Linear transformation of multivariate normal distribution to a higher dimension?











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Suppose I have transformation defined as $Y_{qtimes 1} = C_{qtimes p}X_{ptimes 1}$, where $X sim N_p(mu, Sigma)$. If $q > p$ how do I compute the distribution of $Y$, since I think the standard result $Y sim N_q(Cmu, CSigma C^T)$ will fail as $CSigma C^T$ will be rank deficient. Please help?










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    Suppose I have transformation defined as $Y_{qtimes 1} = C_{qtimes p}X_{ptimes 1}$, where $X sim N_p(mu, Sigma)$. If $q > p$ how do I compute the distribution of $Y$, since I think the standard result $Y sim N_q(Cmu, CSigma C^T)$ will fail as $CSigma C^T$ will be rank deficient. Please help?










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      Suppose I have transformation defined as $Y_{qtimes 1} = C_{qtimes p}X_{ptimes 1}$, where $X sim N_p(mu, Sigma)$. If $q > p$ how do I compute the distribution of $Y$, since I think the standard result $Y sim N_q(Cmu, CSigma C^T)$ will fail as $CSigma C^T$ will be rank deficient. Please help?










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      Suppose I have transformation defined as $Y_{qtimes 1} = C_{qtimes p}X_{ptimes 1}$, where $X sim N_p(mu, Sigma)$. If $q > p$ how do I compute the distribution of $Y$, since I think the standard result $Y sim N_q(Cmu, CSigma C^T)$ will fail as $CSigma C^T$ will be rank deficient. Please help?







      probability normal-distribution






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      asked Nov 21 at 9:38









      sh10

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          I'm not sure if this answer is satisfying.
          What you have written I right up to the place where you concluded that the above form cannot represent covariance matrix.
          Covariance matrix can indeed be not of a full rank.



          Look answer of this question for example.



          111.



          Anyway, you can consider simple example with $p=2$, $q=3$. The last row of $Y$ then can always be represented with first two. (If rows of $C$ are linearly independent.)






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            1 Answer
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            1 Answer
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            active

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            up vote
            0
            down vote













            I'm not sure if this answer is satisfying.
            What you have written I right up to the place where you concluded that the above form cannot represent covariance matrix.
            Covariance matrix can indeed be not of a full rank.



            Look answer of this question for example.



            111.



            Anyway, you can consider simple example with $p=2$, $q=3$. The last row of $Y$ then can always be represented with first two. (If rows of $C$ are linearly independent.)






            share|cite|improve this answer

























              up vote
              0
              down vote













              I'm not sure if this answer is satisfying.
              What you have written I right up to the place where you concluded that the above form cannot represent covariance matrix.
              Covariance matrix can indeed be not of a full rank.



              Look answer of this question for example.



              111.



              Anyway, you can consider simple example with $p=2$, $q=3$. The last row of $Y$ then can always be represented with first two. (If rows of $C$ are linearly independent.)






              share|cite|improve this answer























                up vote
                0
                down vote










                up vote
                0
                down vote









                I'm not sure if this answer is satisfying.
                What you have written I right up to the place where you concluded that the above form cannot represent covariance matrix.
                Covariance matrix can indeed be not of a full rank.



                Look answer of this question for example.



                111.



                Anyway, you can consider simple example with $p=2$, $q=3$. The last row of $Y$ then can always be represented with first two. (If rows of $C$ are linearly independent.)






                share|cite|improve this answer












                I'm not sure if this answer is satisfying.
                What you have written I right up to the place where you concluded that the above form cannot represent covariance matrix.
                Covariance matrix can indeed be not of a full rank.



                Look answer of this question for example.



                111.



                Anyway, you can consider simple example with $p=2$, $q=3$. The last row of $Y$ then can always be represented with first two. (If rows of $C$ are linearly independent.)







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 21 at 9:57









                kolobokish

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