 आत कै आद भागा भी लागा की लेखा। उनी वेर्येट नाम्मल रीस्रिबॉशन, देखा काप पड्याई दिलाई से बलागा दीगा आदू दोस्काईदिया आप को अप. उसी उनी लेगा बी यद के अब हम जलने लागा ली नाद मर्टी बी रीट के रंद. तुब कास भीही स्वरृट के लिए, तुक के स्वरृट लें, और वो लेँ सागटा कर नापुद. तुक भी बागर रेस्ग तूpe रेस्पूरुपतरूई है. तो उनीवेडिएट आप लोग ने पले पड़ा है, now here the univariate normal distribution x with mean mu and variance sigma square. We know that, के उनीवेडिएट में normal distribution के पारामीटर्स क्या है, with mean mu and variance sigma square. The pdf of the univariate which is equals to this, we know this is the pdf of the univariate distribution. अब उनीवेडिएट को हम ने जनलाइस करना है, इसको हम ने अप कन्तिनुग करना है, for 2 or more than 2 variables के लिए, the p-dimensional random variable x. तो मुल्टीवेडिट हम p-dimensional के लिए चेख करें, with random variable x is said to have multivariate normal distribution, if and only if every linear compound of x has the univariate normal distribution. अब मुल्टीवेडिट को अब में आगे फर्दड लेके जा दिए मुल्टीवेडिट की अंदर. तो उनीवेडिट का ये pdf का idea है आप को यहाप को पता है कि उसके पारमीटर से, ये उसके पारमीटर से, और हमाने पस x is a random variable. तो तो तो अब मुल्टीवेडिट, नाँँ़ी लेके अगर tired of normal distribution, a random vector x is said to have a मुल्टीवेडिट नाँँ़ी लेके आचबॉशाग, a multivariate normal distribution का देंसटी फुंच़्न क्या है? If its density function is given by this, येमारे पस क्या? का? मुल्टीवेडिट का? देंसटी अब आप देखु अप देश्टी फंक्षन में या मैं आप पर से एक वालू आर ये के विर के इस कोंस्टेंट, के कोंस्टेंट है, अब उसकी वालू हमें दिटार्मिन करनी होगी के इस के एक वालू क्या है, मू, the mean vector, which is equals to P into 1 vector of means and sigma is a P into P positive definite variance covariance matrix. ये देखु आप के पास P, multivariate में आप के पास mu किस के एक वाल है, जिस को हम के रहें P cross 1, it means P rows and one column, this is the P into 1, and sigma variance covariance matrix अमार पास किस के एक वाल है, P cross P k. अब ये के की वालू, बेसे कर लिए ब हमने एस में क्या देटार्मिन करने है, the value of K. देटार्मिन आप के पास में आप के पास में आप के वालू, now this is the property, we know that using the property ये आप देखु ये इंटिगरल है, युनी वेर्येट में हम जस वाल इंटिगरल लेरे है, because this is the multivariate case, so i varies one limit minus infinity to infinity upto so on minus infinity to infinity integral K of f of x into dx which is equals to 1. We know that the total area under the normal curve which is equals to 1. First of all we have to find the pdf, so what is the first property of pdf is that total area under the normal curve is 1. So integral we have taken minus infinity to infinity K times exponential of this. Now this is the factor we have from previous, this is the exponential of this value. So here we have entered it further. So because K as it is, we have given density function, we have done the value of density function. dx, dx, i.e. for first we have x, x1, x2 upto so on, xp which is equals to 1. Then we have K. Now what we have done with K is that we have one of the equalities, so K will go to the equalities, it was multiplying here, so it is divided here. This is called the equation number one. Now let us see what we have done next. Since sigma inverse, this is the sigma inverse value which is equals to A is a positive definite symmetric matrix. This is the property of the positive definite symmetric matrix. We have the property of positive definite symmetric matrix. There exists a non-singular matrix C. All the determinants which is greater than 0, should be 0. Of matrix C such that C prime AC which is equals to identity matrix. So this is the property of the positive definite symmetric matrix. Now find the determinant. We have taken its determinant. We have the determinant equal to 1. Further we have done it, so the determinant is equal to 1. Now here let, let what we are doing? X minus mu, this is the term which you have. In exponential this term is X minus mu. So let X minus mu which is equals to CY. We have C non-singular matrix as CY. Here we have basically applied transformation. We have transformed the X minus mu variable. Next new variable which is equals to Y and C is the non-singular matrix. From here we have found the value of X. We have the value of X which is equals to CY plus mu. Now we have this with X. What we have to do further? J modulus. J means here you have Jacobian. Also J modulus. Now how we have to take out J determinant, how we have to take out Jacobian which is equals to mod determinant of curly X over curly Y. Means here we are taking derivative. For single variable we take its derivative. Two or more than two variables