 ठीट्उधान तोstalk 있으ों आप आप मोझनेखा, मुत्वेरीट नामाल्स्टी पूश्छन्द. मुत्वेरीट मांजि़ी लिग। एक वैदिसे, मुत्वेरीट मांजि़ी च्छंदेख, PDAF drive, PDAF mean of the PDAF Variance of the multivariate normal distribution Mathematically we have derived We have seen the PDAF of multivariate normal distribution Its mean is equal to Its variance is equal to We have checked all that mathematically We have checked the multivariate normal distribution Basically we have checked it numerically That we know How we have to see the results So this is the example of the multivariate normal distribution And here we suppose We have the multivariate normal distribution With three variables Like X1, X2 and the X3 The variable name is the X1, X2, X3 But behind it you have X1 is the name of some variable Which we have quoted as X1 X2 is also a variable Which we have quoted So to understand this We call it X1, X2 and X3 Basically X1, X2 and X3 You will have three variables And the following mean vector And the covariance matrix are the given Now the mean is given Variance covariance matrix Sigma is given To calculate the PDAF Of this distribution at the given point of this Point is given 1.5, 3.2 and 4.6 Three variables say X1, X2, X3 Okay We have to calculate the PDAF Of the distribution On these points Okay this is the mean vector Variance covariance matrix Now recall the PDAF Of the multivariate normal distribution What was the PDAF? Now here is the solution We can use the formula Of the multivariate normal distribution This is the PDAF of the multivariate normal distribution 2.2 Minus P by 2 P is the dimensions How many dimensions in the particular question So P which is equals to 3 We don't have the dimensions Sigma Variance covariance matrix is given Model S Minus 1 by 2 Exponential minus 1 by 2 X minus mu transpose Minus mu This is the PDAF of the multivariate normal distribution Where P is the number of variables Or you can say that the dimension X is the vector of the Variable In case of the X transpose Which is equals to 1.5, 3.5 These are the particular points Given Mu is the mean vector And sigma variance covariance matrix Given We have to find here To solve this With numerical data, we need Sigma inverse Sigma inverse is the inverse of the covariance matrix Okay PDAF is here, we also found out What are all the variables Exponential EXP is basically Exponential We have to use these all To find the value of F of X X minus mu First we have this portion X minus mu Okay, so we have X 1.5, 3.2, 4.6 And mu is 2, 3, 4 So X 1.5, 3.2, 4.6 And mu is 2 So then we have X minus mu Calculate it directly So what will you do 1.5 minus 2 3.2 minus 3 And 4.6 minus 4. So then first We will look at the dimension How much is the dimension of X Here is 3 rows 1 column And how much is the dimension of mu 3 rows 1 column So what will be the final result 3 cross 1 Means 3 rows 1 column 3 cross 1 So X minus mu Minus 0.5, 0.2 And 0.6 This portion is here Okay, this portion is here Because this is transpose This portion is here Next what we have to do First we need to calculate The variance covariance Matrix This is the determinant value Raise to the power minus 1 by 2 And then Variance covariance matrix This is the inverse We have these values We need the determinant Of variance covariance matrix And the inverse of variance matrix Using a matrix calculator We find this Now with matrix calculator We can easily find this And next I am telling you That we will find this with excel So we will easily find this With excel We will find the inverse And we will find the determinant Of variance covariance matrix Here is the value of the sigma That is the 2 We enter the variance covariance matrix 2 1.5 And 0.8 0.8 And next is the 1.5 3 And 1.2 Final is the 0.8 1.2 And the 2.5 This is the variance covariance matrix Of sigma We need the determinant And the inverse For the determinant Equal to m Determinant m stands for matrix I have to calculate Determinant of this matrix Highlight Racket close And m determinant We have to enter shift control Enter Now here is the value Of the determinant 7.4 0.5 Next up we have to do Sigma inverse For sigma inverse I don't need the inverse Equals to 3x3 inverse I have highlighted 3x3 Because 3x3 inverse So equals to m inverse This is the m inverse Sigma Variance covariance matrix I need inverse Then bracket close Shift control Enter So here is the value Now we have the determinant Of sigma And next we have the inverse Okay now we will enter this In pdf and after entering We will have a solution Next what we have to do What will we have? Determinant value Sigma inverse value This is the x- mu And transpose And this is the x- mu This is the sigma inverse We have seen the order 3x3 is the order 3x1 is the final order We have 3x1 Inverse we have x- mu So I will write the value of x- mu In excel x- mu What was x- mu? I have minus 0.5 Then 0.2 Then 0.6 Okay now we need the multiplication So equals to Now what is the result First we have to see the result 3x1 So here I am taking 3x1 3x1 column Equals to Multiplication is m m Multiplier Metrics of multiplication Metrics of multiplication Highlight this Then Comma Multiplier by this call Okay Close Shift Control Enter So these are the values We have multiplied the mu This is the value Now let us check the final Now you have the value 3x1 Now see the order This is the order Order 1 row 3 columns And next we have 3x1 Final result is You have a single value 1x1 Means single value is coming Now we have to take this Multiply it on excel Or you have Transpose what will happen to us Convert the rows to columns I have the values Here minus 0.5 Then 0.2 Then 0.6 Now I have to multiply this Again Equals to M Multiply M Multiply Then Comma Array Next Multiplication Enter Shift Control Enter So we have the value 0.4 9 0.2 0.5 Now we have Particular point in PDF Now I have entered 0.4 9 0.7 Its PDF's value This factor This factor's value Determined Then multiply by minus 1 by 2 Further we have multiplied it Minus 1 by 2 Say we have taken its exponential 2 pi pi You know that pi which is equals to 22 by 7 Its value has been entered 2 multiplied by 22 by 7 Race to the power 3 by 2 Now we will find it on excel Now write 2 into 22 by 7 On excel And its power It has a power button Shift If I press it Then the power button will come Multiply by 3 by 2 Then its determinant Its power half And after that After all the calculation I have the result 0.001689 This value is particular What is this value This is the PDF What is its interpretation What are we showing it The probability At the point X1 1.5 X's value was given We had to find the PDF on that particular point What is the interpretation The probability at the point Of X1 1.5 X2 3.2 And X3 4.6 Is approximately Equals to 0.001689 On this particular point Probability of X Is determinant 0.001689 This is the whole scenario Of this example This is the multivariate normal distribution That we have checked the PDF Of normal distribution Mathematically Then we are checking it numerically We solved it with 3 by 3 Metrics We have determined its probability point Dear students What we have determined with this particular example We have 3x3 Multivariate data And on that We have determined the probability point Of the PDF When we have given X's Different values We had 3x3 We solved it according to the three dimensions Example multivariate normal distribution Mathematically we solved it We have also removed its mean Now on this particular point We have to determine its probability Numerically We have solved it with the multivariate normal distribution With the PDF On this particular X's values Our probability determinant X is 0.001689 Something is coming