Success with Random Variables

So this afternoon, I finished grading a quiz focusing on random variables associated with a sequence of Bernoulli trials. Based on these quizzes, I'm trying to understand what makes a probability course so difficult for students to understand. The other night, I was at a department social and talking with a professor who has taught Math 318 many times. He came right out and stated (without me giving any prompting) that he is always surprised at how students have such a hard time with the course, even though the actual mathematics involved in the course are fairly straight-forward.

In the last entry, I noted that at least part of the challenge is that there are so many different types of functions that appear. But I think that a significant issue comes back to confusion about what random variables represent and how they relate to questions that are posed. Recall that a random variable summarizes some aspect of a random experiment as a single number. (We will soon generalize to the ability to summarize with multiple random variables.) Typical questions in probability focus on the probability of events and the average of certain quantities. Almost always, we answer such questions by identifying an appropriate random variable. We then decide how to characterize events in terms of that random variable, or how to express the quantity being averaged in terms of that random variable.

The examples of random variables related to a sequence of Bernoulli trials provide our first real example of this type of reasoning, and I think this transition is part of why the topic was difficult for many of you. First, we must remember that the random variable is not the same as the random experiment, but simply one of many different ways to summarize an aspect of that experiment. The experiment itself is characterized completely by the sequence of Bernoulli trials, each of which is going to be either a success or a failure. Random variables will be measurement that are related to these successes or failures, and we must choose an appropriate measurement that will allow us to answer the questions of interest.

On our quiz last week, I gave the following scenario: "Apples are packaged in 3-pound bags. Suppose that 4% of the time, the bag weighs less than 3 pounds. You select bags randomly and weigh them in order." In principle, there is an unlimited supply of bags of apples, and we indefinitely select a random bag and weigh it. If we were to weigh enough bags over a long enough time, we would find that the fraction that are underweight appears 4% of the time. However, for any particular number of weighings, we could have found any number of underweight bags. The random experiment is described by an infinite sequence of F's and U's, with an F if the bag was full and a U if the bag was underweight.

So now consider the first question: "What is the probability that you must weigh at least 20 bags before you find an underweight bag?" One method that we could use to do this is to directly understand the event in question. That is, we could create a tree diagram that would eventually give enough information to answer the question. Unfortunately, this tree diagram would be much to cumbersome to answer a question about the first 20 bags weighed: 2^20 different outcomes after 20 weighings. So we try to see if there is a random variable for this random experiment that has enough information to answer the question instead.

There are actually multiple ways to choose a random variable to answer the question. The best random variable is a geometric random variable, which counts the number of Bernoulli trials until we see the first success. Since our question considers when we find the first underweight bag, we define success for our purposes as weighing a bag as underweight. Thus, X counts the number of weighings until we find the first underweight bag. If the first bag is underweight, X=1. But if the first 5 bags are full and the 6th bag is underweight, X=6. Every path on our hypothetical tree diagram is associated with a particular value of the random variable. Since an underweight bag will be chosen with probability p=0.04, we use X~Geometric(p=0.04). Now, having chosen a random variable, we must determine how we answer the question in terms of that random variable. The event of interest, "weigh at least 20 bags before you find an underweight bag," corresponds to an event, "X is greater than or equal to 20". Thus, we wish to compute P[X≥20].

Another way that we could answer the question is to consider that finding an underweight bag for the first time on or after the 20th weighing means that the first 19 bags must have all been full. Consequently, we could answer our question using a Binomial random variable with n=19. So let X count the number of underweight bags out of the first 19 that are weighed. Again, a "success" is finding an underweight bag, so that X~Binomial(n=19,p=0.04). Our event can now be restated as saying that "X is equal to 0". But remember that the X in this paragraph is different from the random variable in the previous paragraph. For this random variable, our question will be answered by computing P[X=0].

The third question on the quiz asked, "What is the probability that you find the 5th underweight bag before you weigh 80 bags?" The best choice for a random variable that will answer our question is to count the number trials until you do find the 5th underweight bag. If we say that a trial is a success if the bag is underweight, then we are counting trials until the 5th success. That is, our random variable X is a negative binomial random variable with r=5 (the number of successes needed to stop counting) and p=0.04 (the success probability). We write this: X~Neg.Binom.(r=5,p=0.04). To answer the question, we must realize that the event of interest is to say that X<80. (We stop counting before we reach 80.) Thus, the answer would be P[X<80].

And of course, as I mentioned earlier, we could actually compute the probability using another type of random variable. For this problem, we must find another way to describe the event of interest. One way to do this is to realize that we find the fifth bag prior to the 80th weighing if there are at least 5 underweight bags by the time we weigh the 79th bag. This can be represented using a binomial random variable. Let our random variable X count the number of underweight bags in the first 79 weighed. Thus, X~Binomial(n=79, p=0.04). The probability we wish to compute is P[X>5].

