In this week, we are going to learn about the important concept of a random variable. So what have we have learned so far is we have learned about what we call a random experiment. After a random experiment, we defined what was a sample space. A sample space is what we defined as a set of all outcomes of this random experiment. Then we define the notion of an event which is a subset of a sample space and then we define probabilities of the event.
When we define probabilities of an event, we approach that through the axiomatic approach to define probabilities. Now what are we going to do? After this, then we introduce the notion of conditional probability, we introduce what was Bayes theorem.
But basically what we have done is we have started introducing this entire framework of probability through the notion of. random experiment and sample space. So, by now all of you should be really comfortable in computing probabilities of events. To help you compute the probabilities of events, we also introduced you to permutations and combinations. So, today and the next 2 weeks, we are going to focus on the following.
So, we are going to learn about random variables. In particular, after the end of these two weeks, you should be able to know what is a random variable and then you need to understand what is a discrete random variable, what is a continuous random variable. When you are talking about discrete random variable, which is going to be the focus of the first two weeks, we will be introducing notions of a probability mass function, how the graph of a probability mass function would look.
And then afterwards we will also introduce the notion of a cumulative distribution function. Now when we talk about both the probability mass functions and the cumulative distribution functions we are going to give a lot of examples which might be very useful. Finally we are going to introduce important concepts namely the expectation and variance of random variables and we are going to see how we can apply this in our day to day lives.
So, we begin with the notion of a random variable. So, what is the notion of a random variable? Let us start by revisiting a random experiment. So, recall when we talk about a random experiment, for example, when I toss a coin, I have a random experiment, the sample space is head and tail. When I roll a die once, I can have any one of these outcomes 1, 2, 3, 4, 5 and 6. Any one of these 6 outcomes can happen.
So, we say that every time I can use the notion of a sample space to describe the outcomes of the experiment. But many a time I might not be just interested in knowing about the outcome of an experiment. When a random experiment or a probability experiment is performed, we are not interested in the actual outcome, but we might be interested in the value of some numerical quantity determined by the result.
I repeat, we might not be interested in the outcome itself, but we might be interested in the value of some numerical quantity that is determined by the result. What do we mean by this? For example, when I roll a dice twice, I know there are 36 possible outcomes. For example, I could have a 1 in my first toss, I could have 1 in my second toss, I could have 1 in my first toss, 2 in the second toss, I could have a 1 in my first toss and a 6 in the second toss.
I notice that my sample space would have 36 such outcomes. This is something which we have already seen in our earlier lessons. But suppose I do not care about the individual outcomes, but I am interested only in the sum of the outcomes. So what is the numerical value or quantity I am associating?
I am associating and I am only bothered about the sum of the outcomes. In this case, for example, I am not I am only bothered about whether the sum of outcome is 7, but I am really not concerned about whether that 7 has a reason because of outcome 1 6 or 2 5 or 3 4 or 4 3 or 5 2 or 6 1. I am not concerned about it, but I am interested in knowing that the sum is 7. Now, whenever I am talking about such a numerical quantity. These numerical quantities of interest more formally they are actually functions defined on the sample space.
I am not going into the mathematical rigor of defining such functions. In advanced courses you will be subjected to the mathematical rigor. But what I want you to understand is with every outcome of the sample space and I am associating a quantity, a numerical quantity.
So, this quantity of interest is what I refer to as a random variable. Since these random variables are again the values of the random variables are determined by the outcome of a random experiment and I can actually assign values to these outcomes of the random experiment, I can assign probabilities to the possible values of the random variable. So, let us look at an example to understand what we mean by numerical quantities. Let us revisit the experiment of rolling a dice twice.
So, when I roll a dice twice that is I have a 6 sided die and I am rolling it twice, we have already seen I could have 36 possible outcomes and those 36 possible outcomes are listed in the form of my sample space which is given here. I have S, I have 36 possible outcomes, it could be a 1, 1, 1, 2 up to 6, 6. By I, J, I mean I in the first toss or the first roll of the die and J is the outcome of the second roll of the die. So I am throwing a dice twice or I am rolling a dice twice. So this is something which you have already seen.
