 Dear student, here we are going to learn the properties of the Vishat distribution. Now, we have the property number 1. Let A is Vishat with parameter n and sigma and L. What is L? L basically we have done transformation on it and L be a P into 1 fixed vector. Then L prime A L. L prime A L is distributed as sigma square chi square with n degree of freedom. So, sigma square L which is equals to this one. What we have done? We are saying that we have to take transformation on Vishat. We have taken transformation L and L is equal to P into 1 fixed vector. So, we will convert Vishat into Univariate. Okay, let us prove it. By definition of the Vishat distribution is defined as here A equals to this one. Now, we have taken Z. We have Z as standard normal distribution with mean vector 0 and variance covariance matrix, sigma. So, chi square chi square which is equals to this. Now, L prime A L. Sum as it is L prime L prime A. We have Z i Z i prime into L. So, L prime A L sum as it is. This is the L prime Z. And further, if we take its prime, we have total L prime Z i square cake. Since Z i is a P into 1 vector and L prime is a 1 into P vector. L cake is equal to P into 1. So, L prime 1 into P vector. So, the order checker them the order of L prime Z i L which is equals to 1 into P. Okay, now L prime which is equals to 1 into P and Z i order of Z i which is equal to P into 1. Now then, we have last care 1 into 1. So, 1 into 1 which is equals to scalar quantity. It means that we have this multivariate convert. Univariate because it is about scalar quantity. Now, L prime A L which is equals to this. Sum of square of normal variate. Sum of square of normal variate. So, basically you have this sum of square. So, sum of square of normal variate. Z i tends to this. What do you have in univariate? In univariate, we say sum of Z i square divided by sigma square. Which is equal to sum of Z i square which is equals to sigma square which is equals to the kai square. So, Z i which is equals to this. L prime Z i equals to this. And we know this is the scalar quantity equals to sigma square L. So, what we have done further? In L prime A L, we have multiplied L prime sigma L and divided. Now, by multiplying, whose term is equal to this? Whose term is this? Sum of Z i square over sigma square. This is the sigma square. We have said this is the L prime sigma L. This is the scalar quantity and which is equal to sigma square of L. So, this is the kai square with n degree of freedom. And this is the term as it is. So, L prime A L, L prime A L which is equals to L prime sigma L into kai square. So, L prime Z i is normal with mean 0 and variance sigma square. So, what have you done? In this property, we have converted the vishat into univariate. So, what is equal to the univariate? Sigma square which is equals to this. Now, this is the property number 2. In property number 2, what have we done? What is the property basically? If A, A is a vishat with n sigma and m. Now, m has come. Because m has come, we have come to know this is the non-central distribution. So, what is the vishat we have? This is the non-central. Then, expected value of A which is equals to this. We have to show this. Where m, m is the mean of the vector. So, we have to show this property. If A is a vishat with these parameters, it is defined as this. What is the vishat we have? Some i varies 1 to n xi, xi prime. And, who is following x? With normal mean mu i and variance covariance matrix. What do we have? Then, expectation of both sides. We have to take its expectation. Taking expectation on both sides. Expected value is applied. Expected value of inside the bracket. Expected value of xi, xi prime. Now, by definition which is equals to this, Now, where is this coming from? We know that the covariance of x, y, in the univariate case, you know that the covariance of x, y, which is equal to expected value of x, y, minus expected value of x, into expected value of y. And, expected value of x and expected value of y, which is equals to mean mu, you have expected value of x, y. So, expected value of x, y. What is here? Expected value of xi, xi prime. We have called covariance of x, y as sigma. Basically, whose value do we want? Expected value of xi, xi prime. So, here is the value of expected value of xi, xi prime. We have the covariance of xi, we have expected value of xi, xi prime. This one. Now, where do you want to enter this value? Here. Now, the sum, sum as it is, we have determined its value. We have entered that value here. Now, multiply this sum inside the bracket. This is the constant term. The constant, if you have this sum, if you multiply it with a constant, you know that which is equals to n times of this constant, plus sum of mu i, mu i prime. What do you have? Sum of mu i, mu i prime, which is equals to mean of the vectors. So, which is equals to m prime, m. So, we have proved the second property. This is equal to this. And, whose property is this? This is the central Vishat property. Now, here is the property number three. If a1, a1 is a Vishat with n1 sigma m1. Now, from m, you have come to know that this is the non-central Vishat. Now, we have partitioned a total a in two parts. Now, in two parts, we have first kantak and one kantak. And second part, we have n2. Are distributed independently. Then, a1 plus a2, that is, when we have summed a1 and a2 which is also equals to the Vishat with n1 plus n2 sigma m1 m2 which is equals to m. Where m prime which is equals to this. That is, who do you have m? m which is equals to m1 prime into m2 prime. You have partitioned, right? You have half part here, n1 terms here and n2 terms here. So, m1 prime. Now, m1 prime which is equals to mean of n1 we have taken. And next, after n1, what will be next term? n1 plus 1 n1 plus 2 up to 0 on kantak goes to n2. So, this partition is done in two parts. This is the whole property. That means, if we split the Vishat in two parts, then whose addition will be equal which is also equals to the Vishat. So, let's start the proof. This is the a1 is Vishat with n1 sigma m1. We know a1 is equal to some i varies from 1 to n. Now, how far will it be? Some i varies from 1 to n1. x i, x i prime. Now, where is x i going? Which is normal with mean vector mu various covariance matrix sigma and i's variation kantak which is equals to n1. That means, 1, 2 up to 0 on n1. Also, a2. Similarly, a2 which we have is also Vishat. Then, Vishat n2 sigma m2. Now, what should we do here? This is the a2. We have also written a2 as the value of n1 plus 1 and 2. x i prime of this. x i is following you with normal with mean mu i and i varies from 1 to n1 plus 1 to up to 0 on n2. Now, what will we follow with a1 plus a2? a1 plus n a2 is equal to a1. Where is a1? Second is a2, a2. Since x1, x2 up to 0 on xn then n1 plus 1n 1 plus 2 up to 0 on n2. So, whose portion is this? A2 and this portion is a1. Are distributed independently and depend upon some covariance matrix. When there is a lot of covariance matrix no one will be there. Now, we have that covariance matrix of x. Here is the a1 plus a2. This is the addition. Now, how did we write this sum? Look, you have a i varies from 1 to n1. Next, where is n? n1 plus 1 to n2. So, we will write this sum. i varies from 1 to n1 plus n2. When we open it, it becomes this form again. Now, a1 plus a2 this is follow the with n1 and 2 and variance and the mean of that is the same. That is, it depends on if we split the total vishat in two parts the remaining part will also follow the vishat distribution. And this is the mean vector.