 Assalamu alaikum dear student. Today we will discuss the linear combination in the multivariate analysis. So, what is the linear combination? Today we will discuss what is the linear combination? Let we have Y. Y which is equals to the amount of money you have spent in a shopping trip which is equals to X1 plus X2. Where X1 equals to how many books you want to buy and X2 how many bag of chips you want to buy. So $30 is using for amount of books you want to buy, the money amount of books and the $5 is how many bag of chips, the amount of money how many bag of chips. So this is a linear combination. Just for your understanding what is the linear combination? The linear combination is the mixture of two or more than two random variables. Basically, what is the linear combination? We call it transformation. Transforming our variables means we transformed X1 and X2 and we introduced new variable which is equals to Y. So, its basic definition now I will tell you how to do it. Let X, X is a vector be a p-dimensional random variable with expected value of X equals to mu. Previously, we have done expected value of any random variable which is always equals to the population parameter that is mu. And variance of X equals to sigma and sigma stands for variance covariance matrix. And A is a vector of constant of order p into 1. A is a vector of order p into 1. Then mu is a scalar random variable called linear combination of X with expected value of X A transpose mu and variance of X A transpose sigma A. Basically, what we have is Y. Y is a random variable which we are transforming Y which is equals to A transpose X. And we have A which is equals to the order of p into 1 random variables. So, this is the transformation A which is equals to A transpose X. Now, taking expectation on both sides, so expected value of Y which is equals to A A transpose into expected value of X. And we know that, so what is the value of expected value of X which is equals to mu. So, expected value of Y which is equals to A transpose mu. You have also seen that here we have expected value of X which is equals to A transpose mu. Similarly, variance of X which is equals to A transpose sigma A. Now, this is the equation Y on both sides. So, the variance of Y which is equals to A transpose variance of X A. Now, how will we write this? Which is equals to A transpose variance of X which is equals to variance of variance matrix sigma and A. As you know that variance is in square form. We write variance in square form. So, here you have A transpose A. We are taking variance of Y which is equals to variance of Y which is equals to A transpose sigma A. We can take A transpose sigma A according to multiplication. We will check the multiplication of matrix or vector. So, in general, consider the Q linear combination of P random variable X1, X2 up to so on, XB. Set of linear combination is called the linear compound. Now, here we have a linear combination of Z1, C11. Suppose C11, we have said 3. C12, constant value is C11, C12 constant value which is equals to 2. X1 is a random variable. X2 is also the random variable. As I gave you previous example of books and bags of chips. Similarly, we are multiplying it. Up to so on, we have so many variables. P random variables with constant C1P. So, basically, you have a vector. I will tell you this. We have Z which is equals to Z1, Z2 up to so on, ZP. We have a vector which we have lit up which is equals to Z. Now, what is the order of the vector? How many rows do you have? How many rows? So, P into 1. P rows and one column. Similarly, the constant value C11, C12 up to so on, C1P, 1P, C21, C22 up to so on, C2P. Now, the last is CQ1 up to so on, CQ2 and CQP. As you have C11, C12, C1P. I have written all these constant values in matrix form. Now, what is the order? How many rows and columns in this? So, we have Q rows and P columns. We have a matrix of Q rows and P columns. So, X1, X2 up to so on, XP. How will we write this further? This is the vector of X1, X2 up to so on, XP. How many rows and columns in this? So, here is the P rows into one column. So, we have a vector of P cross 1. Now, how can we write this in the journal form? Z equals to CX. So, Z is the vector, C is the matrix and X is also the vector. You know that this vector form is in small letters. And we have matrix form in capital letters. So, what is the idea of linear combination? In general, consider the Q linear combination