 Hi, I'm Zor. Welcome to a new Zor education. Continue talking about random variables and its characteristics. We have introduced in the previous lecture two new characteristics of the random variable, variance and standard deviation. And I would like to present just a very simple problem where I'm basically calculating these characteristics. Just to make, you know, certain practical maybe usage of these. And our task for today is, we have the following problem which we would like to solve. The random variable we are talking about is a sum of two numbers which we get when rolling two dice. Right? So we have two regular perfect dice and we have a sum of the two numbers which are rolled on these dice. This is a random variable because depending on certain circumstances this sum of two numbers takes different values, obviously. And what I'm interested in is what's my expectation, what's my variance and what's my standard deviation of this random variable. Alright, I think the very convenient way to represent the values of my random variable is in the matrix where I have one, two, three, four, five, six, one, two, three, four, five, six. So my roll number would be the first dice and my column number would be my second dice. On the crossing I will have the sum, obviously. So it's two, three, four, five, six, seven, three, four, five, six, seven, eight, four, five, six, seven, eight, nine, five, six, seven, eight, nine, ten, 6, 7, 8, 9, 10, 11, 7, 8, 9, 10, 11, 12. Now, since every dice has six different numbers, it can actually show the two dice which are independently rolled together will have six times six different pairs. So, all these are represented in this square. Now, we are interested in a particular random variable, which is a sum, which is inside of this. So, to represent the variable, we need to know which values it takes, and apparently the values are from 2 to 12, right? And what's the probability of each value? Now, in the previous examples, we usually use something like equal probabilities. In this case, it's not equal. Why? Because every combination of two numbers has the same probability. There are 36 combinations, so we have 136 as a probability of each square, if you wish, in this matrix. Now, the number two occurs only one, which means that the probability of two is equal to 136. The probability of three, there are two different squares in this matrix, which means either the first dice rolls two and the second one, or the first one rolls one and the second two. Each one of these pairs has 136 probability, but we have two pairs. Right? Two one and one two. So, the probability of the number three is 236. Number four occurs in three times. So, the probability of four is 346. Probability of five is equal to one, two, three, four. Four 36. Probability of six is equal to one, two, three, four, five. So, five different combinations, each one of them has 136, so there are five of them, so it's five 36. Probability of seven is equal to one, two, three, four, five, six. Six 36. Probability of eight is equal to five, 36. Probability of nine is equal to four 36. And probability of ten, three 36. Probability of eleven, two 36. Probability of twelve, one 36. There is only one case, six and six, when you can get twelve. So, now this is a full specification for our random variable. It takes the value two, three, four, five, six, seven, eight, nine, ten, eleven, twelve. And these are probabilities with which these values are taken by our random variable. That's all we have to really know about the random variable. Which values it takes and what's the probability of each variable. It's a full description and that's a sufficient description to calculate expectation, variance and standard deviation. Now, if I will tell somebody, okay, this is my random variable. It's a full description. Could you evaluate approximately what you will win or lose and what will be your risk if you play for money some kind of a game? Let's say you have a game that the sum of numbers will be greater than five, for instance. If I will ask the question right now, it will probably be very, very difficult to answer just looking at this picture. And especially difficult will be what's the risk involved, right? However, using the expectation and standard deviation and variance, we can answer these questions in a manner which is sufficient to intuitively feel these particular characteristics of the random variable. And that's what I'm going to do right now. So I don't need this square anymore. And let me just address directly the issue of expectation. So this is my random variable c. So what is its expectation? Well, if it takes value two with a probability of 136 and value three with a probability 236, etc., etc., etc. and value 12 with a probability of 136, that would be my expectation, right? So I have to make a weighted average of these values from 2 to 12, from 2 to 12. And the weights are the corresponding probabilities. For two, the weight is 136. For three is 236. For 10 is 336. It would be somewhere here. And for 12 is 136 again. So this is my expectation. Now, I calculated it, and it's equal to 7, which is kind of intuitively obvious. Why? Because if you remember in the previous lecture, I was using just one dice and I was calculating its expectation of the value from 1 to 6. Remember? Oops. From 1 to 6, 1, 2, 3, 4, 5, 6, 7. No, there is not 7. Only 6. And the expectation in this case was right in the middle, which is 3.5. And if you remember, the mathematical expectation is an additive function. So if you have two random variables, and now we have basically two random variables, one is the result of one dice and another result of another dice. Each one of them has 3.5 as a mathematical expectation, right? So the sum of them, considering the expectation is an additive function, should be 3.5 times true, which is 7. All right. Now I would like to evaluate the variance of this particular random variable. And the way I'm doing it is I'm using the weighted average of a square of a difference between this value and the expectation, which is 7, with weights equal to, again, probability. So it's actually exactly the same thing as with expectation in terms of this is a weighted average, where weights are the probabilities. But in this case, in case of expectation, I'm using directly the values. In case of variance, I'm using squares of a difference between the value and expectation. So it would be 2 minus 7 square times its probability, plus 3 minus 7 square times its probability, plus 4 minus 7 square times its probability, etc. Up to the very last one, 12 minus 7 square times its probability. And I have calculated this. It's equal to 5.8 approximately. So that's my variance. Now finally, my standard deviation is just the square root of variance. Okay, so now we see that our behavior of the random variable can be described in terms of expectation and standard deviation. So the sum would be somewhere around number 7. That's the expectation. That's the weighted average of all the different values of this sum. Or if you wish, that's the approximate frequency the sum will take if you have many, many experiments of rolling a couple of dice. But now I can also say that the diversion around this 7 would be different, obviously. It can be from 2 to 12. But on average, the diversion would be from 7 to minus 2.4, which is what? 4.6 to 9.4, right? So these points represent like grouping, if you wish, of the values to the left or to the right of number 7. So my spread around, if you wish, 7 would be different, obviously, but the average spread would be between these and these. Okay, basically that's it. This is just an example of how in this particular case expectation variance and standard deviation can be calculated. And sort of intuitive understanding how we can use these particular characteristics of the random variable to, well, not really predict, but at least to expect something, some behavior of our random variable as we progress in our experiments with it. So the values would be somewhere around number 7, more or less evenly distributed around it, and the average distance from the 7 should be either to the left or to the right, should be by 2.4. That's the average, weighted average difference. That's it for today. Thanks very much. I recommend you to read the notes for this lecture, and they are obviously Unisor.com. I basically recommend to watch the video also from the Unisor.com website, just because you have basically the whole curriculum over there in front of your eyes, so you can go to some other lecture, etc. If you forgot something, you can go to the previous lecture, and it gives you some nice perspective of everything the whole course. So that's it. Thanks very much, and good luck.