 यह जूटडड तीज़ामपल लमपट्ज़ी आस्धार अनलतिस यह तो वाई में दिपना सेगटान नाई पनिब अन अद़ चलूँँँग करेंगे आत्चार औनाललिसिख को आईक प्रूँँँँँँँँँँँँँँँँँँँँँँँँँँँँँँँँँँ� गी साद करने प्रूयत Nabhis weekly ज़़े है खोरनिन सासाच्टाय। बसेरए नयी सुए एक मिदhe nahde क Bose लेंब्राद 3 is the third value of the eigenvalues, and here is the beta 1, beta 2, beta 2 these are the eigenvector, now the first part is, assume k equals to one factor model, where we have to do one factor solution, and I have explained that we can do one factor mathematically, we have seen one factor mathematically, we have solved it mathematically but more than one factor solution we will solve it on the software. we have to calculate the loading matrix, sigma and the matrix of the specific variance psi using principal component solution method, now we have the restriction that we have to find the loading matrix and the specific variance psi, but with whom we have to use the principal component analysis, we have to solve it with that method, now the second part, what proportion of the total variation explained by the first common factor, so in the second part we have to find total population variance is explained by the first common factor, and the first factor is the total variation which is explained by the first common factor. so this is the question, and here is the solution, now look at this, principal component method for k equals to one, so now what have we done, this is the lambda, lambda prime, now what have we done here, now we have pair of, this is the pair of the eigenvalues and the eigenvector used, so lambda basically we have lambda one, beta one, sk root, so lambda prime we have beeta one, lambda one sk root, beta one transpose, in general we have lambda one sk root, lambda one sk root when we have multiplied and then when we have cancelled sk root, it will turn into sk form and we will cancel, so lambda lambda prime, multiply by 0.567, then second value, previous we solved this in all the matrices. Now look at this, we have the matrix of 3x3. Now we know that, lambda, lambda prime is not there, now we have lambda. So what we have lambda, lambda 1 1, lambda 2 1, lambda 3 1. So lambda prime, lambda 1 1, lambda 2 1, lambda 3 1. So what happened, if I first do lambda 1 1 into lambda 1 1, lambda 1 1 square, second, lambda 1 1, lambda 2 1, lambda 1 1, lambda 3 1. So what will it be, 3x3 has come to you. Now what have we done with this, we can compare it easily. So lambda 1 1 square, lambda 1 1 square is 0.766, its square root. So after taking square root, we have lambda 1 1, 0.88. Now lambda 2 1, lambda 2 1, 0.69, its square root, 0.83, lambda 3 1 square, and we have 0.629, its square root. So after that, we have lambda 1 1, lambda 2 1, and lambda 3 1. Now lambda prime, which is equal to 1 1 2 1, 3 1, values are lambda prime. For k equals to 1, now what have we done with this, we have done it with eigenvalues and eigenvector, this is the solution. We have used eigenvalues here, and then we have used eigenvector. That way we have solved the principal component, basically the principal component factors. So through the principal component, we have solved the factor analysis accordingly. For k equals to 1, 1 factor solution with 3 variables, we have this, this is the journal, lambda prime, you know that, sigma lambda lambda prime plus psi, if you don't put this bracket here, then it doesn't matter, because there is no such thing here. Sigma equals to lambda lambda prime plus psi, then we have value of sigma. Like we were doing in the previous video, we have values of sigma. Now lambda 1 1 square plus psi 1, compare with this, and where do you have this? So from there we have the values of sigma, compare it, and you know that we will have 6 equations. Further we have to find the values of lambda 1 1, lambda 2 1, lambda 3 1, then we have to find the values of psi. So specific variance is given as this. After simplification, now there is no simplification here, because in the previous two examples, we were solving this. After simplification, we will get the answer of the specific variance, which is equals to this one. So psi, we have psi 1 1, psi 2 and psi 3, this is psi 1, not 1 1. Now specific variance we have found earlier, because what we had, assume k equals to 1, calculate the loading matrix, loading matrix we have found earlier, and the specific variance by using the principal component solution method. A part we have completed here, this is the part A. Psi has also come, and we have the values of lambda, further we have lambda b. Now next is the B part, proportion of variation is explained by the first common factor. A proportion of variation is explained by the first common factor, we have to check the proportion of the first factor. So which is equals to lambda 1 over P. Now lambda 1, what is it? Eigen value is given as lambda 1. So lambda 1 is given as which is equal to 1.96 divided by P, dimensions or variable, how many variables are there? 3. And which is equals to 65.33%. It means that first factor, first variable is explaining 65.33% variation. That means one variable is explaining so many variables. So remaining where will it go? We will go to the other factors. We will tell the other factors. So what we have got first? The proportion of the variation is explained by the first common factor, which we have lambda 1. Because you have the highest eigen value of lambda 1. So according to the highest eigen value, we have seen that it is explaining 65.33% variation. Now then, we have to find the communalities in it. The communalities, as I told you in the previous video, what we have got is some of the squares. So we have got some of the squares. lambda 1, 1 square, lambda 2, 1, lambda 3, 1. And total communalities. Total communalities, we have added them. After adding them, we have total communalities. The value of lambda is 2.087. And you know that these values cannot be greater than 1. Now total variation of X. Now that total variation of X, that is how many we have now, which is equal to trace of sigma. And trace of sigma, the diagonal values of this sigma, which is equal to 1 plus 1 plus 1 equals to 3. The proportion of total variation of X, explained by the common factor. Now as I told you in the previous video, one factor. Now here we have to see the total variation explained by the common factor. Common factor, how many variations are being explained? lambda divided by trace of sigma. So what was lambda? Now we do not have lambda 1 or 2. This is the total communality. This is the total communality, lambda 2.087. The 2.087 divided by trace of sigma divided by 3, which is equal to 69.5%. It means that total variation of X, explained by the common factor, 69.5%. Remaining, how many will you have? 30.5 factor, where will it go? Specific factors? Proportion of the total variation of X, explained by the specific factor, which is equal to this. This is the formula. To obtain the proportion of the total variation of X, sum i varies 1 to p, psi i divided by trace of sigma. And psi's values we have defined. So psi which is equal to this, psi i, we have also defined previously. Here is the psi i, psi 1, psi 2, psi 3. This is the psi 1, psi 2, psi 3. Divided by trace of sigma, which is equal to 3. And after calculation, the value is 30.5. Now how many total do you have? You have 69.5. So how many remain? 30.5. So 30.5, which variation you have explained? Specific factors you have. 69.5, variation is being explained. Common factor. So this is the example number 3. In example number 3, we have used the concept, principal component analysis. We have used the principal component. This is the principal factor method. This is the principal factor method. In this, we have solved all these solutions with eigenvalues and eigenvectors. So this is, you also have that we have given total variation. Common factors, but we have found common factors and specific factors. Through which, eigenvalues and eigenvectors. With principal component. So this is the example number 3 of factor analysis.