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1 Probability and Random Variables/Processes for Wireless Communications. Professor Aditya K Jagannatham. Department of Electrical Engineering. Indian Institute of Technology Kanpur. Lecture -15. Special Case: IID Gaussian Random Variables. Hello, welcome to another module in this massive open online course on probability and random variables for wireless communication. So, in the previous module, we started looking at Gaussian random variables and the various key properties of Gaussian random variables. One of the important properties of the Gaussian random variable we said is the following thing that is, (Refer Slide Time: 0:36) if I have L Gaussian random variable X1, X2 and XL are Gaussian random variables with expected Xi equals mu i expected Xi minus mu i that is the mean of the Gaussian random variable i is mu i. The variance of the Gaussian random variable i is Sigma i square, that is expected Xi minus mu i square. Further if I look at the covariance, that is we said, if I look at the expected value of Xi minus mu i times Xj minus mu j, we said this is equal to the this is equal to the quantity sigma ij and this we said is the this we said is the covariance.

2 (Refer Slide Time: 1:46) Now, we said if I generate a new random variable X which is generated as a linear combination of these Gaussian random variables a1 X1 plus a2 X2 plus so on up to al XL, so this is a Gaussian random variable which is generated as a linear combination or a linear transformation, a linear combination of X1 X2 up to X L, then we said this X is a Gaussian random variable. Very interestingly, the Gaussian random variable which is generated as a linear combination of a group of random variables is in turn because in random variable and we also calculated the mean and the variance of this new Gaussian random variable. (Refer Slide Time: 2:44)

3 So we said this X, the most important of the key properties, if we said is X is, this is internal Gaussian random variable and the mean and variance of X are given as N, this is the mean is summation ai mu i and the variance is summation i summation j ai A or summation I summation over I ai Sigma I square plus summation over I summation over J J0 equal to I ai aj Sigma ij, so this is the mean mu and this is the variance Sigma square of this Gaussian random variable X which is generated as the linear combination of X1, X2 XL, using the weighting coefficients that a1, a2 up to al. (Refer Slide Time: 4:07) Let us now consider a special case of this linear combination, let us consider a special case, so we are going to now consider a special case of this linear combination. So, let us consider a special case of this linear combination where all the Gaussian random variables are 0 mean that is mu i that is expected Xi equals mu I equal 0. That is all the Gaussian random variables that were considering are 0, that is all the Gaussian random variables are 0. Further let us consider that all of them have Identical variance, that is expected Xi minus mu i square equals Sigma I square which is equal to Sigma square, so this is the variance. All the Gaussian random variables have Identical mean and in fact the mean is identically equal to 0. So, all the Gaussian random variables X1, X2, XL are Identical 0 mean Gaussian random variables and further, all the Gaussian random variables Xi have variance, that is expected value of Xi minus mu square is equal to Sigma i square which is equal to Sigma square that is the variance of all the Gaussian random variables is the variances are equal to Sigma square.

4 (Refer Slide Time: 5:54) So, these Gaussian random variables are basically identical Gaussian random variables. So, these are basically say because the mean and variances of these Gaussian random variables are Identical, so we say these, all the Gaussian random variables have 0 mean and all of them have variance Sigma square. (Refer Slide Time: 6:14) So, these are bunch or a group of identical RVs and further these are identical Gaussian RVs. These are identical Gaussian random variables.

5 (Refer Slide Time: 6:41) Further we are also going to assume that the covariance, that is expected value of Xi minus mu i Times Xj minus mu j (since mu i and mu j are equal to 0), this reduces to expected value of Xi Times Xj which we are going to assume equal to 0, so the covariance is basically equal to, the covariance of any 2 Gaussian random variables, covariance, so we are going to assume that the covariance is equal to 0. Such random variables are known as uncorrelated random variables, which means the covariance of 2 random variables is 0, they are known as uncorrelated random variables. So, we are assuming that all Gaussian random variables are identical and further, they are uncorrelated. So, these 2 random variables,, so this implies that Xi, Xj are uncorrelated. And specifically for the case of Gaussian random variable, uncorrelated also implies independence. This is not true for any random variable but specifically for the Gaussian random variable, uncorrelated, the property of uncorrelated implies that they are independent. And remember, this is not true for any random variable in general.

6 (Refer Slide Time: 8:43) This is only for the case of Gaussian s, this implies for Gaussian and only for Gaussian it implies independence. That is 2 Gaussian are uncorrelated, then they are also independent. So, therefore X1, X2, XL, remember that were saying they are because they have 0 mean, all of them have 0 mean, that is mean of all these Gaussian random variables is 0. All of these Gaussian random variables have identical variance Sigma square, further they are all any 2 pair, if you take any pair of Gaussian random variable X1 Xi and Xj, they are uncorrelated and besides specifically for the case of Gaussian random variables, they are uncorrelated, it also implies that they are independent. So, therefore were considering a group of L independent and identically distributed Gaussian random variable.

