welcome their input.

1. Common Abbreviations

Many insurance concepts have generally known abbreviations that do not need to be defined unless an explicit definition is requested. On most questions, it would be safe, for instance, to assume that the grader will know what ALAE, ULAE, IBNR, IBNER, PDLD, GAAP, SAP, and terms of similarly common usage stand for.

There are also commonly used general abbreviations, such as “&” for “and”, “b/c” for “because”, “w.r.t” for “with respect to”.

2. Uncommon Abbreviations

It is also possible to define uncommon (even self-invented) abbreviations once, and use them thereafter. For instance, one could refer to “the Bornhuetter-Ferguson method (B-F)” and then subsequently state that “B-F assumes…” or “according to B-F…”.

As long as the grader understands what the abbreviations mean in the context of one’s answer, full credit should be possible.

Here is a non-exhaustive list of abbreviations that may be useful for the 2010 CAS Exam 6 in particular:

B-F: Bornhuetter-Ferguson method
B-S: Berquist-Sherman method
Cat.: Catastrophe
CL: Chain ladder
C-N: Conger-Nolibos generalized approach
Co-part.: Co-participation
Cov.: Coverage
Dev.: Developed or Development (depending on context)
G-B: Gunnar Benktander method
GL: General liability
Inc.: Incurred
Lim.: Limit
M-A: Mango-Allen adjustment
O/S: Outstanding
QS: Quota share
S-B: Stanard-Bühlmann method (“CC” for “Cape Cod method” can also be used).
SS: Surplus share (definitely define that one before using!)
U/W: Underwriting
WC: Workers’ compensation
XOL: Excess-of-loss

3. Shortcuts for Repetitive Calculations

It is possible to save time in cases where one must perform multiple calculations using the same basic formula or approach. Instead of displaying every single calculation, one could simply display (1) the formula used, (2) a sample calculation, and (3) the final results of all the other calculations.

As a non-insurance illustration, suppose you were faced with the following problem:

Find the hypotenuses of the right triangles with the following legs:
(3, 4)
(8, 15)
(9, 40)
(20, 21)

The long way to answer would be to display all four calculations. A shorter way would be the following:
Formula: c = √(b2 + a2)
Sample: √(32 + 42) = 5
Answers: 5, 17, 41, 29

The only possible drawback to this approach is that, if one makes a mistake in a calculation other than the sample calculation, the specific nature of the mistake will not be visible to the grader. It is possible that the grader will simply assume a mechanical error and therefore be lenient in giving partial credit, because the formula and sample calculation demonstrate an understanding of the ideas involved. However, it is impossible to offer any guarantees here.

4. Alternatives to Complete Sentences

While, in academic settings, answering in complete sentences is a requirement for most exams and assignments, the sheer time pressure of an actuarial essay exam renders this approach sub-optimal. A review of past exam answers that have received full credit suggests that graders do not remove points from responses that convey a candidate’s knowledge of the tested content but are written in sentence fragments.

Instead of writing in complete sentences, there are many possible alternative ways of answering, depending on the question. For instance, a question asking the candidate to compare and contrast certain aspects of Method X and Method Y might be answered as follows:

Method X: (List features of method)
Method Y: (List features of method, preferably using language parallel to what was used for Method X.)

Using a bulleted or numbered list to answer some questions may not only save time but may make it easier for the grader to identify the substance of the answer.

Chains of causation or implication may be expressed via an “→” symbol (e.g., “Writing new business → acquisition expense recognized immediately, premiums earned over time → decline in policyholders’ surplus → need for surplus relief.”

It is also acceptable to omit certain articles and to omit stating the premise of the question in the answer’s first sentence, as long as the meaning is clear. Furthermore, some instances of expressions like “that”, “then”, and “in order” may be omitted without compromising the answer’s intent.

As an illustration, I present two ways of answering my Problem S6-9-3(b): “What effect should be removed in order to evaluate development patterns correctly (Statement of Principles, p. 16)?”

Complete-sentence answer (my original): “The effect of discounting should be removed in order to evaluate development patterns correctly. If a reserve is established as a present value of future costs, then upward development may occur simply as a result of paying claims, and this may send a misleading signal.”

Condensed answer: “Effect of discounting should be removed. If reserve is set as present value of future costs, upward development may occur simply as result of paying claims → misleading signal may result.”

Again, I welcome input on these ideas and other ideas for facilitating conciseness on actuarial essay exams.