when we have data or we have two or more than two variables if we want to see the mathematics. So how we will take its derivative with Jacobian. Now here is the Jacobian mod of curly X over curly Y. Here mod. Curly here we have the value of X. Now we have let X which is equals to CY plus mu. We have to take its derivative with respect to Y. And for the above transformation Jacobian which is equals to mod of C. Now here you see its derivative with respect to Y. So Y with C is involved. When we take its derivative we have Y1 and we have the result of mod of determinant of C. Now why we have mod covered? We have mod covered because here we are considering only positive value determinant. We are considering positive value. So this is why we have used mod covered. Now you see we have these terms in exponential. What is written further? X minus mu transpose sigma inverse X minus mu. Now X minus mu we have let which is equal to CY. Transpose sigma inverse sigma inverse we have sigma inverse which is equals to A. A is equal to then we have X minus mu which is equals to CY. Now from here you know what we have said C prime AC which is equal to C prime AC which is equals to identity matrix. This is equal to identity. So Y prime Y we have the final result of this particular exponential part of Y prime Y. So equation 1 becomes now we have equation 1 in equation 1 you have this where we have 1 over K. Now equation 1 we will put those values in equation 1. Now limit minus infinity to infinity up to so on exponential minus 1 by 2 and we have its value basically Y prime Y and we have the value of D of X mod of C determinant Y1, Y2 up to so on Yp because it is transformed value variable we have transformed variable Y which is equals to 1 over K here mod of C it is integral to Y then the integral values have come expected value of this we are writing this further Y prime Y because it has quadratic form Y prime Y which is equals to square form sum i varies 1 to p Y i is square we have written and this term as it is which is equals to 1 over K now we will take all the integral as it is so what we are doing here this is equals to pi pi i varies 1 to p capital pi i.e. this is product see this is product so we are writing it one time we have taken pi i varies 1 to p integral minus infinity to infinity i.e. write this integral or we will take it in pi exponential what we have done minus Y square over dy into 1 over K now we know that this integral which you have this integral the value of this integral what we have lambda 2 pi square root this is equal to after taking integral when we solve it integrate after solving it result is 2 pi square root i.e. now we have to enter its value so equation 2 becomes mod of C pi into 2 pi square root which is equals to 1 over K so mod of C now you solve it further now we will take the value of pi or if we take the value of pi if we take the value of pi then what will happen 2 pi is under root 1 over 2 so when pi how many times it is multiplying then its power will be P by 2 from here because it comes to you 2 pi its power 1 by 2 so it is multiplying so it is the power of 2 pi P by 2 which is equals to 1 over K to find mod C on the value of mod C we need the value of mod C is equal to so C prime AC which is equals to identity and when we take its determinant which is equals to 1 now we have separated its determinant which is equals to 1 C prime C C determinant is square A as it is so C square which is equals to A what happened to you here see how we have brought it to equality A determinant so it will be divided from 1 we have brought the divide so it is multiplied we have brought the divide so C square which is equals to A determinant minus 1 so mod of C mod of C we have equal to A power minus 1 by 2 now minus 1 by 2 what you have taken its square root when we are taking its square root so it is minus 1 by 2 square term has been divided so we know that A inverse which is equals to sigma so mod of C mod of C its value is A which is equals to A inverse which is equals to sigma so A inverse see here it is inverse A inverse we have put value instead of A which is equals to 1 by 2 because it is minus it is minus plus so it is 1 by 2 as 1 over K which we have here 1 over K mod of C 2 pi this was the value now we have put value in this K is equal to mod of C 2 pi B by 2 okay so K now you have K is equal to here we have taken 2 pi minus B by 2 determinant of sigma minus 1 in this K's value in that PDF which we had a journal where we were entering K constant's value we have put its value there so final multivariate normal distribution we have this PDF 2 pi minus B by 2 sigma determinant minus 1 by 2 exponential of this value