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Okay, now the recording starts. You should see the slide. Now a big prediction. Okay,
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as I said, the first chapter is a review of probability. There are many, many textbooks
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which give you basically the same type of information that I supply here. And I have
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just cited one reference, which is Woolridge's introductory econometrics. You can consult
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this or any other textbook, basically covered almost everywhere, where you find introduction
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into statistics. The first thing I actually want to do is that I want to give you a motivational
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example to show you that a probability theory is not just something abstract, but that can
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be really useful in real life with some knowledge of probability theory. So the example that
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I give you is something which actually occurred to me with a colleague of mine who was married
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and his wife was pregnant. I think for the fifth month of pregnancy, and she went to
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a doctor to have a regular test that everything was okay. And among the tests, there was a test
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for the severe mental illness of the fetus. And to the woman's horror, this test turned out to
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be positive. So the nurse, which gave her this information that the test was positive, tried to
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be helpful and also provided the information that the test result is correct in 90% of the
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cases. And she more or less already asked the woman to consider the question of whether she
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wanted to have an abortion, given that with a high likelihood, the fetus had a severe mental
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illness. Now, this wife, of course, was terrified, shocked by this information, was the first child
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to be born to the couple. And she went, of course, back home to her husband to inform of this test
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result and discuss the question with him. And fortunately, the guy had some knowledge and
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probability theory. And the first question he was asking was, what exactly does it mean when the
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nurse says correct 90% of the cases? And he picked up the phone and actually called the doctor
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because he said, well, a nurse probably doesn't know much about a probability theory. So he talked
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to the doctor, but it turned out the doctor also did not know much more than that the test was
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correct in 90% of the cases. But the doctor supplied some additional information. And this
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information was that in this particular woman's age class, which was this case older than 30, the
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prevalence of the illness is relatively high, meaning that about one in 100 births of women
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of age older than 30 are affected by this illness. So the probability of the fetus being ill, even
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without a positive test result, was 1% or 0.01. So the question that the couple had to answer or the
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woman had to answer for herself was given that the test is positive, given that the prevalence of
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the severe mental illness of the fetus is relatively high in her age class, and given that the test is
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90% accurate, should the woman have an abortion? Now, as I said, my colleague used some statistics
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and he analyzed the case the way I will do it with you now. He said, suppose we randomly test a high
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number of pregnant women aged older than 30. What do we know? We know that the prevalence of the
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illness is 1%. So out of the 1,000 pregnant women of this particular age class, 10 will bear a fetus
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which is severely mentally handicapped, and 990 will bear a healthy fetus. Now, suppose that the
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accuracy of the test, so what the nurse described with correct 90% of the cases, applies to all
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fetuses. If we interpret correct 90% of the cases as applying to all fetuses, then this means
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those 10 women with a fetus which is mentally handicapped will have nine positive tests
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and one false negative test. So if we test all the 1,000 pregnant women with this particular test,
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then we know among the 1,000 women will be the 10 women which have the ill fetus in their womb,
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and nine of them will be identified by the test. One will not be identified.
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Then there are 990 women with a healthy fetus, and assuming that the test is correct in 90% of the
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cases would mean that there will be 99 false positive tests. So 10% of the 990 women will have
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a positive test result even though the fetus is healthy. 90% of the cases the test will correctly
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indicate that the fetus is healthy, so the test result will be negative. This means 891 women will
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get a negative test and will ensure that their fetus is healthy, but 10% of the women given a 90%
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accuracy of the test, 10% of the women will receive a false positive test, which in total
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resides in 108 women out of 1,000 receiving a positive test result.
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So we will have 108 women whose test result will be positive given the information we have.
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108 women with a positive test result out of 1,000 tests that we do on the 1,000 pregnant women
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means that the probability of a positive test is slightly above 10%, 0.108.
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Now what are we interested in? We are interested in the probability of the fetus being ill given
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that the test result is positive. So we are interested in something which we call a conditional
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probability. We will later formalize the concept. It should be clear here what we are actually
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interested in. We are not interested in the probability of the test being positive. We are
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not even interested in the probability that the fetus is ill because this is just one percent as
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we know. We are interested in the probability of the fetus being ill severely mentally handicapped
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given that the test result is positive. So how high is this probability?