We have defined events on this. Now suppose I am asking two questions. So I am asking the questions.
out of these outcomes, how many outcomes do we have? We have 36 outcomes. How many outcomes will result in a sum of 7? Notice I am just interested in the final value of 7. I am not interested in the individual outcomes as such.
The other question I am interested in knowing is how many outcomes will have the smaller of the outcomes as 3? What do I mean by smaller of the outcome? If I have an outcome 1, 5, the smaller of these outcomes is 1, whereas in 4, 3, the smaller of the outcome is 3. And for convenience, if it is 3, 3, I am going to take it as 3. If both the outcomes are the same, we say the smaller of the outcome is itself. So, these two are the questions I am interested in answering.
So, how do we answer these two questions? So, these questions are the notice the experiment is the same, the sample space is the same, the questions are actually based on the outcomes I am associating some numerical quantity, one is the sum of outcomes and one is what is the smaller of outcomes. So, it is important for us to notice that the random experiment and the sample space that I am going to use to answer both these questions are the same. So, what is a random variable?
So, now let us start with the same example. Let me denote the sum of the 2 rolls by x. Let me denote y to denote the lesser of the 2 outcomes. So, for example, I know this is my sample space, I have 1 1, 1 2. to 1, 6 and I have up to 6, 6. This is what we have listed earlier.
This was our sample space given here. So, I am just writing the sample space again here and I am telling x is denoting the sum of outcomes. So, if I have 1, 1 as my outcome, the value x will take is 2, which is 1 plus 1. If I have 1, 6 as my outcome, the value x would take is 7 which is 1 plus 6. If 6 is my outcome, the value x would take is 12 which is 6 plus 6. So, x is denoting the sum of outcomes. Similarly, y is denoting the lesser of two outcomes.
So, what is the lesser of these two outcomes? I said if the outcomes on both the roles are the same, I am going to take it as itself. The lesser of 1 and 6 is again a 1. The lesser of 6 and 6 is a 6. Suppose I had 3 and 4, the value of X would be a 7 and value of Y would be a 3. So this is how we are defining our X and Y here. So for the entire problem, we can see that these are, I have 36 outcomes.
For 1 and 1 the value is 2, 1 plus 2 is 3, 1 plus 3 is 4. So, for 2 plus 6 is 8, the minimum of 1 and 1 is 1, 1 and 6 is 1, 2 and 1 is again 1, 2 and 2 is 2. So, you can see that 4 plus 1 is a 5, 4 plus 2 is a 6, 4 plus 6 is a 10 and here you can see the outcome 5 plus 1 is a 6, minimum of 5 and 1 is a 1. So, minimum of 5 and 6 is a 6, 5 plus 6 is 11, 6 plus 6 is a 12 and the minimum of 6 and 6 is a 6. So, you can see that for each one of these 36 outcomes of my sample space, I have a value of x and I have a value of y which is associated with it. Now, why? What next?
Now, these outcomes are outcomes of a random experiment and associated with each one of these outcomes is a value of x and value of y. So, when I have the outcomes and these values, I can talk about associating or I can talk about the following is, can I talk about probability of the values, x taking the values? Now, let us go back here. X is taking set of values, Y is taking set of values. On closer inspection, I can see X takes the value 2, it takes the value again.
Does it take the value 1? It does not take the value 1. It takes the value 2. It takes the value 3. I have an X taking a value here. I take X again taking a value here.
Then I have, yes. 4, yes, it takes a value 4 here, it takes a value 4 here. 5, yes, 5, 1. So you can see that x is taking the value 5, yes, 6, 7, 8, 9, 10. So, it is taking 7, 8, 9, 10, 11 and 12. So, if you look at the values this X is taking, it is taking the values 2, 3, 4. 5, 6, 7, 8, 9, 10, 11 and 12. These are the values this X is taking.
I can see that X is taking the value 2 when the outcome is 1, 1. It is taking the value 3 when the outcome is 1, 2 and it is taking the value when it is 2, 1. It is taking the value 4 when the outcomes are 1, 3. Then it is also 2, 2 and it is taking when it is 3, 1. So, this way you can see that it is taking a particular value where I have more than I could have only one outcome corresponding to the value as in this case and in this case or I could have more than one outcomes corresponding to the value. So what are the values X is taking? X is taking the values which are which is denoted as the sum it is taking the value 5, 6, 7, 8, 9, 10, 11 and 12. These are the values x is taking and however defined x, I have defined x to be the sum of the outcomes of both the throws or both the rolls of the dice.