What You Need to Know for Actuarial Exam P (2007) – Article by G. Stolyarov II

What You Need to Know for Actuarial Exam P (2007) – Article by G. Stolyarov II

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The New Renaissance Hat
G. Stolyarov II
July 9, 2014
******************************
This essay, originally written and published on Yahoo! Voices in 2007, has helped many actuarial candidates to study for Exam P and has garnered over 15,000 views to date. I seek to preserve it as a valuable resource for readers, subsequent to the imminent closure of Yahoo! Voices. Therefore, this essay is being published directly on The Rational Argumentator for the first time. While it has been over 7 years since I took and passed Actuarial Exam P, the fundamental advice in this article remains relevant, and I hope that it will assist many actuarial candidates for years to come. 

***
~ G. Stolyarov II, July 9, 2014
***

This is a companion article to “How to Study for Actuarial Exam P Without Paying for Materials“.

If you desire to become an actuary, then passing Exam P on Probability is your opportunity to enter the actuarial science profession and get a starting salary ranging of about $46,000 to about $67,000 per year. But the colossal number of topics listed on the syllabus may seem intimidating to many. Fortunately, you do not need to know all of them to get high grades on the exam. In May 2007, I passed Exam P with a the highest possible grade of 10 and can offer some advice on what you need to know in order to do well.

Of course, you need to know the basics of probability theory, including the addition and multiplication rules, mutually independent and dependent events, conditional probabilities, and Bayes’ Theorem. These topics are quite straightforward and do not require knowledge of calculus or any other kind of advanced mathematics; you need to be able to add, multiply, divide, and think logically about the situation presented in the problem — which will often be described in words. Visual aids, such as Venn Diagrams, contingency tables, and the use of union and intersection notation can be eminently helpful here. Try to master these general probability topics before moving on to the more difficult univariate and multivariate probability distributions.

Next, you will need to know several critically important univariate probability distributions, including some of their special properties. Fortunately, you do not need to know as many as the syllabus suggests.

The Society of Actuaries (SOA) recommends that you learn the “binomial, negative binomial, geometric, hypergeometric, Poisson, uniform, exponential, chi-square, beta, Pareto, lognormal, gamma, Weibull, and normal” distributions, but in fact the ones you will be tested on most extensively are just the binomial, negative binomial, geometric, Poisson, uniform, exponential, and normal. Make sure you know those seven in exhaustive detail, though, because much of the test concerns them. It is a good idea to memorize the formulas for these distributions’ probability density functions, survival functions, means, and variances. Also be able to do computations with the normal distribution using the provided table of areas under the normal curve. Knowledge of calculus, integration, and analysis of discrete finite and infinite sums is necessary to master the univariate probability distributions on Exam P.

Also pay attention to applications of univariate probability distributions to the insurance sector; know how to solve every kind of problem which involves deductibles and claim limits, because a significant portion of the problems on the test will employ these concepts. Study the SOA’s past exam questions and solutions and read the study note on “Risk and Insurance” to get extensive exposure to these applications of probability theory.

The multivariate probability concepts on Exam P are among the most challenging. They require a solid grasp of double integrals and firm knowledge of joint, marginal, and conditional probability distributions – as well as the ability to derive any one of these kinds of distributions from the others. Moreover, many of the problems on the test involve moment-generating functions and their properties – a subject that deserves extensive study and practice in its own right.

Furthermore, make sure that you have a solid grasp of the concepts of expectation, variance, standard deviation, covariance, and correlation. Indeed, try to master the problems involving variances and covariances of multiple random variables; these problems become easy once you make a habit of doing them; solving them quickly and effectively will save a lot of time on the exam and boost your grade. Also make sure that you study the Central Limit Theorem and are able to do problems involving it; this is not a difficult concept once you are conversant with the normal distribution, and mastering Central Limit problems can go a long way to enhance your performance as well.

Studying the topics mentioned here can focus your preparation for Exam P and enable you to practice effectively and confidently. Remember, though, that this is still a lot of material. You would be well advised to begin studying for the test at least three months in advance and to study consistently on a daily basis. Practice often with every kind of problem so as to keep your memory and skills fresh. Best wishes on the exam.

How to Study for Actuarial Exam P Without Paying for Materials (2007) – Article by G. Stolyarov II

How to Study for Actuarial Exam P Without Paying for Materials (2007) – Article by G. Stolyarov II

Comments 24 comments
The New Renaissance Hat
G. Stolyarov II
July 9, 2014
******************************
This essay, originally written and published on Yahoo! Voices in 2007, is my most-viewed article and second-most-viewed work of all time, at over 81,600 views to date. I seek to preserve it as a valuable resource for readers, subsequent to the imminent closure of Yahoo! Voices. Therefore, this essay is being published directly on The Rational Argumentator for the first time. While it has been over 7 years since I took and passed Actuarial Exam P, the fundamental advice in this article remains relevant, and I hope that it will assist many actuarial candidates for years to come. 

