 now here is the another example, which is the mean corrected components code, what is the mean corrected components code? basically if we go to standard dice, then sometimes we need to get the value of the mean. now how can we get the value of the mean in the principal component? so we are checking here. since y equals to beta prime x, you know that this is the principal component, we generally have zero mean, non-zero means, we generally have non-zero means but we have to do some transformation from where we can make our mean zero. therefore a transformation on y is made such that the mean vector y is zero. now we have to take some transformation from which we have the mean vector y equals to zero. if the sample mean vector x bar and the b is the matrix of the eigenvectors, you know that of the sample covariance matrix s, this is the notation of the sample covariance matrix s, then the transformation to get the zero mean vector. now we are taking the transformation, transformation is the y which is equals to b prime x minus x bar. we have done transformation on it and after transformation we have the mean vector zero. for the keyth principal component, the component score is given this, this is the general equation. in case we find the components from the sample correlation matrix, we have given the notation from the r. the component score would be the computed from the standardized observation. we use the notation z where z equals to capital V diagonal, we have capital V 1 by 2 raised to the power minus 1. i.e. capital V to take the square root and raise to the power inverse. which is equals to x minus x bar. so v diagonal is the diagonal standard deviation matrix. i.e. you have v basically, we are taking the diagonal and we have standard deviation in the diagonal. basically, we have experiences in the diagonal. now why is the standard deviation here? because its power is 1 by 2. i.e. we have taken the square root. now here is the equation, this is the general equation. now here is the equation, this is the general equation. you can determine the value of x from here. how will we get x? v is in the transpose, v is here on the side of equality. then you have minus x bar on the side of equality, positive x bar. this is the general scenario. we are using this scenario in the example. consider the random vector, this is the random vector. has the mean vector and the covariance matrix. this is the mean vector, sample mean and this is the sample covariance matrix. find the principal component and the mean corrected component score. what we have to find? principal component and the mean corrected component scores. now this is the solution. we have given the covariance matrix. this is the sample covariance matrix. now eigen values and are the solution of this. now we will take the eigen values from here. first in the previous example, we will take the population values. now we have given the sample covariance matrix. we have used the sample covariance matrix. which is equals to zero. the solution of the quadratic equation is this. just we have directed this. you know how is the value of x? minus lambda into identity. plus 49964 minus lambda and next is the identity matrix. after this, after this we have checked the solution. we have found the quadratic form. now we will generate the lambda 1 and lambda 2. lambda 1 and lambda 2 are the eigen values of the matrix s. lambda 1 and lambda 2 are the eigen values of the matrix s. now lambda 1 is against the eigen vector. then the first principal component. lambda 2 is against the second eigen vector. and the second principal component. now we have generated the first eigen vector. so lambda 2 is against the second eigen vector equation. and lambda 1 is against the first eigen vector. and lambda 2 is against the second eigen vector. so principal components of this. this is the principal component. further, now we find mean corrected. mean corrected component scores using. this is not the correlated. this is the corrected. y which is equals to b prime x minus x bar. now you have b value. what is b? this, this, this. we have b 1. so this is the b 1 1, b 1 2, b 2 1 and b 2 2. so b prime. you have taken the prime. how do you know the transpose? we have converted it to column. first row 1 point this. and you have converted it to column. this is the b transpose. next is the x minus x bar. you know that the value of the x 1. and the x 2 minus x bar is given. we have generated x minus x bar. so mean corrected component scores are this. which is equals to this equation. we have entered it. and further you have 2 equations. here is the y 1. and here is the y 2. now what do you have? now if we check further. how do we get the mean 0? so 0.145 x 1. so x 1. what will you put here? x 2. we have 7. then minus 11. in this you have to enter the values. that we have the value of y 1. 0 is coming. similarly y 2. x 1. we have put x 6. x 2. we have 11. after putting we have the value of y 2. 0 is coming. so this is the mean corrected component scores. sometime when we want the mean 0. so how do we solve it? this is the method of the mean corrected component scores.