of P random variables X1, X2 up to so on, XP. As you have full linear combination. Now, what is linear combination? We transformed our variables. Like here, we transformed variable X1, X2 up to so on. Similarly, you have a full multivariate analysis of how linear combination is being introduced or how to explain it. I have solved this for you. Now, let's look at the examples of this linear combination. How do we understand it? Now, example one is the determine and expression of the mean and variance of the following linear combination in the terms of mean and covariance of the random variable X1, X2 and X3. Now, we have the three random variables. X1, X2 and X3. What we have to find for this? Mean and various covariance matrix sigma. We have to find that. Now, the number one is X1 minus 2X2. First part we are solving. But the condition is if X1 and X2 are independent. First, we have to solve all these parts. But the condition is if X1 and X2 are independent, we have to solve it according to this. Determine expression of the mean and the variance. Now, what is the first part? X1 minus 2X2. Now, we have the mean vector and the covariance matrix. Mean vector means mu1, mu2, mu3. Mean vector means mu. And how many variables? X1, X2, X3. So, X1, X2, X3 means mu1, mu2 and mu3. So, we have the mean vector. Now, look at this. This is the variance covariance matrix sigma. What is the first value of sigma 11? This is sigma 12. But the condition is X1 and X2 are independent. It means when you have variance covariance matrix of sigma, then we have these diagonal elements. These are the variances. And off diagonal are the covariances. So, covariance, when you have independent X1 and X2 are independent, it means that their covariance will be zero. That is why we have sigma 12's value is zero. Similarly, sigma 21's value will also be zero. Because the cross product terms that we have are independent, it will be zero. So, according to this condition, we have to solve our 1 to 5 parts. Now, we find the mean and variance linear combination RS. We have to find the mean and we have to find the variance for linear combinations. Now, we have the first question is X1 minus 2 X2. And we named it by ourselves, which is equals to Z. That is, we have taken a variable candidate. Now, Z which is equals to X1 minus 2 X2. Now, what you have? The constant A that we were taking, we have constant values with A. What is the constant time with X? Which is equals to 1 X2. What is the constant time with second? Minus 2 third X3. X3 is not in a particular example, that is why its value is zero. Now, X1 X2 X3, this is the vector. So, we have the expected value of X. Now, this is equals to the A prime X. So, A which is equals to 1 minus 2 zero. According to the particular example, we have A minus 2 zero. So, A transpose, we have converted columns to rows. A transpose is 1 minus 2 zero X1 X2 X3. Now, further we have expected value of X. What is the expected value of X? The expected value which is equals to A transpose A. A transpose X, you know A is constant. We will take it by constant expectation. So, expected value will be applied on random variable. So, expected value of X which is equals to mu. So, we have A prime mu which we have transformed. So, what is the expected value of Z? A transpose mu. So, A which is equals to 1 minus 2 zero. And mu 1 mu 2 mu 3. We have a particular example. Now, if you solve this, we will multiply rows with columns. So, 1 multiplied by mu 1 minus 2 multiplied by mu 2. This is the final answer of expected value of Z. So, we have to find variance of variance. Variance of Z. I have told you how variance of Z is coming. A transpose variance of X. Into A variance of X. What are we writing? Sigma variance covariance matrix sigma. So, A transpose A. A transpose sigma A. What is the expected value of A? 1 minus 2 zero. Sigma. Sigma. We have just defined that what is sigma? Sigma 11. Covariance terms are zero. So, you have sigma matrix. Into A. This is your vector. And here we have taken the transpose of that vector. So, first we can multiply into 2. So, sigma 11. 1 multiplied by f. You have rows multiplied by column. So, sigma 11 multiplied by 1. Sigma 11. Minus 2 multiplied by zero. Zero. Zero multiplied by sigma 13. Okay. Look at it once again. What is this you have? It is the matrix of 3 into 3. And this vector you have. What is the order? It means 3 into 1. This is the order you have. So, what is remaining you have? When I will multiply it. So, 3 into 1 means. That you have 3 rows and 1 column. It is a vector generator. So, zero multiplied by 1. Zero. Minus 2 multiplied by sigma. 