7 (Refer Slide Time: 9:57) So, we are considering group of X So X1, X2, XL, in this context, in this scenario, X1, X2, XL these are independent, remember we have seen this nomenclature before, independent and identically distributed random variables. That is each of these has mean 0, variance Sigma square and the covariance between these 2, the covariance between any 2 random variables is 0, which means they are uncorrelated and since they are Gaussian, this also means they are independent. So, were considering L independent identically distributed Gaussian random variables. Let us now again consider X which is generated as a linear combination of these Gaussian random variables. (Refer Slide Time: 10:59)

8 So, similar to before, let us consider the new random variable X which is generated as a1 X1 plus a2 X2 plus al XL which we can now write using vector operations as a1 a2 up to al, X1 X2 up to XL. Which I can write as, let me write this as vector A bar transpose this I can write as a vector X bar, so this I can write this as A bar transpose X bar where A bar is the vector a1 a2 al and X is the vector or X bar rather is the vector as we can see above X1 X2 XL. So, I can write this as the vector, so I can write this linear combination a1 X1 a2 X2 up to al XL as the row vector a1 a2 al times the column vector X1 X2 XL which is basically A bar transpose X bar. This is the new random variable X which we saw previously is Gaussian in nature. (Refer Slide Time: 12:34) Now what is the mean of this Gaussian random variable? The mean of this Gaussian random variable X expected value of X which is equal to mu which is equal to summation i ai of ai Times mu i but mu i we are saying each mu i is equal to 0. So, this reduces to summation over i ai times 0 which is equal to 0. So therefore mu is equal to 0. Since each of the Gaussian random variables Xi have 0 mean that is mu i equal to 0 for each random variable Xi, summation ai mu i is summation ai times 0 which is 0, therefore mean mu of the Gaussian random variable X is 0. Now let us look at the variance, we have also derived an expression for the variance, the variance expected value of X minus mu square and we have shown that mu is equal to 0 is equal to the expected value of X square which is equal to summation over i Sigma i square

9 plus summation over i summation over j j not equal to i ai aj Sigma ij and remember we said Sigma ij, this is equal to this is equal to 0 because the covariance is equal to 0. (Refer Slide Time: 14:46) Further, each Sigma i square is equal to Sigma square, therefore we have the variance, this is equal to to Sigma square, therefore we have the variance expected value of X minus mu square which is equal to Sigma square which is equal to as we are seeing above summation of i ai square Sigma square which is equal to Sigma square summation over i ai square but summation over i ai square is nothing but the norm. So, summation over I ai square is a1 square plus a2 square up to al square which is nothing but the norm square of the vector A bar. So, this is basically equal to Sigma square norm of A bar whole square.

10 (Refer Slide Time: 15:42) Therefore for IID Gaussian random variables therefore for IID Gaussian random variables X1 and X2 up to XL we have X which is defined as A bar transpose X bar, this is Gaussian. (Refer Slide Time: 16:17) Further X can now be represented as we have calculated the mean, the mean we have said is 0, the various is Sigma Square norm of A bar square where A bar is nothing but the weighting vector, remember we have define A bar as the weighting vector a1 a2 up to al. So, what we have done is we have considered the special case, that is all the Gaussian random variables X1, X2 up to XL are 0 mean, they have identical variance Sigma Square and further they are uncorrelated, that is their covariance is, Sigma ij the covariance is equal

11 to 0. And for the Gaussian random variables, we said uncorrelated also means that these Gaussian random variables is are independent, therefore were considering a group of IID, that is independent identically distributed Gaussian random variables and then we said if we generate a new Gaussian random variable, if we consider a new random variable X which is a weighted combination of these IID Gaussian random variables, that is a1 X1 class a2 X2 so on up till al XL, then the mean of X is 0, that is X is 0 mean and the variance is Sigma square times norm of A bar square, that is Sigma Square Times a1 square plus a 2 square so on uptill al square and X is a Gaussian random variable in turn with this 0 mean and variance Sigma square nor of A bar square. (Refer Slide Time: 17:44) So, this is an interesting property which will in fact come very handy, which is in fact very useful when we consider, when we analyze various communications, the behavior and the performance of various communication system as well as wireless a medication systems. So, we have seen a special case of linear combination of a group of random variables. So, we will end this module here and proceed with other topics in subsequent modules. Thank you very much.