Now given this, now let us do the following. I see what is the value of x. Now let me take the value of x to be Now, what is the relevant event?
I know x will take the value 2 when the outcome is 1, 1. So, the relevant event that is corresponding to 2 is just the event 1, 1. There is no other way x can take the value 2. Now, x can take the value 3 if my outcomes are, the first outcome is a 2 and second is a 3. outcome, first outcome is a 1 and second outcome is a 2, first outcome is a 2 and second outcome is a 1. So, this is again the relevant event which will give me the value of X equal to 3. So, if I am going to look at each one of them, I get 2 with the event 1, 1, 3 I have the event which is 1, 2 and 2, 1, 4 I have the event 1, 3. 2, 2 and 3, 1 this the sum of these outcomes which I can define as a relevant event will give me the sum of 4, 5 it is going to be 1, 4, 2, 3, 3, 2 and 4, 1 and so forth I keep going and then you can see 9 it is going to be 3, 6, 4, 5, 5, 4 and 6, 3 all of them add up to 9 for 6 for. 10 I see the events are going to be 4, 6, 5, 5 and 6, 4. For 11 it is going to be 5, 6 and 6, 5 and 12 I see the relevant event is again just 6, 6. So, what we have done is first we have associated value with each of the outcomes and then I have just mapped I have seen that the x takes the value 2 through 12. x takes these values. I have mapped the relevant event for each value x takes. Now, what is the point of mapping this relevant event?
We recall what we said is because these random variables are defined on my probability sample spaces, what I can actually see and what we have defined earlier is the following is because the value of a random variable is determined by the outcome of an experiment, we may assign probabilities to the possible values. Now, come back to the example, what are the possible values of the random variable? I see that the possible values of the random variable are 2 through 1. These are the values, it can take the values 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12. So the question is can I assign a probability of X taking the value 2? Can I assign the probability of X taking a value 3?
Can I do this for all the values X is taking? That is the question we are asking. To answer this question, we are finding and we are mapping to each value that x is taking what is the relevant event.
We have already seen how to come up with probabilities of events. Once we do this mapping, We can see that the probability of X is the same as the probability of the relevant event. So, what is the X value of X?
I am looking at value of X, X takes the value 2. I am interested in knowing what is probability X is equal to 2. I know X equal to 2, the relevant event is just this set. I need to know what is the probability of this set happening. And we know the probability of this happening is. 1 by 36. This is something which we have already seen in our earlier discussions. Similarly, if x takes the value 3, I am interested in knowing what is the probability of x is equal to 3. So, I know what are the outcomes that give me x equal to 3. So, again I go back and I see it is 1, 2 and 2, 1. These are the, this is the irrelevant event and I know the probability of this event is 2 by. So, if I continue in this way, I get probability X equal to 2 is 1 by 36, probability X equal to 3 is 2 by 36, X equal to 4 is 3 by 36, 5 is probability X equal to 5 is 4 by 36. I can keep continuing it this way and I can verify that probability X equal to 9 is 4 by 36, X equal to 10 is 3 by 36. Probability of X equal to 11 is 2 by 36 and probability X equal to 12 is 1 by 36. So, what we have established is I know that this X, I say X is a random variable which takes the values 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12 and I can also assign a probability to the where x taking a particular value which I get from recognizing the probability x taking a particular value I will find out what is the probability of the relevant event and I have assigned probabilities to each of the values x takes.
Now let us look at the other. So, I asked two questions. The first question I asked is what is the probability of the sum equal to 7. That was a question. So, I am not interested in anything else. I just need to check whether X takes a value 7. I see that X takes a value 7. I need to know what is the probability X would take the value 7. I know probability X would take a value 7 is same as probability 1, 6, 2, 5, 3, 4, 4, 3. 5, 2 and 6, 1 which is same which would be 6 by 36. So, I am not interested in the individual outcomes but the probability that the sum is equal to 7 and I know that that happens with the probability of 6 by 36. So, what we have done is we have defined what is a random variable and we have assigned probabilities to that random.
Now on the same sample space, what do I mean by same sample space? I have again rolled 2 die, I have the same sample space. Now I am going to define y to be the lesser of the two outcomes.
