 So, we already know what is a Gaussian distribution and what is a Gaussian random variable and now we want to say when we have a random vector, when we are going to call it a Gaussian random vector, what is the properties it needs to satisfy. So, we already have for a single random variable the PDF is given by, so this means I have a Gaussian random variable x that has mean mu and variance sigma square. And what was its characteristic function at a function u, how did it's characteristic function look like, exp minus u mu, I do not know where was minus, j u mu plus sigma square by 2. This happened to be a Gaussian random it's a characteristic function. Now, suppose I have a set of such random variables which are independent, but not necessarily identical. So, what I mean by independent, we already know what I mean by independent of set of random variables, right? We know that if they are joined, their distribution, the joint distribution splits into their marginal distribution, then we are going to call them independent. And I am going to say set of random variables to be identical if all of them have the same distribution, okay? So, now suppose I have a collection of random variables and all of them have and they are independent, then it so happens that you take any linear combination of this random variables, it is still going to be again Gaussian, maybe with a new mean value and a variance, okay? So, this we are going to write it as a lemma. So, take any m random variables that are independent and all of them are Gaussian random variables. So, each may have its own mean and variance different values. Now, we are going to say if you take the linear combination like this, you understand linear combination, right? I will just take different weights, multiply each one of them with their associated weights and then add up. Then I got this linear combination, we are going, we are saying that this is also Gaussian random variable. So, why this is true? So, to see this, we are going to use the property that if a set of random variables are independent, then if you are going to look at the characteristics function of their joint distribution, that also splits into characteristic function of individual random variables, right? So, that we have said. So, we have said that a set of random variables are independent, if and only if their characteristic function also splits, not only their CDF splits, that also implies that their characteristic function also splits. So, let us use that property just to quickly verify this. Suppose, no, I am not saying that, I am just saying that they are independent. Yeah, question, but Gaussian, they could be with different mean and variance, right? They, for a given mean and variance, it is going to define different Gaussian distributions. It is a parameter. So, now let us say I want to define characteristic function of my random variable X. Here, X is now the joint collection of these random variables. So, X is a vector here now. It is not simply a scalar. So, now, okay, no, what I mean is, okay, now I am going to define this X not, I am going to define this X to be this linear combination. Okay, X is now again another random variable for me. It is not a vector as I said earlier. It is just a linear combination of this and I want to compute the Gaussian distributions, sorry, the characteristic function of this random variable X. Now, what is this going to be? It is going to be expected value of e to the power j u and all this, right? a1 x1 all the way up to, so this is nothing but expectation of e to the power j u a1 x1 into e to the power j u a2 x2 all the way up to e to the power j u a n xn. Now, that x1 x2, they are independent. Do you think a1 x1, a2 x2 and a n xn, they are also independent? If x1 and x2 are independent, suppose x1 are multiplied by a1 and x2 multiplied by a2, will a1 x1 and a2 x2 will be also independent? So, these random variables are independent. Now, if they are independent, now can I write this expectation as the product of expectation of each of these terms? If they are independent, I should be able to do this. Now, x1 is Gaussian, I know. Is a1 x1 is also Gaussian? a is a constant, right? So, what is this quantity now? This is a characteristic function of what? a1 x1 and what it will be? So, we already know that for a, if x is a Gaussian random variable, its characteristic function is going to just look like this, j u mu. Now, so it only depends on your mean value and the variance, right? So, what is the mean value of a1 x1? Suppose, let us say xi, expectation of xi is equals to mu i, for i random variable, which is Gaussian has mu i. So, then what is this mu of this going to be? And it will be, what will be the variance of a1 x1? So, if variance of let us say x square, I am going to denote it as sigma square, what will be the variance of axi? So, then what is then u square and then I am going to replace sigma square here by 2, right? Or you can alternatively think of this as like instead of this as a random variable, you can think as this is u1, u a1 is the point at which you want to evaluate this characteristic function and for this random variable x. So, in that case also like it is like same like you are replace, you are going to replace u by u a1 here, u by u a and u square by u square a square. Or alternatively you can also think of a1 x1 as another random variable with new mean and variance. Now, all the way to what is its value going to be? So, now if I further simplify this, what this going to look like? It is going to look like j u summation a1 a i mu i minus u square summation a i square sigma i square divided by 2, right? Now, if you are now go back and we have already said that every distribution has a unique characteristic function, right? Now, you should just look at this characteristic function, what is this distribution corresponds to? What is this characteristic function corresponds to which distribution? What are parameters? Just suppose let us go, if I want to map it to this, suppose let us say this I have to if this is some mu and this is some sigma square, okay? Anyway, this is constant, right? And it will be exactly in this format. So, then this is going to be a Gaussian width. So, this corresponds to a Gaussian with what mean? And variance. So, this is going to be another. So, this means my x here which is a linear combination of independent Gaussian distributions is going to be another Gaussian distribution with mean like this and variance like this. So, what