 Today, we will continue discussing CDF's. So, yesterday we said. So, we have a sample space omega a probability space omega f p and we said a random variable is a measurable function on this space, which is just saying that if you take any Borel set on the real line, the pre image of Borel set of that Borel set must be an event. That is what a random variable is. So, we also define. So, we define the probability law of X as. So, p X of B for any Borel set B, p X of B, we defined as probability that X of omega belongs to B. I should write again curly braces here. I will abuse notation and just write this as p of X in B. This is. So, the probability law define gives the probability that the random variable takes values in a specific Borel set B and is defined for all Borel sets. Now, we also said that the Borel sigma algebra is generated by the pi system of semi infinite intervals minus infinity X. So, the probability measure p X is also defined on the generating class of the Borel sigma algebra. So, we said the CDF, we gave it a name CDF. So, f X of X was simply the probability law of sets of the form that and. So, that was simply the probability of omega for which X of omega less than or equal to little x. Again, I will abuse notation by writing this as probability of X less than or equal to little x. So, this actually has this bad notation for sure, but when I write this I mean this and similarly when I write that I mean that. This is just shorthand you like. So, this is the cumulative distribution function. We stated a theorem last class, we did not prove it, but we said that if you specify the probability measure actually this is true for all finite measure spaces. So, if a finite measure is specified on a pi system it gets uniquely specified on the sigma algebra generated by the pi system. And since this minus infinity X is a pi system on R and it generates the Borel sigma algebra. If I specify the probability measure of these semi infinite intervals the probability measure of all Borel sets on R gets uniquely specified. So, that is something we stated without proof. The proof is a bit non-trivial bit technical we just need to know the result that is all. Any questions so far? That was just a recap of what I said yesterday. So, this we will. So, from now on we will just focus more on the CDF, because this probability law although it is the most complete specification it is you have to specify it for all complicated Borel sets. Borel sets can be pretty crazy whereas, f X of X is simply the probabilities of nice sets and specifying this is enough to specify the probability law. So, we will worry about this for the most part. So, we will now give four properties of cumulative distribution functions which hold regardless of what random variable you are talking about. Any cumulative distribution function satisfies certain properties which I will list now. So, as usual so omega f p is a probability space and x from omega to R is some random variable. So, you first property and f X is the CDF of that random variable. First property limit X going to minus infinity f X of X is equal to 0. Second property limit X going to infinity plus infinity f X of X is equal to 1. Property 3 property 3 is monotonicity if X is less than or equal to y, f X of X is less than or equal to f X of y. Fourth property limit f X is right continuous i e for all X limit epsilon going down to 0 of f X of X plus epsilon is equal to f X of X. So, these are four properties that are satisfied by all CDF's regardless of what random variable you are talking about. So, the first says that when X is going to minus infinity f x is going to 0 and when X is going to plus infinity it goes to 1 and everywhere in between it is non decreasing. So, it is something that starts off at 0 at minus infinity and then it ends up at 1 at infinity and in between it has to only increase it can be flat or it has to increase it cannot decrease. It does not have to it does not have to increase in a continuous way, but it is right continuous. Have you heard of right continuity? Well if you have not heard of it you just heard of it this is what right continuity means. So, continuity means that see continuity means that there are no jumps right you can give a more rigorous definition, but it just means that the function does not jump right. So, but CDF's can jump they do not have to be continuous functions of X CDF's can jump, but what they cannot do is if you approach from the right the limit and functional value have to be equal. So, what this notation see I remember I here I did not write epsilon arrow like that I did not write a right arrow right if I do that I just mean epsilon is going to 0 whichever way when I do that epsilon down arrow 0 I mean that epsilon is going to 0 from the positive side. So, some people write this some people write instead of that some people write epsilon going to 0 plus which means it is going to 0 from the positive side it is not going to 0 from the negative side or jumping around or any such thing right it is going to 0 from the positive side. So, what this is saying is that if you have a CDF right. So, heuristically speaking you have if you are trying to plot a CDF FX of X. So, it has to at minus infinity it starts of at 0 right and remember it is a probability right. So, it always has to be between 0 and 1 right and at infinity it ends up at 1 right approaches 1 and everywhere in between it has to increase right, but potentially with jump. So, it could do something like it could increase continuously at some in some place for some places it could jump in some places there is no problem with that right you could do something you could stay constant for some time no problem it cannot go down right you can again jump and then go to 1 for example right I mean this is just one possibility does not have to necessarily look like this, but what this what this right continuity property is saying is that when you have a jump in the function. So, you have a jump in these places right whenever it jumps this is hollow and that is solid what that means is that the left limit is not it may not need not be the functional value, but the right limit is always equal to the functional value right. So, when I approach like that right that is what means right. So, if this is my X FX of X plus epsilon is approaching the