What do I mean by it? If ij is an outcome, if i is lesser than j then y will take the value i. If i equal to j, y will again take the value i. So, I can define that y takes the value i if i is less than or equal to j. So, what are the values this y would take?
For example, if I have 1 1, the value of 1 y is equal to 1, 1 2 the value is again equal to 1, 2 1 the value is again equal to 1, 2 3 the value would be equal to 2. 2 to the value is equal to 2. So, these are the values that I am associating with each of the outcomes. So, I know that for each of these 36 outcomes, I can see that again I can go back here. So, if we go, so we can go back here, you can see here the value is 1. So, y takes the value 1. The second value y takes is 2. So, this is where it takes the value 1, 2. it takes the value 3, takes the value 4, takes the value 5 and 6. So, you can see that on the same sample space, now I am interested in the value y takes. So, y takes the values 1, 2, 3, 4, 5 and 6. These are the values this variable or this y takes. So, you can see y takes the values 1, 2, 3, 4, 5 and 6. So, I can again repeat what I did for the earlier case.
So, I see y with y takes the value 1, the relevant event is when will y take the value 1? It takes a value 1 when my outcome is either a 1, 2, 1, 3, 1, 4 or all. Outcomes in which 1 appears either in the first toss or in the second toss. So, what are the outcomes in which 1 appears either in the first toss or second toss? So, I have 2, 1, 3, 1, 4, 1, 5, 1 and 6, 1. So, these are the outcomes in which 1 appears and for these two outcomes y takes the value 1. So, I can say that y is taking the value 1. The relevant event is this following which is listed all the possible outcomes I have just discussed.
Now, when would y take the value 2? y would take the value 2 when 2 is equal to 2 or 2, 3, 2, 4, 2, 5, 2, 6, 3, 2, 4, 2, 5, 2 and 6, 2. So, you can see that these are the outcomes where. The lesser of the two outcomes is 2 and that you can see is 2, 2, 2, 3, 2, 4, 2, 5, 2, 6, 3, 2, 4, 2, 5, 2 and 6, 2. The third thing, when would, why take the value 3?
Again I have a 3, 3, I will have a 3, 4, I will have a 4, 3, I will have a 3, 5, 5, 3, 3, 6 and 6, 3 which I can list in the same way. I have all these outcomes where y takes the value 3. Similarly, for 4, y takes the value when it is 4, 4, 4, 5, 4, 6, 5, 4 and 6, 4, 5 when it is 5, 5, 5, 6 and 6, 5 and y takes the value 6 only when the outcome is 6 and 6. So, you can see immediately what you notice is the value y is at taking whereas x was the sum which took the value 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12 and I obtained what was the probability x equal to 2, x equal to 3 up to x equal to 12. So similarly here I have y which is taking value 1, 2, 3, 4, 5, 6 and I can find out Based on what is the relevant event, I can also tell what is probability y equal to 1, y equal to 2 up to probability y equal to 6. I can tell this similarly like the way we did earlier. So let us see what are the probabilities. So I have y equal to 1, y equal to 1. And I can see there are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 in the event and the probability of this event happening is 11 by 36. So, probability y equal to 1 is 11 by 36. Similarly, the probability with which y takes the value 2, I have 1, 2, 3, 4, 5, 6, 7, 8, 9 and all of them are equally likely, so it is 9 by 36. Probability y equal to 3, 1, 2, 3, 4, 5, 6, 7 which will give me a 7 by 36. Probability y equal to 4 is 5 by 36, y equal to 5 is 3 by 36 and finally probability y equal to 6 is 1 by 36. So, this was an example where we have defined two random variables which are capturing different numerical quantities on the same sample space which comes from the same experiment.
Now, let us look at another problem. Now, let us look at another example. Again, this is an example which we have seen earlier. I am tossing a coin 3 times.
So, when I toss a coin 3 times, I am observing what is the outcome of each of the tosses. I know the sample space in this is my first toss. So, I record what is my first toss, my second toss and my third toss.
This is what I am recording. So, If I look at the outcome, I could have a head, head, head. I could have a head in my first toss, head in the second toss, a tail in my third toss.