it says is that what we have just showed is any linear combination of independent Gaussian random variables is going to be another Gaussian random variable. So, now we are going to use couple of more definitions. So, what I mean by here is let us take a collection of random variables. You understand this notation x i, i belong to capital I. So, i is the index set here. This index set is going to have finitely many values. They are said to be jointly Gaussian if every linear combination is Gaussian. And they are then if that and they say to have a joint Gaussian distribution. So, what we just what we mean here is you are given a set of Gaussian distribution, sorry set of simple random variable. If you are going to take any linear combination of them and they should be Gaussian, if that is the case then you are going to call this set of random variable to be Gaussian. They are going to you are going to call them as jointly Gaussian and they are said to be have Gaussian distribution. So, we have already said that if x 1, x this set of x i's happens to be each one of to be Gaussian and they are independent. We know this is already two, right. You take any combination of them, they are again going to be Gaussian. So, here these are this set of random if they are independent and Gaussian, they are again already jointly Gaussian. But let us say if some given set of random variables happens to satisfy this property then we are going to call them jointly Gaussian. Then and now when we are given a Gaussian vector, so this vector will have components, right. And then we just treat those components as these components and if they are coordinates. So, when I say x, x is a random vector it will have x 1, x 2 let us say up to x m then these I can treat it as coordinates of this vector as we discussed in the previous class. We can treat them as these components and if these components happens to be jointly Gaussian then we are going to call that random vector as simply Gaussian random vector. So, this was obviously, so here we just said collection of random variables but I can treat all this collection of random variables as a vector, right, in which this vector constitute the components. Now, fine, when I have a collection of random variables how I am going to denote its distribution. So, if I have a random variable x that is Gaussian random vector. So, I am now starts using instead of random variable I am going to use rv for random vector here. It should be clear whether I am just talking about a random vector or I am talking about a random, sorry I am just talking about a random variable or random vector. I am going to denote its distribution by mu and k here and you will see that it just like it only depends on these parameters and here mu is the mean vector and k is the covariance matrix. What is mu here? It is like mu 1, mu 2 all the way up to mu n. So, let us say I have components, right, x is a random vector and it has some n components let us say then this mean vector is nothing but the vector of individual components and what is k here in the covariance matrix it is going to be a matrix of covariance of x 1, x 2, x 1 with itself covariance of x 1, x 2 to covariance of x 1, x n and similarly x 2, x 1 and you can write all the way up to covariance of x 1, x 1. So, what is this here in this covariance matrix? What is this diagonal implies? So, the diagonal contains the variance of each of the components. Now, I will just list down some of the properties of this Gaussian random variables and we are not going to prove any of them, you should verify all these things yourself. Yes, or Gaussian random vectors or like for which we are going to say it is going to be it has joint Gaussian distributions. Now, let us say if x i, I have this collection of random variable has a joint distribution then each of the random variables itself are Gaussian random variables. Is this true? Why is that? So, I have a collection of random variable and saying that if it is jointly it has joint Gaussian distribution then each of the random variables is itself is a Gaussian random variable. Why is that? So, if it is a jointly Gaussian distribution by definition we want for every linear combination. So, in this every linear combination what I could do is like I have to choose a 1, a 2, all the way up to a n these are the weights. I can just choose a 1 to be some nonzero value and set all the others a 2, a 3 all the way up to a n to 0. In this case I have I need to check I need to satisfy that a 1, x 1 is Gaussian, but since a is essentially a constant it is the better that x 1 is Gaussian in this case. So, and similarly for each of the components. So, it must be the case that if the set of random variables are has joint Gaussian distribution each one of them has each one of them itself is a Gaussian random variable. And this property we have already verified I will just write it for the sake of completeness. If right now I am not saying anything like you say you are given a set of random variables and if it satisfies if they are jointly Gaussian distribution for that our lead is every linear combination should be Gaussian. That is the only thing I am not I do not want I am asking anything like they are independent identically distributed on anything just applying definition. So, suppose this xi's are each Gaussian and independent then xi has a joint Gaussian distribution. This we have already shown right the second point what we are saying is if this xi's collection of random variables are such that each are each Gaussian and independent independent. If we show this or not if each collection of random variables they are Gaussian and they are independent and we have already shown that you take any linear combination of them that is again going to be Gaussian and then by definition it has a joint Gaussian distribution. The third property says that say xi joint Gaussian distribution and you are going to construct and let this yj be another set of random variables where j belongs to flat j. So, it is the statement clear here what I am saying is take a collection of random variables that has joint Gaussian distribution. Now what you do is you come up with the another set of random variables call them yj, j taking again value in some set capital J where each of this yj itself is a linear combination of this xi's. Let us say you have x1, x2 all the way up to x10 you have 10 random variable that is jointly Gaussian distributed. Now let us say you make you take one