 In this topic, we are discussing the mean vector of multivariate normal distribution. In previous lecture, we have developed a PDF of multivariate normal distribution. And in this, we will find the mean of the multivariate normal distribution. Now, we have the PDF of multivariate normal distribution, which is equals to this. PDF of multivariate distribution, we have developed, let sigma inverse equals to A. Sigma inverse, which is equals to A. And what we have here is, this is the positive definite symmetric matrix, it's property. There exists a non-singular matrix C, such that C prime A is T, which is equals to identity matrix. And modulus, if we take it, which is equals to 1. Now, let this portion, let X minus mu, which is equals to CZ. What we are doing is, basically, we transformed it. X minus mu, which is equals to CZ. And the value of X here, we have found, which is equals to CZ plus mu. After the Jacobian of transformation, two or more than two variables, when we take its derivative, we have to use Jacobian. Now, Jacobian, which is equals to mod of curly X over curly Z. Because the second variable is Z. Now, mod curly X, X's value enter key, then curly Z, Z, curly Z. With respect to Z, we have differentiated it. When with respect to Z, you have differentiated it, then what value we have here? Mod of C determinant. Since the modulus of C prime A, which is equals to 1, we have separated its determinant. So, determinant separated, which is equals to 1, C prime C, C square, A as it is, equals to 1. We will take A to the side of equality. So, this is equal to minus 1. So, we have taken its square root. When we are taking its square root, then mod of C, which is equals to minus 1 by 2. The density function of Z. Now, whose density function is Z? Why Z? Because we have transformed the variable X into Z. Which is equals to this one. Now, here was X minus mu transpose. So, what value we have for X minus mu transpose? Look, X minus mu, C, Z. We have entered C, Z, A. Now, for A, we have, we know that the sigma inverse, which is equals to A, A, C, Z. And what is this you have? We have A, which is equals to, whose equal A? We know that the A, which is equals to sigma inverse. Now, we have transformed Pdf. Further, we have solved it. Look, minus sigma, it has been cancelled out. 2 pi P by 2 exponential of this value. Now, you have transpose. How will we write transpose? Z first, transpose and C prime transpose. A, C, Z. Now, we know that T prime A, C, which is equals to identity. If we do this in identity, then we have exponential of minus 1 by 2 Z prime Z. Further, now, what did you write? 2 pi minus P by 2 Z prime Z, we have written. I varies sum, I varies 1 to P, Z i square. Now, you have P terms. What did we do with P terms? Because we are multiplying, right? If P terms are multiplying, we can take it as product. And this product is capital pi, I varies 1 to P, 2 pi. Now, it is with P, so here it is 1 by 2. E raise to power minus Z square by 2. For mean, now you know that we have to find the mean, so it means that we have to take expected value. Now, expected value of i-th component of Z is, now we have to take its expected value. For expected value, what you have? Limit minus infinity to infinity up to so on minus infinity to infinity because it is a multivated case. Z i, simply you have expected value of Z in the univariate. Sum, unit into its probability. Similarly, unit into its probability. So, what is unit? We have Z i into F of Z. This is a continuous case. You have it in discrete and it is in continuous. Further, Z i as it is, F of Z. Here, we have entered the F of Z which is expected value of Z i. It is simplifying. We have just entered the value. Now, we will solve it. Here, you see what we have done. We have separated the i-th term and we are separating the remaining terms. Integral minus infinity to infinity Z i. This factor is multiplying with Z i. We have taken it from the integral. We have taken it from one integral and the remaining values are coming from the other integral. So, Z i, we have taken the i-th unit. This value came here. Into Z i. The i-th value is different. Now, this is the i-th. Now, there will be no i in it. Other than i, the remaining values are coming. So, capital pi. Next, we have a variable. H varies 1 to P but H is not equal to i because we have taken the i-th values. So, here, you have the remaining values. H varies 1 to P. Integral minus infinity to infinity. So, this is the whole as it is. But, here, the variable comes from the i-th. Further, we know that we have to solve this part. What is the answer for this part? Further, we have just done it here. 2 pi minus 1 by 2. We have taken it from the integral. So, 2 pi square root Z i into this. We have the remaining factor like this. So, this is not i equals to 1. This is basically H equals to 1. And H is not equals to i. Now, here, we know that this factor, the odd power of it is an odd function of Z is equals to zero. We know that we have done this in normal distribution. That the odd function we have is equal to zero. So, this is the first function of the previous equation. This is the odd function. So, the odd function of Z is equals to zero. That is, the expected value of the first part is zero. Multiply. What is this dot basically? Multiply of the other value. So, zero is equal to zero. So, the expected value of Z i is equal to zero. Hence, the expected value of Z is equal to zero. Now, we have X minus mu equals to CZ. This is what we have transformed, which we had let, we had suppositioned. So, applying expectation on both sides. Okay, we have applied expectation. Expected value of X minus mu which is equals to expected value of CZ. Expected value of X. This is constant. The expectation of constant is equal to constant. So, the expected value will be applied to us in X. And here, you have C constant expected value applied to Z. So, expected value of Z. Z i is zero. So, expected value of Z is equal to zero. Further, expected value of X which is minus mu which is equals to zero. Because, this factor is zero. Now, this is zero. So, where will this mu come from? Now, hence, prove expected value of X which is equals to mu. So, this multivariate mean vector of multivariate normal distribution we have determined. Expected value of X which is equals to mu.