2 to. Zero multiplied by sigma 23. Zero. Sigma 31 multiplied by 1. Sigma 31. Minus 2 multiplied by sigma this. Minus 2. Sigma 32. Zero multiplied by this. You have one. Now, what is the order of this vector? 3 cross 1. It means 3 rows and 1 column. I will multiply it once. So, this is the sigma 11. Minus 2 multiplied by this. Minus minus plus. 22 is a 4. And sigma 22. Zero multiplied by any value which is equals to 0. So, you have the first part. We have the variance of z which is equal to sigma 11 plus 4 sigma 22. Similarly, the second part. z which is equals to minus x1 plus 3x2. Again, as we have solved the previous part, follow the same method. Minus. x1 means the constant is minus 1. x2, 3. And we do not have the third variable. That is why its value is zero. Similarly, x1, x2, x3. We have taken A transpose x. Then, apply expectation on both sides. If you apply expectation, then you have the final result. A prime mu. A prime minus 1, 3, 0. On mu, we have mu 1, mu 2, mu 3. We have solved this equation. This will be the result. Similarly, z. Variance of z which is equals to this one. You have taken out the value of variance of z. I told you in the previous video which method we have to solve. We have the answer of variance of z. Now, next. Let z equals to x1, x2 plus x3. You have three random variables. x1, x2 and x3. So, with x constant 1, with x2 constant 1, with x3 constant 1. x1, x2, x3. Similarly, A prime x. Taking expectation on both sides. So, the expected value of z which is equal to expected value of A prime x. Now, A prime expected value of x which is equals to mu. So, now you have expected value of x with expected value of z. The result is 1 mu 1 plus 1 mu 2 plus 1 mu 3. Okay? Again, variance of z. A prime sigma A. A prime, we know that 1, 1, 1. Sigma, we know that. Sigma matrix and A, 1, 1, 1. Now, further you solve this. So, you have here, sigma, 1, 1. Okay? Zero multiplied by plus sigma, 1, 3. Zero multiplied by 1. 1 multiplied by this, sigma, 2, 2. And 1 multiplied by sigma, 2, 3. 1 multiplied by sigma, 3, 1. Plus sigma, 3, 2. Plus sigma, 3, 3. Okay? Now, you will multiply with this. 1 multiplied by sigma, 1, 1, plus sigma, 1, 3. Okay? 1 multiplied by sigma, 1, 1. Plus sigma, 1, 3. You have 3 rows. Again, 1 multiplied by sigma, 2, 2. Then, plus sigma, 2, 3. Plus sigma, 3, 1. Plus sigma, 3, 2. And plus sigma, 3, 3. Here, you just have to tell that, but, sigma, 1, 1. Plus sigma, 3, 1. Which is also equals to the sigma, 1, 3. That is, we have added up the sigma, 1, 3. Which is equals to the sigma, 3, 1. So, we have written this twice here. Sigma, 1, 3. Similarly, sigma, 2, 3. And sigma, 3, 2. How do we write this? Sigma, 2, 3. You have sigma, 1, 1. Sigma, 1, 1. Sigma, 2, 2. Sigma, 2, 2. Sigma, 3, 3. But, we have added up the sigma, 3, 1. And sigma, 1, 3. We have added up the sigma, 3, 1. And we have written it twice here. Now, this is the part D. Here is the x, 1. Plus 2x, 2. Minus x, 3. Similarly, you have solved this pattern. 1, minus 2, minus 1. x, 1, x, 2, x, 3. So, we have got a prime x. So, expected value of z which is equals to a prime mu. Again, you have to find the variance. We have taken variance matrix. a prime sigma, a. According to that, you have solved this. Now, look here. You will have 2 sigma, 1, 3. It means that you have sigma, 1, 3. And sigma, 3, 1. We have added it up. And we have added up minus 2, sigma, 1, 3. Here, you have sigma, 2, 3. You can check that sigma, 2, 3 is coming twice. Or we have added it up with sigma, 2, 3. With sigma, 3, 2. Now, this is the last part. 3x, 3. Minus 3x, 3. This will be 3x, 1. 3x, 1. Minus 4x, 2. So, 3x, 1. 3, we have constant. Minus 4x, 2. You have this 4 with you. And we don't have the third variable. It means its value is 0. Similarly, you have solved this. So, you have a prime mu. You have value of a prime mu. Minus 3 mu, 1 minus 4 mu, 2. And this is the variance covariance matrix. Variance covariance matrix. You have this matrix multiplying with this. Multiply here. Further, we have solved this. After solving this, we have the final result. The result is 9 sigma, 1, 1 plus 16 sigma, 2, 2. Maybe there is an add-up somewhere. Or you have this value by multiplying it. Now, you can calculate this yourself. So, you will get this result.