Head in my first toss, tail in my second toss, head in my third toss. I could also have a head in my first toss, a tail in my second toss, a tail in my third toss. I could have a tail in my first toss, head in my second toss, tail in my third toss. Tail, head, head. Tail, tail, head.
and tail, tail, tail. So, you can see that this 1, 2, 3, 4, 5, 6, 7, 8 are the set of possible outcomes that is I can have a head in my first, head in the second. So, it is basically head appearing in the, so I have three tosses.
The first toss could be head or tail. So, there is a two ways of happening it. Second toss also could be head or tail. Another two ways of happening, third toss could also be head and tail, another two ways. So, 2 into 2 into 2 which is 2 to the power of 3 which is 8 possible outcomes for this experiment.
And the sample space for this experiment is hence given by these 8 possible outcomes. Now, suppose for this experiment again I am asking two questions. The first question I am asking is I want to know how out of this 3 tosses how many tosses will be heads. In other words, I am counting the number of heads in each of the tosses. The second question I am asking is of the 3 tosses which toss results in a head first?
What do I mean by it? Is my first toss a head? or the second toss ahead or the third toss ahead. For example, in this outcome, my first toss is ahead.
In this outcome, my first toss is ahead. In this outcome, again my first toss is ahead. In this, my first toss is ahead.
In this, my second, the head appears for the first time in the second toss. Here also the head appears for the first time in the second toss. Here it appears for the for first time in the third toss here it does not appear at all.
Now, when I am counting the number of heads, I know I have 3 heads here, I have 2 heads here, I have 2 heads here, I have 1 head here, I have 2 heads here, I have 1 head here, I have 1 head here, I have no head here. With a caution I have not put a number here because here I am not telling it is a 0th toss because I do not know whether what is the 0th toss. So, I just left for now I am calling this nil.
So, you can see that on the same sample space and with the same outcomes we have defined 2 random variables or 2 numerical quantities which we seek to answer. Again observe that the experiment and the sample space are the same. So, now let us define x to be the number of heads that appear and y to be the toss in which a head appears first that is the order of the toss whether it is the first toss or the second toss or the third toss. Again as in the previous example for each outcome let us find out what is the value of x and what is the value of y. So, what do I have here?
What are the outcomes? I have these 8 outcomes. In this outcome, I know the value of head. There are 3 heads.
So, value of x is 3 and the head is appearing in the first toss. So, the value of y is 1. Here, I have 2 heads. Head appears in the first toss.
So, value of y is 1, x is 2. Again, x is 2, head appears in the first toss, 2 and 1. Head appears in the first toss and there is only 1 head. So, both x and y take the value 1. Here head appears in the second toss, y takes the value 2 and there are 2 heads. Here again head appears in the second toss, y value is 2 but there are only 1 head. head appears in the third toss and there is only one head. Here there are no heads so x takes the value 0 whereas here I am writing nil or none because y does not appear head does not appear in any of the tosses.
So, I am just assigning a value nil. Now this value could be a very high value or none but what I want you to see is the y takes a value corresponding to this What is the value I need to give to this nil is something it could be any real value. For now I want you to understand that these are the values that y takes.
So y and x are defined on the same sample space and for each outcome I have a value. So now let us look at the x that is number of heads in the outer. So, what are the values this X is taking? So, again go back here, you can see that X takes the value 0, 1, 2 and 3. So, X takes 4 values and what are the values X is taking?
X is taking the value 0, 1, 2 and 3. So, now let us look at the relevant. events where x takes the value 0. I know x takes the value 0 means that all the 3 tosses result in a tail. So, the relevant event is t, t, t.
x takes the value 1 when 2 tosses. So, 1 of the tosses is a head. So, I have the first toss, second toss, third toss.
So, the head in the first toss or head in the second toss or head in the third. toss. So, the possible ways it can happen are the following.
So, what are the possible event, the relevant event of X taking the value 1 is HTT, THT or TTH. Similarly, X is taking the value 2. So, I can have my first, second and third toss, H can be the first two tosses or the first and the third toss. or the second and the third toss.