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Well we know we have 108 positive tests and of these nine in nine cases the fetus is ill.
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Therefore the probability of illness given that the test is positive is in nine cases
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out of 108 cases is the fetus truly ill. So since we look at the probability given the test is
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positive we just now have to look at those cases of women where the test turned out to be positive
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which are 108 women and we know of these 108 women 99 have received a false positive test.
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Only nine women have truly a fetus which is severely mentally handicapped. So the probability
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of the fetus being ill given a positive test result is 9 over 108 and this is approximately 8 percent.
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So this result I think is quite interesting because it more or less turns the impression
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of the initial test result on its head. The initial test result and the way it was explained
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by the nurse in the hospital and the doctor in a later telephone call seemed to suggest that
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given the test result there's a 90 percent chance that the fetus is mentally handicapped.
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But applying proper probability theory we see that this is absolutely not true
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but it is it is precisely the converse almost precisely the converse which is true.
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The probability that the fetus is healthy given that the test result is positive
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this probability is 92 percent. It's just the complement of the 8 percent here. The probability
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that the fetus is healthy is 92 percent even though we know that the test result was positive
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or in other words the probability of the fetus being ill prior to the test was 1 percent.
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By the positive test result this probability of 1 percent has increased to 8 percent.
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It has not increased to 90 percent not increased to what the doctor and the nurse supposed to be
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the accuracy of the test. So after having analyzed the case program probabilistic terms
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the woman decided not to have an abortion as I said it's a true case and a couple of months later
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she gave birth to a perfectly healthy baby. So she took the risk of the 8 percent
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or she made some moral considerations that it is inappropriate to kill an unborn
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not having any strong evidence that it is mentally actually mentally handicapped.
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So she gave birth to the baby and a couple of months later turned out that the baby unfortunately
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was perfectly healthy. The important thing here to note is that the positive test result is by far
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not as informative as doctor and nurse seemed to think it was. The risk of the illness increased
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certainly by the indication of the test result but only from 1 percent to 8 percent and not
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to 90 percent. Now this was my motivational example which helps show you that it's some
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circumstances in real life which actually may also occur to you sometime I hope not but maybe.
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A proper analysis of probability may really be life-saving at least for the unborn and also for
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the parents of course who in this case were really I mean they had been waiting for a baby a long time
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and they were shocked by the prospect of either having to have an abortion or having to live with
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a mentally handicapped child. So fortunately probability theory could give them some
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relief there and help them in taking the right decision. For those of you who remember their
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statistics education in the bachelor program I suppose that you will have heard of Bayes theorem
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because what we did here was actually nothing else but as an application of Bayes theorem.
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In short a trained statistician or econometrician would say well of course the case is easy the
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probability of ill of the health of the fetus being ill given that the test result is positive
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is by Bayes theorem the probability of a positive test result given that the fetus is ill
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this is the 90 percent times the probability of the fetus being ill this is one percent the
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unconditional probability divided by the probability of the test being positive this was the 10.8
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so at 0.9 times 0.01 divided by 0.108 and this is approximately 8 percent. So simple
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illustration of Bayes theory or theorem or simple application of Bayes theorem.