***
~ G. Stolyarov II, July 9, 2014
***
Exam P on Probability, offered by the Society of Actuaries (formerly in conjunction with the Casualty Actuarial Society, which referred to it as Exam 1), is the gateway to the actuarial profession. Those who pass the exam can obtain entry-level jobs as actuaries, with salaries ranging from about $46,000 to about $67,000 per year. After some rigorous studying, I passed this examination in May 2007 with a grade of 10 – the highest possible. Here are some study materials that can help you obtain top marks on Exam P without paying a cent.
***

The breadth of material listed on the syllabus for this test is extensive, and many of the topics are tremendously complex in themselves. Fortunately not all of the topics listed are actually tested, and the kinds of questions that are asked on the exam are generally more reasonable and straightforward than the ones present in the recommended readings.

As I found out through personal experience, you do not need to spend money at all in purchasing study materials for this exam. Virtually everything you need can already be found online. The most crucial study aid is the list of sample questions from past exams, generously provided by the Society of Actuaries. Along with these questions, you will also find a list of step-by-step solutions which will enable you to check your work. For successful performance on the test, it is essential to be able to successfully solve these problems on your own and to know why you obtained the solutions you did. The problems on the exam are remarkably similar to the ones in the sample questions, so you should do well on your exam if you can solve the problems from prior tests.

In the course of my own studying, I made the mistake of purchasing Michael A. Bean’s Probability: The Science of Uncertainty: a book which does an extremely poor job at explaining the mathematical concepts required for the actuarial exam, because it already presupposes the reader’s expert knowledge of such concepts. Too often, crucial explanations and proofs are omitted from this book, to be left as “exercises to the reader”– quite a challenge for a reader who simply seeks a basic grasp of the subject!

Furthermore, the exercises in Bean’s book are not conducive to learning the essentials of the probability concepts discussed; these problems are instead so convoluted and laden with unnecessary complications as to baffle even the expert mathematician. Exam P itself is much more reasonable than that; the problems often require some thinking and multiple steps, but you will not be required to pull brilliant, esoteric insights out of thin air, as Bean’s exercises require you to do. To add to the trouble, Bean does not provide answers in the back of the book for most of his problems — thus disabling you from checking your work.

The best textbook by far for actuarial students preparing to take Exam P is Marcel B. Finan’s A Probability Course for the Actuaries, which you can download for free in pdf format. It starts with the very basics of set theory and probability and gradually works its way toward the most advanced concepts to be tested. Every section focuses on a different key idea and builds on the previous topics; there are also instructive detailed proofs, examples, and practice problems to guide you along the way. This book is available to the public and has an immense wealth of excellent resources. In preparing for the exam, I did some work for Dr. Finan in making the answer keys to the exercises in this book, which I made available in 2008.

With focus, determination, and discipline, you can do well on Exam P and achieve entry into the fascinating and lucrative world of actuarial science. And with these excellent free study materials, all you will need to invest into your education is the $225 exam fee, your time, and your effort.

How to Solve Problems of Integration by Parts Using the Tabular Method (2007) – Article by G. Stolyarov II

How to Solve Problems of Integration by Parts Using the Tabular Method (2007) – Article by G. Stolyarov II

Comments 3 comments
The New Renaissance Hat
G. Stolyarov II
July 6, 2014
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Note from the Author: This essay was originally written and published on Associated Content (subsequently, Yahoo! Voices) in 2007. The tabular method is one of my favorite mathematical techniques, capable of saving tremendous amounts of time in completing certain integration problems. While the method is not commonly emphasized in conventional calculus curricula, the enthusiastic reception of this article showed that there is significant interest among students in this concise, elegant approach. This article earned over 30,000 page views on Associated Content/Yahoo! Voices, and I seek to preserve it as a valuable resource for readers, subsequent to the imminent closure of Yahoo! Voices. Therefore, this essay is being published directly on The Rational Argumentator for the first time. 
***
~ G. Stolyarov II, July 6, 2014
***
Doing calculus problems which require integration by parts can be a lengthy and tedious process, even for someone with experience in finding the integrals of functions. Fortunately, for many of the most common types of integration by parts, there is fast, simple shortcut available. I have used the tabular method to great advantage on exams and math contests; this technique was one of the tools that enabled me to earn a perfect score of 40 on the Continental Mathematics League’s nationwide calculus competition in 2005.
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The tabular method can be applied to any function which is the product of two expressions, where one of the expressions has some nth derivative equal to zero. For instance, the tabular method can be used to find the indefinite integral of x4e3x, but not of sin(x)e3x. This is because the 5th derivative of x4 is equal to zero, whereas sin(x) does not have any nth derivative which always exhibits zero values.

