Open course ware

PROFESSOR: OK. We'll go through these together. Repeat after me. Civil. AUDIENCE: Civil. PROFESSOR: Civility. AUDIENCE: Civility. PROFESSOR: Civilized. AUDIENCE: Civilized. PROFESSOR: Civilization. . AUDIENCE: Civilization. PROFESSOR: Equal. AUDIENCE: Equal. PROFESSOR: Equality. AUDIENCE: Equality. PROFESSOR: Equalize. AUDIENCE: Equalize. PROFESSOR: Equalization. AUDIENCE: Equalization. PROFESSOR: Fertile. AUDIENCE: Fertile. PROFESSOR: Fertility. AUDIENCE: Fertility. PROFESSOR: Fertilize. AUDIENCE: Fertilize. PROFESSOR: Fertilization. AUDIENCE: Fertilization. PROFESSOR: Final. AUDIENCE: Final. PROFESSOR: Finality. AUDIENCE: Finality. PROFESSOR: Finalize. AUDIENCE: Finalize. PROFESSOR: Finalization. AUDIENCE: Finalization. PROFESSOR: General. AUDIENCE: General. PROFESSOR: Generality. AUDIENCE: Generality. PROFESSOR: Generalized. AUDIENCE: Generalized. PROFESSOR: Generalization. AUDIENCE: Generalization. PROFESSOR: Hospital. AUDIENCE: Hospital. PROFESSOR: Hospitality. A...

We will now go through a beautiful example, in which we approach the same question in a number of different ways and see that by reasoning based on the intuitive properties of a Poisson process, we can arrive quickly to the right answer. The problem is as follows. We have three lightbulbs, and each light bulb is being lit at time zero, it starts working, and the light bulb lasts for a certain amount of time, which is random. So this light bulb lasts so long, this one lasts so long, this one lasts that long. The lifetime of a light bulb, the time until it burns out, will be a random variable, and we make the following assumptions. The lifetimes of the three light bulbs, which we denote by X, Y, and Z, will be independent random variables, each of which is an exponential random variable with the parameter lambda. We're interested in the question of calculating the expected time until a light bulb burns out for the first time. So in this picture, the light bulb that burn...

We will now go through a sequence of examples that illustrate the different types of questions that we usually answer using a normal approximation based on the central limit theorem. In general, one uses these approximations to make statements of this type. That the probability of the sum of n, i.i.d. random variables being less than a certain number, that this probability is approximately equal to some other number. Notice that this statement involves three parameters, a, b, and n, and you can imagine problems where you are given two of these parameters, and you're asked to find the third. And this gives us the different variations of the questions that we might be able to answer. So we will go through examples of each one of these variations. Our setting will be as follows. We have a container, and the container receives packages. Each package has a random weight, which is an independent random variable that's drawn from an exponential distribution with a parame...

Let's now go through another example, which will be a little more challenging. We're going to revisit an old problem. We have a coin that has an unknown bias, Theta. And we have a prior distribution on this Theta. We fix some positive integer, n, we flip a coin n times, that has this unknown bias. And we record the number of heads. On the basis of the number of heads that have been observed, we wish to estimate the bias, Theta, of the coin. To make things more concrete, we're going to assume a prior distribution on Theta that is uniform on the unit interval. Now, this is a problem we have considered before. We have calculated the expected value of Theta given X. And we did find that the expected value takes this particular form. Now, notice that this is a linear function of X. And if it turns out the least mean squares estimator is a linear function of X, then we're guaranteed, since this is the best, that this is also the best within the class of linear e...

We will now go through an example that involves a continuous unknown parameter, the unknown bias of a coin and discrete observations, namely, the number of heads that are observed in a sequence of coin flips. This is an example that we will start in some detail now, and we will also revisit later on. And in the process, we will also have the opportunity to introduce a new class of probability distributions. This example is an extension of an example that we have already seen, when we first introduced the relevant version of the Bayes rule. We have a coin. It has a certain bias between 0 and 1, but the bias is unknown. And consistent with the Bayesian philosophy, we treat this unknown bias as a random variable, and we assign a prior probability distribution to it. We flip this coin n times independently, where n is some positive integer, and we record the number of heads that are obtained. On the basis of the value of this random variable, we would like to make inferences ...

We will now go through another example to consolidate our intuition about the content of the law of iterated expectations and the law of the total variance. The example is as follows. We have a class, and that class consists of 30 students in total who are divided into sections-- the first and the second section. Let xi be the score of students i, let's say the final grade in the class. We consider the following probabilistic experiment. We pick a student at random, uniformly, so that each student is equally likely to be picked. And we define two random variables-- X is a numerical random variable that gives us the score of the selected student. So if student i is selected, the value of the random variable capital X is xi. And capital Y is defined as the random variable, which is the section of the selected student, so that y takes values 1 or 2. We're given some information. For the first section, the average of the student scores is 90. For the second section, t...

We will now go through an example that brings together all of the concepts that we have introduced. We have a stick of length l. And we break that stick at some random location, which corresponds to a random variable, X. And we assume that this random variable is uniform over the length of the stick. So its PDF has this particular shape. And for the PDF to integrate to 1, the height of this PDF must be equal to 1 over l. Then we take the piece of the stick that we are left with, which has length X, and we break it at a random location, which we call Y. And we assume that this location Y is uniformly distributed over the length of the stick that we were left with. What does this assumption mean? It means that if the first break was at some particular value, x, then the random variable Y has a conditional distribution, which is uniform over the interval from 0 to x. So the conditional PDF is uniform. A conditional PDF, like any other PDF, must integrate to 1. So the height ...