 In this lecture, we are checking theorem 1, multivariate normal. Now theorem 1 is basically what we have, we have to check it. If x is the multivariate normal with parameter mean mu and variance sigma, variance covariance matrix sigma, where sigma is a positive definite symmetric matrix, then z, now what we have z which is equals to variance covariance matrix minus 1 by 2 x minus mu. We have transformed our original variable. We are transforming our original variable. What was our original variable basically? We have pdf of x minus mu transpose sigma inverse x minus mu. Now what we have done here is that half is on this side, sigma inverse and half is here. So we have let this portion that we are transforming in a new variable. Why are we doing transformation? Because when we have to transform the variable, what is the result of that? Like we convert rupees in dollars, we are transforming them. So if we want to transform the original variable, then what is the effect in multivariate normal? We will check it. We have the distribution of x is multivariate normal. So what will be the distribution of the transformed variable? We are looking here. We are distributed as multivariate normal with mean vector zero and variance covariance matrix sigma. When we have standard normal multivariate normal? Also each component of z is a standard normal distributed independently from each other. That is z follows the multivariate normal distribution and each component of z follows the univariate standard normal distribution. So z follows multivariate standard normal distribution. We have the probability density function of multivariate normal distribution. This is the probability density function of multivariate normal under the transformation. Now that transformation which we have given in transformation, we will solve it further. This is the transformation. Now how to remove the jacobian? So we will get the answer of jacobian further when we apply it. We will find its jacobian. So its result is sigma minus 1 by 2. i.e variance covariance determinant of variance covariance matrix power minus 1 by 2. So this is the x minus mu. You have got this term. X minus mu transpose sigma inverse x minus mu. Whose equation is it? z transpose sigma inverse z is sigma minus 1 sigma minus 1 by 2 z. When we enter the values in it, basically how it is coming from here? If you have to remove the value of z of x minus mu, then this factor will go here. So the value of x minus mu will come. If we take it here, then what will happen? Sigma 1 by 2. So here you have z transpose sigma 1 by 2. This is sigma inverse as it is. X minus mu. We have seen its value. So sigma half into z. So from here you have basically removed the value of z. We have removed the value of x minus mu. We have entered it here. So sigma minus 1 sigma half sigma half sigma half sigma half variance covariance matrix. Half and half 1 is done. Minus 1 is the factor you have. This factor is 1. So here the z prime z is the result we have. Under the transformation of this. So f of z is the transformed variable. f of x is f of z. f of z which is equal to 2 pi p by 2 sigma minus 1 by 2 is the exponential minus 1 by 2 is the value we have. z prime z. And when we take the derivative with this, mod of z as it is. Now further f of z which is equal to this one. How did this happen? We have cancelled out the answer with this. So we have the remaining answer. But this is the density function of multivariate normal distribution with mean vector 0 and variance covariance matrix identity matrix. So x is the multivariate normal distribution with mean vector mu and variance covariance matrix sigma. When we have transformed it. See we have transformed it. After transformation what we have is that z is the standard normal multivariate with mean vector 0 and variance covariance matrix identity. Further f of z. We have found the value of f of z. Now how will you solve this further? We will write this. z i square sum and z i square. Now further we have capital pi. What will be the value of this? 1 by 2. Because p terms are here. Expected value of this. So each component of z is a standard normal variate distributed independently from each other. So what did we do in this? What did we follow? Multivariate normal distribution. So what we have transformed which also follows the multivariate normal distribution if x is following a univariate then what we have transformed we go to the variable univariate. If x follows the multinorm?? transform variable follows the mille normal distribution. That standard normal is following and we have also followed the standard normal multivariate distribution. इज अख आख पूलुस दी श्टन्दर नालमल वेर्ये दिस्टिभूँटे अंदेपनेडली फ्रुम इज अख़ा.