In summary, to do well in probability, you will need to think probabilistically. Before you can actually compute a quantity (which usually involves fairly straightforward mathematics), you must identify a random variable and understand how to answer the question in terms of that random variable. Then you can use one of the appropriate functions to finally answer the question, whether it is a probability or an expectation.

Random Variables

There seems to be a lot of confusion in relation to random variables. Part of this has to do with students being pressed onto new subjects without necessarily understanding the previous material adequately. Part of this has to do with the plethora of new functions that are introduced in relation to these random variables (e.g., a random variable X is itself a function in the probability space, it has a p.m.f., a m.g.f., and a c.d.f.). And part of this confusion has to do with a variety of ways to compute things: probabilities, expectations, and moments. Before you allow yourself to be engulfed by the onslaught of waves of probability theory crashing down on you, take a breath (of air) and consider the following.

Probabilities most fundamentally describe random experiments. The probability space (or sample space) describes the possible outcomes of the experiments in as much detail as necessary to completely characterize that experiment. A random variable is a single numeric summary of the nature of the experimental outcome. There are typically many different outcomes of the experiment that can lead to the same value for the random variable. Furthermore, a single experiment can provide the information for many different random variables, each of which summarizes a different aspect of the experiment.

For example, suppose our experiment consists of rolling two standard 6-sided dice, one of which is red and the other is green. There are 36 distinct outcomes. We might be interested in the value of the red die, which value we could assign to a random variable R. We might instead be interested in the value of the green die, which value we could assign to a random variable G. Other values of interest might be the largest of the two values, the smallest of the two values, the sum of the values, the difference between the dice, the greatest common factor between the values, etc. The list could go on forever, with each summary value corresponding to a different random variable.

The random variables R and G are independent because the roll of one does not influence the other. They are also identically distributed. The probability mass function (p.m.f.) for the random variable describes the probabilities of individual values for R. We either need a formula for an arbitrary value or a table showing all of the values to describe this function. For this random variable, we have R(x)=1/6 for each of the values in the support S={1,2,3,4,5,6}. The expected value is defined as a sum over all possible values in the support
For a single die roll (R or G), we have E[R] = 1/6(1)+1/6(2)+...+1/6(6) = 3.5.

Let us now consider another random variable, S, which represents the sum of the two values. That is, S=R+G. The support for S is the set {2,3,...,12}. We could compute the p.m.f. for this random variable: f(7)=6/36, f(6)=f(8)=5/36, f(5)=f(9)=4/36, f(4)=f(10)=3/36, f(3)=f(11)=2/36, and f(2)=f(12)=1/36. And we could compute the expected value of the random variable using the definition of mathematical expectation:

E[S]=2(1/36)+3(2/36)+...+10(3/36)+11(2/36)+12(1/36).

However, since S=R+G, we have a lovely little theorem that allows us to use a sum rule:

E[S] = E[R+G] = E[R]+E[G] = 3.5+3.5 = 7.

This is much easier than calculating using the definition.

So, we learn a lesson: if we can express a random variable as a sum of easier random variables, expected value may be more effectively calculated using these individual terms.

To continue, let us consider the distance between the two rolls. That is, we introduce a random variable X = |R-G| to represent the distance between the rolls. The smallest possible value for X is 0, which occurs when the dice are the same. The largest possible value for X is 5, which occurs when one die is 1 and the other is 6. So the support for X is the set S={0,1,2,3,4,5}. In this example, a table for the p.m.f. is much easier, computed based on the number of ways to obtain each distance:
f(0)=6/36, f(1)=10/36, f(2)=8/36, f(3)=6/36, f(4)=4/36, f(5)=2/36

For this problem, the random variable X is not easily expressed as a sum. So we must compute expected value using the original definition:

E[X]=0(6/36)+1(10/36)+2(8/36)+3(6/36)+4(4/36)+5(2/36)=70/36.

In summary, a random variable is a single number that summarizes some aspect of a random experiment. The p.m.f. of the random variable gives probabilities of individual outcomes. The expected value (or mathematical expectation) computes the average of a random quantity weighted by appropriate probabilities of those values.

Statistical Independence

I was talking with a student about independence the other day and realized that the student was thinking of independence as being unrelated. In usual speech, we probably think of two outcomes as independent as being that they are not connected. For example, we say that the American states won their independence from England when they broke the governing ties with England.

