 Hi, I'm Zor. Welcome to Unizor Education. I would like to present a lecture about independent random variables. This is part of the entire course of advanced mathematics for high school students. I do recommend you to watch this lecture and any other actually from Unizor.com website because it contains notes for every lecture as well as exams for certain topics, not all of them, but almost. So, today's topic is independence of random variables. Now, we have actually learned what is an independence among certain events. So, events independence we basically have learned in one of the prior lectures on conditional probabilities. Basically, let me just remind you very quickly that if you have the sample space, let's use the letter omega about and it contains certain elementary events and in this course we are assuming only finite number of elementary events. Now, there are certain events which are subsets of this set of elementary events. So, let's say you have a and you have b two different events. They are subsets of the set of old elementary events. Now, we call them independent if conditional probability of event a under condition of event b occurring is equal to unconditional probability of a. So, if you forgot what conditional probability is, I do recommend you to go to the prior lecture which it's in the it's among the lectures on conditional probabilities. Alright, so we know about events. Now, we have to transpose this particular definition to random variables. Now, by the way, a couple of consequences from this. Once a is independent of b actually I proved in one of those lectures that b is independent of a. So, here b is a condition and a is an event which we are basically examining. Here is a in the condition and b is an event. So, if this is true then this is true and also the simultaneous occurrence of two events a and b, the probability of this simultaneous occurrence is equal to the product of their independent probabilities. That's yet another property of conditional probabilities which we did address before. So, now we're trying to transpose all these definitions into the realm of random variables. Okay, fine. Now, first of all, let me just remind you what the random variable is. Well, random variable is basically a function on the elementary events. For each elementary event, our random variable takes certain value. Well, let's say x1, x2, etc. xn. So, the probability of each elementary event is basically the probability of our random variable xc of taking one of these values. By the way, if we are talking about the random variable which is basically a function, a numeric function, these are numbers, real numbers. So, it's a numeric function defined on the set of elementary events. Now, all x1, x2, etc. xn might not necessarily be different. I mean, you obviously have something like an example of parabola which takes the same value of 1 if argument is equal to minus 1 or if argument is equal to 1. So, any function can take more than once the same value in different elementary events, which is fine, in which case we can say that, for instance, x1 and x2 are exactly the same number. Well, we can say, let's call it y. So, we can say that random variable y takes value of y, x3, x4, etc. But the value of y would be taken with the probability of p1 plus p2 because this is the combined probability of two different events when the value of the random variable is the same. But this is actually doesn't really, you know, it's not really significant. So, we can always say this. There is some kind of random variable which takes some kind of real values with certain probabilities. Okay, fine. Now, what we are interested in is how can we define the independence of random variables. So, let's consider this is our one random variable and this is another random variable. Well, let's use the letter m first and then the letter n second, since alphabetically m precedes n and x precedes y. Okay, and let's call this again something like q. Different probabilities, different values, everything is different. Now, let's return back to the events. So, let's just think about under what circumstances our random variable xi takes value, let's say xi. Well, it's basically the combination of all the elementary events of our initial sample space omega, the combination of all these elementary events where the value of xi is the same. We just chose any particular value. So, we can always say that this is an event, the random variable xi taking a value of xi, it's an event which is a combination of certain elementary events from our initial sample space. So, we can basically call this event ai. So, ai is a combination of let's say e2, e17 and e23. If my omega is this e1, e2, etc, ek. So, it's a combination of certain elementary events where our random variable xi takes one of these values. And I do allow for multiple events, elementary events, to be mapped to the same value. Now, similarly, I can talk about elementary events which are combined together make my random variable eta equal to yj. Let's call it bj. So, this is a set of elementary events on which xi is equal to xi. And this is a set of elementary events from the same omega where a random variable eta takes yj. Now, what we can now say that taking by random variable any particular value is actually transformed into the language of events. And what I can say right now that I can basically define the conditional probabilities of the random variables through the conditional probabilities of events, namely the conditional probability of random variable c taking xi under condition of random variable eta takes yj is basically, by definition, a conditional probability of event ai under condition of bj. So, this is a definition. Since xi taking certain value is an event and eta taking certain value is an event, then I can define the conditional probability of xi taking certain value under condition of eta taking certain value as conditional probabilities of event. That's it. Definition is finished. And now I can say that now I can define using this the independent random variables. I can say that if my variable xi takes certain values x1, x2, xm, whatever, with certain probabilities, and if, sorry, this event conditional on this event equals unconditional probability for all xi and yj. So, all different pairs of xi taking certain value and eta taking certain value. If all the combinations of values really result in this particular equality, if my conditional probability of xi taking certain value under this condition is equal to unconditional probability for any variable which xi can take and any variable which eta can take, then only then my random variables can be called independent. And as in the case of elementary events, all these nice properties, like for instance, if this is true, then the probability of eta taking any value under condition of xi taking any value. If that is true, so if xi is independent of eta, then eta is independent of xi. And also the probability of combined and eta equals yj equals to the product of their corresponding probabilities. Now, the proof of this is basically nothing but using the ai and bj events instead of this. So, everything just follows from the definition quite nicely. So, we have defined what is the independence among two random variables. That's fine. What else is interesting here? So, the properties I have considered, now properties of independence this and this, okay. All right, fine. So, that's it about definition. Now, let's just consider a couple of examples. Now, for instance, our first example is, for instance, you have two dice. So, you're throwing one