functional value. So, in all these places I have drawn it continuously it can even go like that right it does not have to be flat. So, what I mean is that here it is flat, but it can also have for example if this is a jump it can do something like that right. So, what I mean here again here that is hollow and that is solid meaning that when I when I when I approach X like that from the right the limit is equal to the functional value, but I approach from the left it need not be equal to the functional value right. So, CDFs need not be left continuous, but they are necessarily right continuous. So, if it is both left continuous and right continuous it is continuous right, but it need not be continuous. So, is this I mean so this is just some canonical picture of how it might look it can look more complicated also you will see some fairly bizarre CDFs, but this is just some representation of what you can expect from a typical CDF. So, we have to prove all of this, but before that are there any questions on right continuity monotonicity is very trivial right. So, when I so it is usual convention in plotting functions that. So, when there is a jump in the function you have to specify where the functional value is right. So, the functional value could be here or here or it could be somewhere else in general. See what CDF it cannot be somewhere else right cannot go above for example right, but wherever you put a solid dot is where the functional value is at a jump that is just standard plotting convention for functions. And if you put a hollow dot it means that that is not the functional value right I mean this means that the function if you approach the if you approach x like that the limit of the left limit is not the functional value when there is a solid hollow dot over there any other questions. So, CDFs are always right continuous. So, we have to prove all of this right. So, what shall we prove first? So, this one so you are looking at so if you are looking at have to be careful now. So, if you are looking at the probability limit x tending to minus infinity f x of x that is equal to limit x tends to minus infinity probability of omega such that x of omega less than or equal to x is not it. So, now I have to argue that so can I not so I have to use some kind of continuity of probabilities right, because I have to send the limit inside right correct. So, in particular you can take let us say any sequence let x n be real sequence such that x n goes to minus infinity as n goes to infinity. So, I am just going to make this into some kind of a countable limit. So, I can write that as so this guy I can write as this is equal to limit n going to infinity probability of omega such that x of omega less than or equal to little x n. So, I am still maintaining the good notation, but at some point you throw it away I will just write x x less than or equal to little x. So, now what do I have to do right. So, I am looking at so if you look at this picture again. So, you looking at some x n right such that x n is let us say x n is you can actually take it as sequence not only going to minus infinity, but going monotonically to minus infinity. So, let x n be any real any sequence such that x n tends to minus infinity monotonically. So, x n is something that is monotonically decreasing to minus infinity in whichever way you like any sequence that is going to minus infinity. So, then what happens is that the pre image of minus infinity x n will be a bigger set than the pre image of minus infinity x n plus 1 because x n plus 1 will be somewhere here. So, the pre images will be Russian dolls correct they will be nested decreasing sets. So, maybe I can write this. So, I can probably write this is this right. So, limit have to take the limit inside is not it. So, I have the following omega such that x of omega is less than or equal to x n and if I look at omega such that x of omega less than or equal to x n plus 1. Let us say I look at these two sets this is true right this this containment is true. So, if I do that this containment is true right. So, if I call these sets like so, if this I call a n and that call a n plus 1 these are nested decreasing sets correct. So, I am trying to invert continuity of probabilities. So, what I know I do know is that probability of. So, limit n tending to infinity probability of a n is equal to. So, these are nested decreasing sets right what is the continuity probability result for that probability of. So, limit of that is equal to probability of intersection n equal to 1 to infinity i equal to 1 to infinity a i is that correct correct no. So, what is that set. So, I what I have here is the probability limit n tending to infinity probability of a n right correct. So, this is nothing but a n in my notation and that has to be equal to the probability of this countable intersection and what is the what is this countable intersection now tell me it is null set correct. So, this countable intersection will be the null set right because x n is going to minus infinity right the set of all omegas that map to something that is arbitrarily going to minus infinity right that is the null set because everything maps to the real line right. So, beyond some x n your essentially this intersection you have to argue is null. So, you can show that this is in fact 0 because this guy is null I am using continuity of probability. So, property number 2 is also very similar except now you will have you are going to infinity. So, you will want to send a sequence x n which is going to plus infinity and you will realize that instead of null it will become. So, you will have unions instead of intersections right and then you will have this countable union that case will be the sample space itself right after all the pre image of the entire real line is the sample space. So, you have to use to you have to show this properly using continuity that is the only thing I want to emphasize you can know that you know this you can do the second property also is it the other way round is it not true that I am saying that x n see this x n is a sequence that is monotonically going to minus infinity right. So, x n plus 1 is less than or equal to x n correct. So, the pre image of that set has to be contained in the pre image of that set no see you look at all those omegas that map here right that is some