However, statistical independence should be viewed in a different way. It refers to events A and B so that A provides no information on whether the event B occurs or not. If A and B (remember, these are sets) have no overlap (mutually exclusive or disjoint), then if you knew that the outcome was in A, then it would be impossible that the outcome is in B. Thus, the event A is providing information regarding event B and these are not statistically independent. Similarly, if I knew that the outcome was not in A (i.e., the outcome is in A'), then we know that B is a larger portion of the remaining possible outcomes. Again, that the outcome is in A' is giving information whether the outcome is in event B.

We thus draw the conclusion that independent events must overlap. In fact, they must overlap in a very significant way. Suppose that A and B are independent and that P(A)=0.25 while P(B)=0.4. By the definition of independence, we must have P(A∩B)=P(A)P(B)=(0.25)(0.4)=0.1. To make this concrete, imagine that the sample space Ω has 100 possible, equally likely outcomes. Then A includes 25 outcomes and B includes 40 outcomes. Our calculation then requires that the intersection A∩B includes 10 outcomes.

Now, notice what this means about conditional probabilities. If we restrict our attention to the set A (i.e., we are given that event A has occurred), then the newly restricted outcome space has 25 outcomes. If we ask the probability that B occurs given this information, we know that 10 of these outcomes belong to B. So P(B|A)=10/25=0.4. Voila! This is exactly the same as P(B). Similarly, if we are given that event B has occured, then 10 of the 40 outcomes available belong to A so that P(A|B) = 10/40=0.25=P(A).

How do we summarize this idea? Well, if events A and B are independent, then each event must have a restricted but proportional representation of the other independent event. One way that we can do this is imagine that A and B are at right angles and overlap. For the above example, we can arrange the 100 items into 5 rows and 20 columns.

Then A might represent the first two columns (5×2=10) while B represents the first two rows (2×20=40). Being given information that A occurs collapses the larger picture into a reduced picture consisting of only two columns. However, the fraction of rows represented by B remains the same and P(B|A)=P(B).


This idea expands to more than two events. To have three independent events A, B, and C, we must imagine a three-dimensional grid corresponding to the outcome space so that A represents a simple division in one directions, B represents a second direction, and C represents the third direction. Conditional probabilities given event A corresponds to collapsing the space in A's direction. But the B and C fractions of A remain exactly in the same proportion as they were originally.

Simplifying with Factorials

As I was grading the second quiz, I see that I should point out that you should simplify as much as possible whenever you see a fraction involving factorials. For example suppose that you saw a fraction: 10!/(4!6!). You should know that immediately you can cancel the last 6 numbers to get: (10*9*8*7)/(4*3*2*1). But instead of multiplying out, you should cancel the 8 with 4*2 and the 9 with 3 to get: (10*3*7)=210.
If you just multiply things out and do not simplify as you go, you will find that you get some awful numbers that are hard to find factors. In fact, you probably already were looking at the factors.

Discrete Random Variables (Section 2.1)

My philosophy is that class time should be used to facilitate learning and not simply to reiterate concepts that the book already explains adequately. I readily acknowledge that I am still learning how to accomplish such a feat, especially to help students overcome the tendency to avoid reading their textbook.

As we leave chapter 1 where we learned basic ideas of the probability of events, we begin chapter 2 where we will focus on a family of random variables of the discrete type.
Comparing Definition 2.1-1 with the definition I game in my first week slides, you should notice that there is a distinction between the outcome space of the experiment and the space of the random variable. The outcome space should represent the detailed description of the experiment, while the space of the random variable is the range of the random variable (as a function of the outcome or sample space). Pages 58 and 59 provide an important philosophical guide for what we are trying to accomplish and point out that observations might help us to estimate the probabilities associated with the random variable. However, we can often use basic assumptions to create a mathematical model for these probabilities. This chapter introduces a number of models that describe discrete random behavior.

Definition 2.1-2 is very important, introducing the definition of the probability mass function. Problem 3 in the textbook helps test if you understand the basic ideas. One of the major points you need to remember is that for discrete type random variables, properties (b) and (c) compute probabilities using summations. When we get to continuous type random variables, the corresponding properties will replace summation with integration.

In addition to basic principles (probability mass function (mathematical model' prediction) vs relative frequency (statistical estimate), bar graph vs histogram), we meet the first model for a random variable---the hypergeometric distribution---which describes choosing n objects from a total collection of two types of objects. The model is created by considering exactly the types of calculations used in chapter 1, by counting how many ways to select n objects from a total of N objects (denominator) and then also counting how many ways to choose x of the first type and n-x of the second type (numerator).

Spring 2008 Introduction

Welcome to my math blog. Last semester, I created a blog that would be specific to one course. But it seems silly to keep creating new blogs for each course. So I'm going to experiment with using one blog for a sequence of courses.
This semester (Spring 2008), I am teaching Math 318, an introduction to probability and statistics. Most of the blog entries will correspond to that course for the next few months.