So, the relevant event is all these three outcomes put together which is HHT, HTH and THH. Similarly, X takes the value 3 if all the three tosses result in head and that can happen only in one way. So, the relevant event is again HHH. So, what we have done now is I know x which is the number of heads in an outcome that appear is taking the value 0, 1, 2, 3, x takes the value 0 the relevant event is all my 3 tosses are tails and for x equal to 3 the relevant event is all the 3 tosses are heads and for 1 and 2 I have listed what are my relevant events.
So, as in the earlier case, I can find out what is the probability of X equal to 0, X equal to 1, probability of X equal to 2 and probability of X equal to 3, which are nothing but the probability of the relevant events. So, what is probability of X equal to 0? Probability of X equal to 0 is same as probability this even t t t happening and I know that that is equal to 1 by 8 because my sample space has 8 equally likely outcomes. Similarly, probability of X equal to 1 is probability of this happening which is equal to 3 by 8. Again, in the assumption of equally likelihood probability of X equal to 1 is 3 by 8. What is probability of X equal to 2?
Again there are 3 outcomes, probability of X equal to 2 is also 3 by 8 whereas probability of X equal to 3 is just 1 by 8 because there is only 1 outcome that satisfies the condition that X is equal to 3. So, what we have done here to see that from a sample space of tossing a coin or an experiment of tossing a coin 3 times. we list down the sample space, define the random variable to be the number of heads we have got, what are the values this head can take and what is the probability with which X takes those values. Now, for the same sample space, we are interested in knowing which toss results in a head first.
To understand this, let us go back to our earlier table. So, if you look at this table, you can see that y takes the values. What are the y values y takes? y takes the value 1, 2, 3 and nil. These are the values y is taking.
As again now I am not assigning a numerical value here but I could assign a numerical value. I could assign a value 0 but for now I am just writing nil because y equal to 0 would physically I can interpret it as the 0th toss which does not make meaning. So, I am just writing that there is no toss corresponding to this outcome where the first head appears for the first time. So, you can see that y takes these 4 values that is what I can see that y takes these values. And now again let us see when y takes the value 1, when will y take the value 1?
Again y would take the value 1 is same as first toss is head. that is in which toss the head appears first. So again I have my first toss, my second toss, my third toss that is 1, 2, 3. My first toss is a head, then my second toss could be a head or a tail, tail or a head, head or a head or tail or tail.
Is there any other possibility that can happen other than this? So you can see that if my first toss is a head, The relevant events that could happen is head with both the tosses following as a head or head tail head or head tail head or head t t. So, these are the relevant events. So, now let us look at y taking the value 2. So, head y is denoting the toss in which head appears first. So, again I will look at this h h h h h t.
hth, htt, tth, tht, tht, thh and ttt. These are my possible outcomes. So here I have head appearing for the first time in the first 4 tosses. Now in this among these 2 tosses I have head appearing.
first in the second toss for these two. So the outcomes that satisfy that head appearing first for the first time in the second toss is THH and THT. For the third time again we can see for the third time it would be only this outcome which is TTH and that outcome X head appearing for the first time in the third toss is given by TTH.
And then of course in the nil which corresponds to this outcome because head does not appear at all and the way head does not appear. So, head appears first does not happen in this outcome that the head appears first. So, these are the values the y takes, y takes values 1, 2, 3 and nil.
As earlier I can also associate probabilities with the values y takes and what would be these probabilities? Again, You can see that the probability of y taking the value 1 is equivalent to the probability of this event happening. Remember, there were 8 outcomes in my sample space.
I have 4 by 8. Probability of y equal to 2, 2 items, so 2 by 8. y equal to 3 is just 1 by 8. Probability y equal to to nil again we will come back to what is this nil is again 1 by 8. I can represent this nil with any real valued number which makes sense but for now I am just telling that y takes the value nil or if you say it takes a value 0 you should qualify by saying that what you mean by y taking the value 0, y taking the value 0 means that my outcome the head never appears. first in any of the tosses. So, probability y equal to 1 is 4 by 8, y equal to 2 is 2 by 8, y equal to 3 is 1 by 8 and y takes the value nil is 1 by So, what we have seen in the earlier two examples is given a random experiment and sample space, I am associating some numerical quantity to each outcome and I am using this concept of a numerical quantity to answer some things about the experiment.
For example, it could be sum of the dice or the lesser of the two dice. or the number of heads that appear in each outcome or what is I count the number or the order of the outcome will for which the head appears for the first time.