f measurable set there, but this does omegas that map here must be a subset of that right. So, that nested containment works you have nested containments you use continuity of probabilities. So, you see how crucial it is right it is it keeps coming up all the time continuity. And so property number 3. So, property number 3 is absolutely trivial actually property number 3 is. So, you have the following right. So, x is less than or equal to y right. So, you have that the set of omegas for which x of omega is less than or equal to x, but if you just argued right is which way is the containment this way is not it omega for which x of omega less than or equal to y correct again right. So, you x is less than or equal to y. So, you call that x you call that y I just argued that this containment has to hold right any questions. So, now what happens these are all these are both f measurable sets. So, they have probabilities and you know that if a is a subset of b probability of a less than or equal to probability of b right. Therefore, you can put. So, you put p on both sides you have done right. So, if you put p on both sides I should not be doing this, but I should write it again, but this is what happens right correct. Because a is contain in b probability of a less than or equal to probability of b then you get that inequality which means. So, this is nothing, but f x of x which is less than or equal to f x of y for x less than or equal to y correct any questions see all along see these you know these properties already, but what I am emphasizing here is proving them from probabilities of probability measure and continuity and so on right. So, can we prove this continuity of not continuity this is right continuity of c d f. So, right continuity let epsilon in p any sequence. So, epsilon this is a non negative sequence b any sequence such that epsilon in down arrow 0 as n tends to infinity. So, I am epsilon is going to down to 0 right. So, if I show that this is true for every sequence epsilon in which is going down to 0 then I am done right that is what that limit means after all right. So, I consider a sequence like that then I say then consider the following limit n tends to infinity f x of x plus epsilon in right. I have to show that this is equal to f x of x for every sequence epsilon in going down to 0 if this is true then this is true right. So, why am I doing this I have to yeah this is something I did not discuss earlier. So, this epsilon is some continuous thing right this epsilon is some real parameter that is going down to 0, but why am I replacing it with a discrete sequence the sequence epsilon in right. So, the epsilon in this epsilon itself is going down to 0 on the real line, but I am now considering a sequence indexed by n right why am I doing that because I will invoke continuity of probability. I want to somewhere see I know how to handle countable intersection and countable unions right I cannot handle uncountable intersections and unions right. So, I am just that is what I am doing right. So, I am going from a uncountable this epsilon to a sequence. So, that I can get these countable intersections and unions very standard trick in measure theory in probability theory. So, that is simply and I have to prove that for every such sequence only then will that be true right. So, limit n tending to infinity probability of x less than or equal to x plus epsilon in. So, from now on I will use this notation. So, no omega such that x of omega now again I have to be careful. So, this is like these are nested decreasing sets as subsets of omega right and so that will be equal to. So, I am going to skip a step here. So, this will be equal to. So, how do I write this. So, this be omega such that intersection n equals i equals let us say 1 to infinity sorry my bad. So, intersection of sorry i equals 1 through infinity omega such that x of omega less than or equal to x plus epsilon n. So, the intersection should be outside the set obviously right that is the mistake I made agreed. So, I am just taking this is of course, if you want to write it down this should be omega such that x of omega less than or equal to epsilon n you have nested decreasing sets and the limit is outside. So, bring it in by continuity of probability right then I get that epsilon i correct everybody with me. So, I am just saying that this. So, this set. So, if you I mean you have to bring this limit in limit in right for the continuity of probability and that is what you get this is by the nested decreasing property and continuity of probability measures. Now, what is that set equal to the whole intersection remember epsilon i is going to 0 right. So, that whole set will be equal to the omegas for which x of omega less than or equal to x right. So, that will in fact be equal to f x of little x and this is true no matter how I choose epsilon n. As long as it is some monotonically decreasing sequence to 0 this holds right. Therefore, this limit holds. . See for the left limit this argument will not go through because you will be looking at. So, you will be approaching from the right is not it. So, you will have. So, you are looking at for the left limit you are looking at limit. So, you are looking at something like this f x of x minus epsilon n as epsilon n is going from positive or positive right. So, epsilon n is still positive. So, I can do that right, but that will be limit n tending to infinity probability that x of omega is less than or equal to x minus epsilon n is not it right I am making any mistake. Now, this will be what kind of sets of these are they decreasing or increasing sets are not they. So, I should take. So, this will be the probability of the union of i equals 1 through infinity omega such that x of omega is less than or equal to x minus epsilon i is not it agreed. Now, what is that equal to that set is equal to x less than x which is not your CDF right. So, this will simply be this will be equal to probability that capital S is less than little x not less than or equal to little x. So, the left limit is not the CDF right and it will be equal to the probability x less than or equal to x if and only if probability of x is 0. So, you already proved a problem that was going to be homework problem. So, if you have if you have a CDF that is continuous at x that means the probability assigned