 This is the example of multivariate normal, but this example is also related to the previous example. We want to find the mean vector variance covariance metric for the following density. This is the density function of the multivariate normal and f of x which is equals to this one. Now look at this, here the a11, a11 is the coefficient of x square, coefficient of x square which is equals to 2, a22 the coefficient of y square, the coefficient of y square which is equals to 1 and the a12, this is the coefficient of xy. Now you have the value of xy in the given example. Now the coefficient of xy which is equals to 2, now here the 2 a12 which is equals to 2, a12 which is equals to a21, 2 divided by 2, a12 which is equals to a21 which is equals to 1, i.e. you have a21 or a12 its value is equal to 1. Now here is the matrix a2111, also find the coefficient of x. Now the coefficient of x, this is the minus 22, coefficient of y minus 14. The coefficient of x which is equals to minus 22, the same way we have entered the value, so mu of x is the value and mu of y. Now look here, we have mu of x, mu of y is also coming to us and this is the equation. From here basically what we have done, from here we have taken 2 common, then we have divided the 2 here, then you have 2 mu of x plus mu of y which is equal to 11. This is called the equation number 1. The coefficient of y, now you know the previous coefficient of y, we have minus 14, we have entered the same values, so we have minus, minus here you have taken the common, basically minus 2, we have taken the common, minus 2, this is divided, then you have got the answer, 7. Now this is called the equation number 2. Now mu of x and mu of y, we have got the equation, their value is not zero or we have not determined the value. Subtracting equation 2 from equation 1, simultaneously we will solve or we will determine the value of one from the other. Subtracting equation 2 from equation 1, this is equation 2 and equation 1 we have this previous, after solving this, we have the result of mu of x, 4. Put the value of mu x in equation 2. Now mu x, here I have entered the value of equation 2, so the value of mu of y is 3. Now you have mu of x, then mu of y. This is the mu of x and the mu of y. Further to find the sigma square x, sigma square y and the correlation. This is the sigma square x, sigma square y and the covariance is given. Find out the correlation which is equals to covariance divided by variance square, variance square root, now the covariance which is equals to minus 1, sigma square x, we have 1 and sigma square y, 2. After multiplying, finally we have the correlation, minus 0.707. So its range is minus 1 to 1, so it exists minus 0.707. This is the value of the correlation.