dice and then you're throwing another dice. From intuitive kind of understanding of the whole process, it's kind of obvious that no matter what's the result of the first dice, it should not really affect the second, right? I mean, that's kind of obvious. Let's prove it mathematically using our definition of independence. So, what is our omega? Omega is the combination of six different results of the first of the dice, right? But now we have two different dice, which means that we have to really consider the combinations, the pairs of the results. So, it will not be six, it will be six by six. One, two, three, four, five, six. One, two, three, four, five, six. These are results of two different dice rolling one after another. So, let's say these numbers, the number of the rows is the result of the first dice and this is the result of the second dice. Now, I didn't mention it actually, forgot to mention that the very last property, the property of combined value of c and eta, and you take the probability of this event is equal to the product of corresponding probability. So, probability of c is equal to xi and eta is equal to yj is equal to the product of there. Okay. Now, this is a characteristic property. It's very easy to do exactly the same thing as I did with events that all these three equations which I wrote, the conditional probability of this under this condition is equal to unconditional or conditional probability of this under this as a condition is equal to this. And this formula that the probability of the combined event is equal to the product of corresponding probabilities. All these three properties are equivalent. So, if we take one of the definition, other swallows and if they take as a definition, the first one is following. So, basically, all these are called a characteristic properties of the independent. So, let's just think about this particular equation as a characteristic property. And let's just check it. If in this particular case, it holds. All right. So, now, if you take two certain values for c and eta, c is the result of the first, which is roll number, and eta is the result of the second dice. That's the column number. So, if you take their combined value, let's say c is equal to 2 and eta is equal to 4, 2 and 4. So, what would be the probability of this? Obviously, 1 over 36, because from the definition of whatever we are doing with dice, we are rolling the two dice actually, all the chances of all the pairs must be the same, right? So, this is 136. Now, let's consider what these are. So, what's the probability of the c is equal to 2 when we roll two dice? Well, it's this plus this plus this plus this plus this and plus this, right? So, it's 6 times 136. So, it's 636. So, this is 136. This is 636, which is 166, right? Now, what is the probability of eta equal to 4 in this case? So, it's basically this, regardless of the first. We are talking about completely independent value of the second roll, regardless of what the first roll shows. So, the first can be either 1 or 2 or 3 or 4 or 5 or 6. What's important is that the second is equal to 4. And obviously, it's again 1, 2, 3, 4, 5, 6 times 136. So, it's also 636, which is 136 of 16. And obviously, the product of 1, 6 and 1, 6 gives you 136. So, this particular equation actually holds, which proves that throwing, when you're throwing two dice, the result of the first is completely independent of the result of the second. Now, let's consider another example when this independence is not actually the true property. And for this reason, instead of considering our two different results of the first dice and the second dice, I would consider two different random variables, c being the first roll and eta being sum of first plus second roll. Okay. Now, I'm sure you see and you feel that this is not exactly independent thing, that the sum depends on what exactly the first roll is. And it might be different, right? So, let's just choose a couple of numbers. And let's prove that the equation which I wrote about the probability of simultaneous taking certain values is equal to the product of probabilities, that that's not a true equation in this particular case. So, what can we take as a value? Let's say, again, c is, let's say, 2. And eta, well, let it be 4. I don't care actually. Now, what are all our three probabilities which we were talking about? So, the probability of c is equal to 2 and eta is equal to 4. What is this? Well, if my first roll is 2, that's all these guys. But I would like at the same time, sum of them, of the first and the second, to be equal to 4, which means for the second, I have only one choice, which is also 2. So, this is basically this event, 2 and 2. There are no others. And it's equal to 1, 3, 2, 6. This probability, right? Now, let's talk about the product of probability of this times probability of this. Well, probability of c is equal to 2 is obviously 1, 2, 3, 4, 5, 6, 36, which is, we already did this, it's 1, 6. How about probability of eta is equal to 4? Well, let's see. Well, it's this 2 and 2, right? Well, I shouldn't really do this. I should really mark as an intersection. So, this particular cell of my table represents 2 and 2 and the result is 4. Now, also this 1 and 3 and also this 3 and 1. So, all these three results of my throwing a couple of dice result in this. So, it's actually 3, 36, which is 1, 12. And as you see, the product of this 1, 6 times 1, 12 is 1, 7 to 2, not 1, 36. They are not equal. So, the product of these independent probabilities is not equal to the probability of their simultaneous occurrence, which proves that this is not independent variables. Right? And now the last problem, where my calculations will be slightly more involved in this particular case. It's also about 2 dice. I would like to know the following. One event is this and another event is this. Again, since we don't really have any connection between these two roles, then it looks like it should be independent, right? Well, let's just check again. What's the probability of the first one being odd and the second role being even? Well, that's this one. The first is odd is this, this and this. And the second one even is this, this and this. So, on the crossing, I have, let me just wipe out this from the previous problems. So, what I have is this, this, this, this, this, this, this and this. 1, 2, 3, 4, 5, 6, 7, 8, 9. Right? Is that right? So, I have 9, 36, which is equal to 1 quarter. That's the probability of this simultaneously. 1 quarter. Fine. Now, let's calculate separately. The probability of C is equal to odd. So, the first one is odd and we don't care about the second one. So, the first one is odd is all of these, all of these and all of these. Right? 1, 2, 3, 4, 5, 6, 3 times. So, it's 18, 36. So, the probability of C is equal to odd is, what did I say? 1, 2, 3, 5, 6, 18. 18, 36, which is 1 quarter. Now, what's the probability of the second being even? Well, it's all of these, all of these and all of these. So, the second one is even and doesn't matter what's the first one. Right? So, this is also 6 plus 6 plus 6, 18, 36, which is also another one half. And their product is 1 quarter. And that's the proof that these two variables, random variables are independent. Well, I think that's all I wanted to say. I do encourage you to look again onto Unisor.com and take this lecture with its notes, read the notes. It's very useful. I might actually present something slightly differently. And if there are some exams, you obviously are welcome to take exams. That's only for registered students, by the way. All right. Thanks very much and good luck.