to the singleton real number x is has to be 0 if and only if. So, you see what I mean after all this minus that is the probability that x equals x right and you will have continuity of the CDF at x if and only if the probability of the singleton x assigned by the probability law p x is 0 right. So, that is a good question. So, we dissolve the homework already for you right. So, left continuity need not hold and in particular left continuity holds if right left continuity holds the probability of that point is 0 and if the probability of that point is something positive. You will definitely have a discontinuity right you have right continuity and there will be a discontinuity and the jump in the discontinuity will be equal to the probability that x is equal to x. That is actually what we have shown are there any other questions. So, any CDF of any random variable CDF of any random variable has to satisfy these 4 properties. So, what is interesting what is somewhat more remarkable is that there are no other further properties that have to be satisfied by a CDF. Meaning that if I give you any function at all which maps which a function that maps are to 0 to 1 right which satisfies these properties. Namely it is start of at 0 ends up at 1 monotonically non decreasing in between and right continuous. There exist some random variable whose CDF is that function you give me any function that I do not know that it is a CDF of any function any random variable. You give me a function that satisfies all these 4 properties then it is necessarily the CDF of a random variable that also can be shown. So, which means that this is all you need for a CDF there are no further demands that you need to make for it to be a valid CDF. Even if one of these does not hold it will not be a CDF there any other questions have few more minutes. So, I will now introduce very useful random variable called an indicator random variable. So, as usual you have omega f p. So, let a be an event define I a of omega is equal to 1 if omega is in a 0 if omega is not in a. So, I am just looking at. So, it is called indicator random variable because it indicates 1 when your realization is in a it is in the it says 0 otherwise. So, what I have pictorially I have my sample space with some f p on it and I am looking at some other event a and of course, my mother nature picks omega. So, it could fall here or it could fall here right if it falls here I say 1 for I a if it falls here I say 0 it only takes 2 values depending on where omega falls. Now, this is called the indicator random variable of the event a. Now, before I call it the indicator random variable have to prove something I have to prove that it is a random variable right. So, it is see certainly a function from omega to r in fact it is a function from omega it only takes 2 values on the real line. So, it is a function from omega to r, but I have to prove that it is a measurable function f measurable function from omega to r right. So, how do I do that? So, here is my real line right and so, I have 0 and 1. So, these are the only values that the function takes right. So, if you give me any Borel set you give me. So, now what do I have to show you give me any Borel set the pre-image is something f measurable right. So, you give me a Borel set let us say you give me a Borel set like that what is the pre-image of that here null set right, but if you give me a. So, let us see if you give me a Borel set that includes 1 the pre-image will be a which is f measurable right. Similarly, if you give me a Borel set that includes 0 it is pre-image will be a complement which is also f measurable because a is f measurable. Then finally, if I give you so if I give you some Borel set that includes both 0 and 1 the pre-image will be the entire sample space right which is also f measurable. So, you can just proven that the indicator function is in fact a random variable the indicator of a f measurable set is an f measurable function correct. So, this is a legitimate random variable. So, if you want to construct a function from omega to r which is not a random variable you can also do that right what you will do is you will start with some set a which is not f measurable right and then the indicator function will not be f measurable right. So, the for a which is not an event the indicator of that event will indicator of that set will be a function which is not a random variable right. So, the very simple example so what will the CDF of this we have established that this is a random variable the CDF of this will look like what at 0 it is. So, if I try to plot so I am going to call this f I a of x against x right if x is less than or equal to 0 the function will be at 0 right. So, we at 0 and between so if this is 0 and this is 1 so that is 1 right I beyond 1 it has to be 1 right that is also clear and then in between it will be the probability of a or a compliment yes correct here the height of this. So, here this is the probability that x is less than or equal to 0.5 say which is means probability x is equal to 0. So, this height will be the probability of a compliment and this jump will be the probability of a right. So, in all this the functional value will be over there right and here it should be hollow similarly here it should be solid and hollow over there. So, that is how the CDF looks yeah that is at 1. So, this has 2 discontinuities right. So, looking at this itself you can say that the probability of 0 is as I argued there right probability all these points have probability 0 and 0 has probability P a compliment and this jump is equal to probability of a any questions. So, tomorrow there is no class the next class is on Friday we will discuss some important topics we will discuss about sigma sigma algebras generated by a random variable and also we will classify random variables based on the measure they induce on the real line. But we saw that P x is the measure induced on the real line by the random variable x depending on the properties of the measure P x random variables are classified as well under several heads you already know about a couple of heads I guess, but there are more heads right. So, there are several types of random variables depending on the kind of measure they induced on the real line. So, that those topics will be left for Friday I will stop now.