Mixing
Linear transformation of multivariate normal distribution to a higher dimension?
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
asked Nov 21 '18 at 9:38
sh10
12
add a comment |
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
asked Nov 21 '18 at 9:38
sh10
12
add a comment |
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
asked Nov 21 '18 at 9:38
sh10
12
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
probability normal-distribution
asked Nov 21 '18 at 9:38
sh10
12
asked Nov 21 '18 at 9:38
sh10
12
asked Nov 21 '18 at 9:38
sh10
12
asked Nov 21 '18 at 9:38
sh10
12
asked Nov 21 '18 at 9:38
sh10
12
12
add a comment |
add a comment |
1 Answer
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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.)
answered Nov 21 '18 at 9:57
kolobokish
40438
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.)
answered Nov 21 '18 at 9:57
kolobokish
40438
add a comment |
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.)
answered Nov 21 '18 at 9:57
kolobokish
40438
add a comment |
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.)
answered Nov 21 '18 at 9:57
kolobokish
40438
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.)
answered Nov 21 '18 at 9:57
kolobokish
40438
answered Nov 21 '18 at 9:57
kolobokish
40438
answered Nov 21 '18 at 9:57
kolobokish
40438
answered Nov 21 '18 at 9:57
kolobokish
40438
40438
add a comment |
add a comment |
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