 The student, in this example, we are going to learn how to find the mean vector correlation by using the density function. Find the mean vector. A is the variance covariance metric for the following density. Now, this is the multivariate normal density as you know that and this is the joint function f of xy which is equals to this one. Here, sigma inverse. In the density function, what we have here in the multivariate, here you have sigma inverse and here you also have sigma inverse. So, we did it late. Here, sigma inverse which is equals to a and a inverse, a inverse which is equals to sigma and this is equals to a 1, 1, a 1, 2, a 2, 1, a 2, 2. Also find sigma square x, sigma square y and the correlation xy. All we can say that the sigma square x which is equal to sigma 1, 1, sigma square y which is equals to sigma 2, 2. This is the density function of the multivariate normal and we have f of xy. Further, how do we do this? Now, this is the solution. This is the joint function f of xy which is equals to this one given a. Now, open the square a square minus 2ab plus b square. Open the square a square minus 2ab plus b square. Now, f of xy, further we have rearranged it. x square, take the term first, then y square minus 2x, 4y or 1 plus 4 which is equals to 5. Here, we have the final f of xy. Now, a which is equals to this and mean vector. This is the mean vector mu x because we have the variable xy. Now, mu x and the mu y, where a11 is the coefficient of x square. a11, here is the a11. The coefficient of x square which is equals to 1. a22, the coefficient of y square. What is the coefficient we have? Which is equals to 1. 2a12, y2a12 which is equals to 2a12. What do you have? a12 which is equals to a21. So, what have we added to this? This is the 2a12 is coefficient of xy. Now, here you see the coefficient of xy. We do not have the coefficient of xy. So, a12 which is equals to 0. Further, this is not the part of this example just to understand how we have to determine the coefficient. Now, look at this. This is the density function of multivariate norm. Here, x minus mu transpose a into x minus mu. This part, where x minus mu, x minus mu is equal to x minus mu of x, y minus mu of y. We know that the x minus mu of x and y minus mu of y. Here, we have transpose the vector. We know that the a which is equals to this one and x minus mu, x minus mu given, we have taken here. That is, in this function, we have added this. We have put the values. Now, next, multiply. Now, how many rows do you have? How many rows? One row and two columns. Here is the order 2 into 2. So, what will be the order you have? 1 into 2 means one row and two columns. So, multiply this x minus mu multiplied by a11, this row multiplied by this column, this row multiplied by this column. So, x minus mu of x multiplied by a11 plus y minus mu y multiplied by a21. First value. Now, the second, this row multiplied by second column, x minus mu multiplied by a12, y minus mu multiplied by a22. Third function as it is. Now, we have solved the first two. Then, we have multiplied it. You have opened all its calculations. We have opened the calculations from here. So, this result is here. For last, now, see what you have. What is the order of this? One row and two columns. This is one row, one row and we have two columns. Now, we have to multiply it with this. Now, this is the order of 1 into 2. How many orders are there in this? This is the order of 2 into 1. So, what is the final order? We have the final order of 1 into 1. Next, when you have this whole row multiplied by x minus mu plus the second row, the second column you have, this is multiplied by y minus mu. Now, look at this. I have done this in different colors. Now, what did you see in the previous one? What was the coefficient of x square? Which is equals to 1. So, a11 x square, coefficient of x square which is equals to 1, coefficient of y square, y's coefficient which is equals to 1, a22 which we did in the previous one, a22, y square which is equals to 1. Now, the coefficient of xy. This is the xy, coefficient of xy. If we want it or not, then we will take the coefficient of xy. But in the given example, we don't want the coefficient of xy. We have zero. Because in density, xy is not there. Now, the coefficient of x. Now, where are the x's? We have to take them. In purple, you see in purple color, minus a11 x mu x, this is minus and minus twice time of 2 a11 mu x. Okay? Purple, we have x's coefficient. Then, we have y. a21, this is purple color and you will have x color x. This is the x. So, a21 and here is the a12. This is equals to the minus a21 mu yy. So, whose coefficient is this? Coefficient of x which is equals to this equation. Similarly, coefficient of y. Where is y? We have to separate the y coefficients. Here, you have minus a22, y mu of y and this. This is minus, minus, minus twice time of a22 mu y. Whose coefficient is this? y. Then, this one. Now, you have a21 and a12. We took this. So, minus twice time of a12 mu of x. We want coefficient of x and the coefficient of y. But, all these examples, just to understand where these coefficients are coming from, just tell us that. Now, a11 which is equals to 1. Why? Because, coefficient of a, we have 1, a22 coefficient, we have 1 and a12, the coefficient which is equals to the 0. Okay? We have the same value. Also, the coefficient of x. Now, we have to look at the x. We have determined the coefficient of x equation. Now, the coefficient of x. Given example, you see, what is the coefficient of x we have? Coefficient of x, this is equals to minus 2x. So, what we have? Minus 2, the coefficient of x. This is the coefficient of x. Equals to minus 2, as it is, mu of x. We have to determine the value of mu of x and a11, which is equals to 1. We know that, which is equals to 1. Minus 2, we have to determine the value of mu y and a12, which is equals to 0. We have the coefficient of 0. Further, we have solved it. Minus, minus is cancelled out. 2 will be divided here. 2 will be cancelled out. 1 is mu of x, which is equals to 1. Again, the coefficient of y. We have to determine the value of mu of y. This is the coefficient of y. Here, the minus 2, mu a22 minus 2 mu x, a12. We have determined the value. We have determined the whole equation. Now, the coefficient of y. Now, in the density, if you see, the coefficient of y is minus 4, which is equals to 2. We have to determine the value of mu y. A22's coefficient is 1. A12's coefficient is 0. Further, you have minus, minus is cancelled out. 2 is divided on the side of equality. After canceling out, the value of mu y is 2. Now, the mean vector. This is the mean vector. The value of mu of x will be determined. 1 and mu of y's value will be determined. 2. So, we have one part of the solve of an example, that the value of the mean vector will be determined. Now, here, mu x, which is equals to 1, mu y, which is equals to 2. To find sigma x square, sigma y square and the correlation, we proceed as follows. A, we have to say, this is, you have to see what this was. Sigma xx, which is equal to sigma x square, which is equals to sigma xx. So, sigma xx1, sigma xy0, sigma xx, xy0 and the one. Then, find the correlation. Correlation, correlation, which is equals to the covariance over variance square root. Variance of x into variance of y square root. Zero. Covariance, which was xy, xy, which is equals to zero. And the covariance of x, this is the variance of x, which is equals to 1. Variance of y, this is the variance of y, which is equals to 1. After calculation, we have the correlation of the scale. This is the zero. So, what did we find in this? The mean vector is also fine. We also have the correlation matrix. We have also found the correlation further.