[MUSIC] Now, consider a little bit more complicated example of dependent random variables. Let us consider a random experiment that consists of three coin tossings, Of fair coin. Now, let us assume that X is random variable that is equal to number of heads for first two tossings, for first and second tossings. And Y is equal to number of heads, for second and third tossings. Is is true that X and Y independent variable or not? To answer this question, let us construct joint distribution of these two random variables. We know that variable X can take three possible values, 0, 1, 2 and the same is true for variable Y. So, let us draw a table for X and Y. What is the probability that X equals to 0 and at the same time, Y equals to 0? If X equals to 0, it means that first two tossings gives us tail. And if Y equals to 0, it means that second two tossings, also tails. So the only possible outcome that satisfies the condition that X equals to 0 and Y equals to 0 is tail, tail, tail. So the probability of that X equals to 0, and at the same time, Y equals to 0, is equal to the probability of one elementary outcome, tail, tail, tail. And the probability of this outcome for three coin tossings of fair coin is equal to 1/8. So we have to put 1/8 here, what about this cell? What is the probability that X equals to 1 and at the same time, Y equals to 0? Again, we have to find outcomes that satisfy these conditions. If Y is equals to 0, then it means that second and third tossings give us tail. So the only possible for X to be equal to 1 is to get head at the first tossing. So the only possible outcome that satisfies this condition is head, tail, tail. And the probability of this event, again, is equal to 1/8. What about this cell? If X is equal to 2, it means that first and second tossings gives us head, but in this case, Y cannot be equal to 0. Because if first two tossings are head, it is not possible for second and third tossing, give us 0 heads. So this is not possible and the probability of this event is equal to 0. One can see easily that this cell also contains 1/8. But this cell is a bit more interesting, probability that X equals to 1, and at the same time, Y equals to 1 is equal to the probability of event that is satisfied by two possible outcomes. Indeed, it is possible to have two heads at the first and the last tossings and tail here. In this case, here, we have one head and here, we also have one head and this satisfies our condition. But it is also possible to have tail, head, head, in this case, this head is included into both variables, X and Y. So again, we have one head here and head here. So this probability is equal to probability of such event. As we have two outcomes that satisfy this condition, the probability is equal to 2/8 or 1/4 In a similar way, we can see that we have 1/8 here, 0 here, 1/8 here and 1/8 here. To test, is it true that X and Y are independent variables, we have to find distributions of X and Y alone, marginal distribution. We can do it, for example, for Y. To do it, we have to find sums over this row and we have 1/ 4 here, 1/2 here, and 1/4 here. We can see that the distribution of X, marginal distribution of X is actually the same. Now to check if these random variables are independent, we have to answer, for example, is it true that this value is equal to product of two values like this 1/4 here and 1 over 4 here? It is clear that it is not true, it means that these two random variables are not independent. However, if we know that X equals to 0, it doesn't give us precise information about the value of Y. We see that if X equals to 0, then Y cannot take value 2, but it can take values 0 and 1 with equal probabilities. It is different from the distribution of Y, but it doesn't give us exact value of Y. So this is an example of a situation when two variables are not independent, but they are not in deterministic dependence between each other. The dependence between them is a kind of probabilistic. It gives us some information about probabilities of Y depending on the value of X or vice versa. [MUSIC]