The tabular method uses a convenient table with three columns. We can call the first column “Signs,” the second column “u” and the third column “dv.” Under the column called “Signs,” we list positive and negative signs in alternating order for as many times as the problem requires. The first entry in the column labeled “u” will be the part of the function we want to integrate which can be reduced to zero through successive differentiation. In integrating our sample function, x4e3x, we will put x4 in the column labeled “u.” The subsequent entries in the “u” column will be the successive derivatives of the first entry — all the way to zero.

The first entry in the column labeled “dv” includes the other part of the function we want to integrate. The subsequent entries in this column will be the successive integrals of the first entry. For our sample problem, the first entry in the “dv” column will be e3x.

This is how the table for finding the integral of x4e3x will look:

Signs u dv
+ x4 e3x
– 4x3 (1/3)e3x
+ 12x2 (1/9)e3x
– 24x (1/27)e3x
+ 24 (1/81)e3x
– 0 (1/243)e3x

Using this table to find the indefinite integral of the function requires taking the sign from each row in the column except the last, applying it to the entry for “u” in the same row, and multiplying the result by the entry for “dv” in the next row. Doing this problem on paper, you would simply draw arrows between the following expressions:

+ x4 and (1/3)e3x
– 4x3 and (1/9)e3x
+ 12x2 and (1/27)e3x
– 24x and (1/81)e3x
+ 24 and (1/243)e3x

Now multiply each of the expressions linked by “and” (or the arrows on paper) and add them together to get the final indefinite integral:

(1/3)x4e3x – (4/9)x3e3x + (4/9)x2e3x – (8/27)xe3x + (8/81)e3x + C

Remember to include the constant C if you wish to leave the integral in indefinite form without evaluating it.

The great advantage of using the tabular method is the ability to do integration by parts mechanically, without needing to exert a large amount of thinking about the special circumstances of the problem. The procedure is fairly quick to memorize and easy to retain. After you learn it once, it will always be at your disposal as a tool for quickly and easily determining indefinite integrals that would otherwise take an immense amount of time to find.

Abstract Orderism Fractal LVIII – Art by G. Stolyarov II

Abstract Orderism Fractal LVIII – Art by G. Stolyarov II

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Abstract Orderism Fractal LVIII - by G. Stolyarov II

Abstract Orderism Fractal LVIII – by G. Stolyarov II

Note: Left-click on this image to get a full view of this digital work of fractal art.

This fractal’s elaborate interwoven pattern is situated within a larger pear-like shape.

This digital artwork was created by Mr. Stolyarov in Apophysis, a free program that facilitates deliberate manipulation of randomly generated fractals into intelligible shapes.

This fractal is an extension of Mr. Stolyarov’s artistic style of Abstract Orderism, whose goal is the creation of abstract objects that are appealing by virtue of their geometric intricacy — a demonstration of the order that man can both discover in the universe and bring into existence through his own actions and applications of the laws of nature.

Fractal art is based on the idea of the spontaneous order – which is pivotal in economics, culture, and human civilization itself. Now, using computer technology, spontaneous orders can be harnessed in individual art works as well.

See the index of Mr. Stolyarov’s art works.

Abstract Orderism Fractal LVII – Art by G. Stolyarov II

Abstract Orderism Fractal LVII – Art by G. Stolyarov II

Comments 1 comment

Abstract Orderism Fractal LVII - by G. Stolyarov II

Abstract Orderism Fractal LVII – by G. Stolyarov II

Note: Left-click on this image to get a full view of this digital work of fractal art.

This plant-like fractal design consists of numerous loops and rounded shapes.

This digital artwork was created by Mr. Stolyarov in Apophysis, a free program that facilitates deliberate manipulation of randomly generated fractals into intelligible shapes.

This fractal is an extension of Mr. Stolyarov’s artistic style of Abstract Orderism, whose goal is the creation of abstract objects that are appealing by virtue of their geometric intricacy — a demonstration of the order that man can both discover in the universe and bring into existence through his own actions and applications of the laws of nature.

Fractal art is based on the idea of the spontaneous order – which is pivotal in economics, culture, and human civilization itself. Now, using computer technology, spontaneous orders can be harnessed in individual art works as well.

See the index of Mr. Stolyarov’s art works.