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So far for my motivational example I will now start with a systematic review of probability
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theory and here we start with distribution functions. Now I suppose all of you have heard
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of distribution functions so I will be rather brief on this. Let us start with a discrete
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random variable. I will not define what a random variable is because this leads us into complex
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territory with sigma algebra and these kind of things but I will rather suppose that you have
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at least some intuitive understanding of what a random variable is and the discrete random variable
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of course is a random variable which takes on just a finite number of outcomes. So the probability
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distribution of a discrete random variable X consists of a listing of the possible values
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that the variable X can take. Obviously since there are only finitely many outcomes it is always
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possible to have such a listing and in this listing we will match each possible outcome
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with the probabilities for this particular value that the variable X may take. So more formally if
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X can take on k different values and we call them small x's x1 x2 all the way up to xk
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then we can define the probabilities small p1 small p2 up to small pk where all these p
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probabilities here are numbers from the closed interval between zero and one so we
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denote probabilities as decimal numbers and not as percentages even though you know you multiply
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by 100 and then you get sort of the percentage magnitude such that pj the probability for outcome
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j the small pj is defined as the probability that the random variable X takes on the particular
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value small x in Xj. On notation in the first set of slides here I will always notationally
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distinguish between a random variable and I will denote the random variable with a capital letter
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so capital X whereas a particular outcome of a random variable is denoted by a small
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letter by the same letter but the small letter then and often with an index like here Xj is just
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the real number some particular outcome of the random variable X itself is a random variable which
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can take on any of the k possible outcomes and we sometimes speak of a realization of a random
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variable when a particular random variable is observed with a particular real number with a
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particular outcome small x so pj is just the probability of the random variable X taking on
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the particular outcome small xj and this can be defined for all the j's from one all the way up to
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k and we of course have the restriction otherwise it would not be a completely defined random variable
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that all the small p's sum up to one. From this we can define a probability density
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function and we call it pdf so this has nothing to do with it. Probability density function or
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brief the pdf the probability density function of a discrete random variable capital X is given by
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a function which maps the space of real numbers on the interval zero one such that this function
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small f for one particular value which is a possible outcome of the random variable in this
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case xj f of xj is defined as the probability that random variable X takes on variable takes on the
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outcome small xj so for all the j of all the k outcomes j runs from one to k for all the k
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outcomes we can define a particular value equal to the probability of this outcome for function f
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and all other x values in the space of real numbers of course have a probability of zero
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so any x being a real number which is not among the possible outcomes x1 x2 to xk has a probability
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of zero probability the probability is zero that the variable the random variable capital X takes
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on this particular value x for instance if you toss a dice die then with the probability of
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having a 3.5 for instance is of course zero whereas the probability of tossing a three or four is
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obviously not zero. Here's a simple example of that a coin tossing experiment so we have a coin
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and the coin has a head and a tail so the two sides of the coin we think that no other outcome
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is possible so the coin will not land on the ground standing upright or this probability is zero
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so it either shows a head or tail and then we denote by x the number of heads
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thrown after tossing the coin three times right so we toss the coin three times and we count
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the number of heads we do not count the number of tails obviously this is the compliment and then
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we can derive the probability distribution of this very simple excrete random variable
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the probability of zero heads is equal to the probability of three tails so the probability of
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x being equal to zero not having thrown single head is the probability of having thrown
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three tails in a row and obviously the probability of throwing one tail is one half times the
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probability of throwing a tail again is another one-half and times the probability of throwing
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a tail again is another one half so one half times one half times one half is one eighth
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and that's the probability of observing zero heads after throwing after tossing the coin
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three times the probability of observing just one head after throwing after tossing the coin
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three times, is obviously the probability of head, tail, tail, plus the probability of tail,
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head, tail, plus the probability of tail, tail, head. These are the three possibilities,
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and it's very easy to see what the probabilities of the single events here are, having head,
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tail, tail, so head in the first toss, tail in the second toss, tail in the third toss,
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has a probability of 1 over 8. And the same thing is true here, and the same thing is true here,
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so adding these up is probability of 3 over 8. The probability of observing two heads is then,
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of course, completely analogous. Analogous, there are three such events, head, head, tail,
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and tail, head, head, and head, tail, head. All three events have probability one-eighth,
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so there are three such events, and therefore the probability of the whole is 3 over 8.
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And finally, there's the probability of X assuming the value of 3, so throwing three heads, tossing the
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coins three times with the event head in each toss, that again has the probability of 1 over 8.