linear combination of this random variables you can take another linear combination like let us say you come up with 100 linear combination of this x1 all the way up to x2. Now this 100 random variables which I have denoted yj. Now this yj's the claim is again has a joint it has a joint Gaussian distribution. Does it make sense? This should be obvious because if this yj's are already linear combination of xi's and now you further take any linear combination of this yj's itself they will be another linear combinations of xi's and that is Gaussian distributed. So any linear combination of yj then has to be Gaussian distributed that is why this property already holds and we are going to call it as jointly Gaussian distributed. So now the pdf random vector n mu k. So let us say I have already said that let us say x is a random vector and its pdf I am going to denote it as it is a let us say I have denoted it by this value right n mu k where mu is the mean vector and k is the covariance vector. So what how this value is going to be like? So this is just denoted a Gaussian random variable vector with parameters mu and k. Now the cdf of this random variable is Gaussian random variable Gaussian random vector is going to look like. So I am assuming that this random vector x here as k m components in this. Just say that x is x1, x2 all the way to xn. So we said that this is a Gaussian random vector where the mean vector is mu and the covariance vector is covariance matrix is k. Now if that is the case then it is cdf one can write it cdf as this. What is here? Mod of k here means it is going to be determined of this matrix k and what I mean by capital T here it is going to be transpose because this x here is a vector and mu also vector. What is this vector? This vector is the mean of all the components and k inverse I mean inverse of the matrix. Now suppose let us see we recover our initial Gaussian distribution for a single random variable with this formula. Suppose x consists of only one component. Now what this formula will reduce to? So here in this case m is going to be 1. This is going to be 2 pi and in that case what is k going to be? So k is going to be a matrix with only one element in this and what is that? That is covariance of x1 with itself. So that is going to be variance and that is also square root. So this is going to be this and now for a single random variable this is like a scalar x is a scalar and x minus mu is now just a mean of that random variable and what is k inverse in that case? Sigma square or 1 by sigma square? Because this is k inverse that is going to be 1 by sigma square and this is going to be x minus mu and we have 2 and this is exactly x minus mu whole square divided by 2 sigma square that we have for a single random variable. So please do take a look into the proof of this that is given in the book. So this proofs comes from the eigenvalue decomposition of your vectors, eigenvalue decomposition of your covariance matrix. From that one is going to derive, do take a look into this. Now this one last thing I want to say if x is n mu k random vector where k is diagonal. So what we are saying is suppose you take a Gaussian random vector which has this parameter mu and k, but this k is special here. It is such that it is off diagonal elements are all 0. Only diagonal elements you allow it to be some values but off diagonals are all 0 that. So off diagonals all 0 means what here? No in terms of covariance. When we say covariance of x i and x j is equals to 0 we say they are uncorrelated. So all pairs of random variables are uncorrelated. If that happens then these random variables are actually independent. So remember in the last class we said independence implies uncorrelatedness but uncorrelatedness does not imply independence. But it so happens that for a Gaussian random variable uncorrelated also implies independence. And this is provided for this is Gaussian random vector. This need not be true for any random vector but provided if it is a Gaussian random vector then that is the case. So you see that like if I already have a model where my random variables are jointly Gaussian distributed and they are uncorrelated then they already implies that they are independent. So this is like a much nicer property help because just by uncorrelatedness I directly get this properties of independence. The last one now how to compute the characteristic function? The characteristic function as we said it has some nice properties. That is it is going to be unique for a given distribution and vice versa. So here for the characteristic functions of, so here the characteristic function of a n mu k Gaussian random variable that is denoted as phi of x u. Here x is a vector and u is also vector because I am talking about random vector here. So that is going to be defined as expectation of e to the power mu transpose x. And then that turns out to be simply e to the power j mu transpose mu. So mu transpose, so what do we are writing it as? Mu is for us, for us the way we are treating it is all column vectors. Mu transpose is going to be a row vector and this mu is a vector of means which is column vector. So u transpose mu is going to be what? That is just one real number. And this is again and what this quantity is going to be? So u transpose k u, k is a matrix, u is a column vector. So k u is going to be one column vector and then u transpose that is going to give just one real number. So this is, you see that it has very much similar to what we get the characteristic of a Gaussian random variable. But in the Gaussian random vector it is just like you have to deal with the vectors. So that is why this transpose and the matrix has come there. So this is about this Gaussian distributions as you see that the Gaussian distributions has some nice properties in terms of this characteristic functions and also if there are uncorrelated it directly implies independence and this helps in many things where when your model satisfies this your analysis become pretty much tractable because uncorrelatedness is directly guaranteeing your independence. When you have independence all you need to worry about is the distribution of each of the random variables. If you have that then joint distributions can be easily computed by just taking the product of this individual random variables. So I do not need to really define the joint distributions there. All I have to worry about is the distribution of each of the component random variables in that case.