Science > Physics > what determines upper bound on eigenvalues of a covariance matrix?
| Topic: |
Science > Physics |
| User: |
"K Yee" |
| Date: |
25 Nov 2004 06:43:37 AM |
| Object: |
what determines upper bound on eigenvalues of a covariance matrix? |
Suppose M is an nxn covariance matrix of n normally distributed random
variables.
Thus, M is symmetric and positive definite.
Let z be the largest eigenvalue of M. In the limit that the size n of
M goes to infinity,
what are necessary and sufficient conditions such that:
i) z/n -> 0?
ii) 0 < z/n < infinity?
iii) z/n goes to infinity?
Thank you in advance.
Roderick
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| User: "Robert Israel" |
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| Title: Re: what determines upper bound on eigenvalues of a covariance matrix? |
25 Nov 2004 01:59:11 PM |
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In article <30d2a0d1.0411250443.3508d2b2@posting.google.com>,
K Yee <kentonyee@hotmail.com> wrote:
Suppose M is an nxn covariance matrix of n normally distributed random
variables.
Thus, M is symmetric and positive definite.
Positive semidefinite, actually.
Let z be the largest eigenvalue of M. In the limit that the size n of
M goes to infinity,
what are necessary and sufficient conditions such that:
i) z/n -> 0?
ii) 0 < z/n < infinity?
That will always be true except in the trivial case where M = 0.
Maybe what you mean is c <= z/n <= d where 0 < c < d < infinity?
iii) z/n goes to infinity?
Your question is not well-defined if you don't tell us how the M for one
n is supposed to be related to the M for the other n's. But I'll
assume that you mean you have a sequence of random variables X_j, and
M(n) is the covariance matrix of X_1 to X_n. Note that for any
v in R^n, v^T M(n) v = Variance(sum_{j=1}^n v_j X_j), so
the largest eigenvalue z(n) of M(n) is the maximum of
Variance(sum_{j=1}^n v_j X_j) subject to sum_{j=1}^n v_j^2 = 1.
Thus z(n)/n -> 0 is equivalent to saying that for every epsilon > 0,
there is N such that for all n >= N and all v in R^n,
Variance(sum_{j=1}^n v_j X_j) <= epsilon n sum_{j=1}^n v_j^2.
Similarly for the other cases.
Robert Israel
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
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| User: "K Yee" |
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| Title: Re: what determines upper bound on eigenvalues of a covariance matrix? |
26 Nov 2004 05:27:38 AM |
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(Robert Israel) wrote in message news:<co5dif$nlv$1@nntp.itservices.ubc.ca>...
Suppose M is an nxn covariance matrix of n normally distributed random
variables.
Thus, M is symmetric and positive semidefinite.
Let z be the largest eigenvalue of M. In the limit that the size n of
M goes to infinity,
what are necessary and sufficient conditions such that:
i) z/n -> 0?
ii) 0 < z/n < infinity?
That will always be true except in the trivial case where M = 0.
I'm afraid not so, Robert. Let e be a random variable with mean zero
and unit variance.
Let X(n) be the n-tuplet whose every component equals e.
Then let M(n) = var(X(n)). In this case, M is the nxn square matrix
whose every
element is 1. The largest eigenvalue of M in this case is z=n. Thus,
z/n=1 is strictly
non-zero for all n.
Roderick
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| User: "Robert B. Israel" |
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| Title: Re: what determines upper bound on eigenvalues of a covariance matrix? |
26 Nov 2004 11:13:41 AM |
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(K Yee) wrote in message news:<30d2a0d1.0411260327.4b8595f5@posting.google.com>...
(Robert Israel) wrote in message news:<co5dif$nlv$1@nntp.itservices.ubc.ca>...
Suppose M is an nxn covariance matrix of n normally distributed random
variables.
Thus, M is symmetric and positive semidefinite.
Let z be the largest eigenvalue of M. In the limit that the size n of
M goes to infinity,
what are necessary and sufficient conditions such that:
i) z/n -> 0?
ii) 0 < z/n < infinity?
That will always be true except in the trivial case where M = 0.
I'm afraid not so, Robert. Let e be a random variable with mean zero
and unit variance.
Let X(n) be the n-tuplet whose every component equals e.
Then let M(n) = var(X(n)). In this case, M is the nxn square matrix
whose every
element is 1. The largest eigenvalue of M in this case is z=n. Thus,
z/n=1 is strictly
non-zero for all n.
How is that "not so"? Last time I looked, 0 < 1 < infinity.
Robert Israel
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
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| User: "K Yee" |
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| Title: Re: what determines upper bound on eigenvalues of a covariance matrix? |
27 Nov 2004 03:43:27 AM |
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(Robert B. Israel) wrote in message news:<a9b60a7c.0411260913.228cedd2@posting.google.com>...
kentonyee@hotmail.com (K Yee) wrote in message news:<30d2a0d1.0411260327.4b8595f5@posting.google.com>...
(Robert Israel) wrote in message news:<co5dif$nlv$1@nntp.itservices.ubc.ca>...
Suppose M is an nxn covariance matrix of n normally distributed random
variables.
Thus, M is symmetric and positive semidefinite.
Let z be the largest eigenvalue of M. In the limit that the size n of
M goes to infinity,
what are necessary and sufficient conditions such that:
i) z/n -> 0?
ii) 0 < z/n < infinity?
That will always be true except in the trivial case where M = 0.
I'm afraid not so, Robert. Let e be a random variable with mean zero
and unit variance.
Let X(n) be the n-tuplet whose every component equals e.
Then let M(n) = var(X(n)). In this case, M is the nxn square matrix
whose every
element is 1. The largest eigenvalue of M in this case is z=n. Thus,
z/n=1 is strictly
non-zero for all n.
How is that "not so"? Last time I looked, 0 < 1 < infinity.
(ii) is true, but (i) is not so.
Roderick
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| User: "Robert Israel" |
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| Title: Re: what determines upper bound on eigenvalues of a covariance matrix? |
27 Nov 2004 10:31:04 PM |
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In article <30d2a0d1.0411270143.15bcc2ba@posting.google.com>,
K Yee <kentonyee@hotmail.com> wrote:
(Robert B. Israel) wrote in message
news:<a9b60a7c.0411260913.228cedd2@posting.google.com>...
kentonyee@hotmail.com (K Yee) wrote in message
news:<30d2a0d1.0411260327.4b8595f5@posting.google.com>...
(Robert Israel) wrote in message
news:<co5dif$nlv$1@nntp.itservices.ubc.ca>...
Suppose M is an nxn covariance matrix of n normally distributed random
variables.
Thus, M is symmetric and positive semidefinite.
Let z be the largest eigenvalue of M. In the limit that the size n of
M goes to infinity,
what are necessary and sufficient conditions such that:
i) z/n -> 0?
ii) 0 < z/n < infinity?
That will always be true except in the trivial case where M = 0.
I'm afraid not so, Robert. Let e be a random variable with mean zero
and unit variance.
Let X(n) be the n-tuplet whose every component equals e.
Then let M(n) = var(X(n)). In this case, M is the nxn square matrix
whose every
element is 1. The largest eigenvalue of M in this case is z=n. Thus,
z/n=1 is strictly
non-zero for all n.
How is that "not so"? Last time I looked, 0 < 1 < infinity.
(ii) is true, but (i) is not so.
Roderick
The "that" I was referring to was (ii), not (i).
Robert Israel
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
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