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And then clearly, when you sum 1 over 8 plus 1 over 8 plus 3 over 8 plus 3 over 8,
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then you have 8 over 8, which is equal to 1. So that's the probability density function of
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X. f of X is 0.125, so 1 over 8. If X is equal to 0, then three times that, 0.375 for X equal to 1,
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the same value for X equal to 2, and then again, 0.125 if X is equal to 3. For all other values
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of X in R, the probability is going to be 0. If we have an actual coin tossing experiment, we can
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write this probability distribution as what we call a histogram. Suppose that we have actually
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found that we have tossed the coin such that X is equal to 0 occurs one time, and the other two events
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here, X equal to 1, X equal to 2, occur three times, and X equal to 3 occurs one time again,
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then we can have such a bar diagram here, which is in some sense a rough approximation of the
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probability density function. Note that we have here on the ordinate the probabilities 0.125,
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0.325, and we don't really have scaling here on the abscissi because, well, this here is not really,
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you cannot write a zero here because zero would be just a point. Basically, you have to interpret
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this here as an interval of 0, this here is an interval of 1, and an interval of 2, and an interval
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of 3, but I think you get the feeling of what is indicated by the histogram representation of this
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random variable. Next, we define the cumulative distribution function, which we abbreviate CDF.
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The cumulative distribution function CDF of a random variable X is given by a function capital
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F, which again maps R into the closed interval 0 to 1, such that capital F for some given real number
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X is defined as the probability for the random variable X taking on a value which is equal to
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or smaller than X, so that's small x. F of small x is equal to the probability of capital X being
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less or equal than equal to capital small x. Now, in the case of discrete random variables,
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the cumulative distribution function f of x sums the probabilities pj over all the indexes j,
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for which xj is less or equal than x. So capital F of x is just the sum over all probabilities,
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and I have indicated now this xj less or equal than x with an indicator function. That's a 1
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with an indication range, meaning that this function here assumes the value 1 if xj is
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less or equal than x, and it assumes the value 0 if xj is greater than x. So that's called an
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indicator function, and it ensures that we in this expression only sum those probabilities,
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which are associated with xj values less or equal than x. So there's a standard notation in the
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statistics, and you may not yet have heard of it. It is probably not often used in bachelor
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statistics lectures, but it's actually a fairly easy concept, easy to understand,
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a function which assumes the value of 1 on a certain range for x and a value of 0 on the
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complement of this range. Now let us go back to our example, to the coin tossing experiment and
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compute the cumulative probabilities. The cumulative probability of x being less than
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or equal to 0 is of course 0.125, because there's just this one event that x takes
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on value 0, and this has a probability of 0.125. Actually, the cumulative probability
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stays at this level for all x values smaller than 1, and there's a jump in the CDF at 1. So f of 1
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is then 0.125 plus the probability of having at least one head plus 0.375, so this is 0.5.
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And then there's a next jump at f of 2, where we also add to the probabilities of x assuming 0
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or x assuming 1, the probability of x assuming 2, so having two heads among the three tosses of the
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coin, then we have a probability of 0.875. And finally, the probability of x being less than
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or equal to 3 is of course the sum of all four probabilities, and the sum is as we know 1.
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So the corresponding CDF is f of x is equal to 0 if x is smaller than 0,
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if 0.125 is x if x is between 0 and 1 with smaller than 1, it is 0.5 for x being greater than or
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equal to 1 and smaller than 2, 0.875 if x is greater or equal than 2, but smaller than 3,
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and for x greater or equal than 3, the value of the cumulative distribution function is just 1.
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But here we can draw it, the cumulative distribution for x with again having on the
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ordinate probability, and then we have a real scaling also on the FCC here on the x-axis.
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So here is 0, and obviously the value of f is 0 before we reach the 0.0 here, then it
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jumps to 0.125, stays constant at this value for all values of x which are smaller than 1,
00:30:21.000 --> 00:30:31.000
jumps again here, stays constant up to 2, jumps at 0.22, 0.875, stays constant until we reach 3,
00:30:31.000 --> 00:30:36.000
there's another dump here, and then we have 1, and actually continues at the level of 1 after
00:30:37.000 --> 00:30:46.000
the value of 4, so stays constant until infinity. So that's the cumulative distribution for x.
00:30:51.000 --> 00:30:58.000
The same thing actually applies to continuous random variables, but given that we only have
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a couple of minutes left, I think I will continue here on Thursday and present the same material for
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the continuous random variables, rather I will stop the recording now and after I have
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stopped the recording. Did I already stop the recording? Oh yes, I'll stop it now.