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2004/057 - entries of the variance-covariance matrix

Posted by: **jmclaugh** (IP Logged)

Date: 06 November 2011 18:20

Near to the bottom of page 5, this paper states that the vector (\hat{c}_1, ..., \hat{c}_m) will be "distributed around" the vector

((-1)^{z_{1}^{*}}c_1, ..., (-1)^{z_{m}^{*}}c_m)

according to a distribution with a variance-covariance matrix M in which:

* All entries not on the top-left - bottom-right diagonal are zero. (This clearly follows from the assumed independence of the linear approximations)

* All entries M_{ii} on this diagonal are equal to 1/sqrt(N).

Now, the paper has already relied on each t_j having variance \approx N/4. Since

(2t_{j}/N) - 1 = \hat{c}_j

we expect \hat{c}_j to have variance

(2/N)^{2} * (N/4) = 1/N.

Given this, I can't understand why the entries on the diagonal are 1/sqrt(N) and not 1/N. Can someone explain?

Edited 1 time(s). Last edit at 06-Nov-2011 18:23 by jmclaugh.

((-1)^{z_{1}^{*}}c_1, ..., (-1)^{z_{m}^{*}}c_m)

according to a distribution with a variance-covariance matrix M in which:

* All entries not on the top-left - bottom-right diagonal are zero. (This clearly follows from the assumed independence of the linear approximations)

* All entries M_{ii} on this diagonal are equal to 1/sqrt(N).

Now, the paper has already relied on each t_j having variance \approx N/4. Since

(2t_{j}/N) - 1 = \hat{c}_j

we expect \hat{c}_j to have variance

(2/N)^{2} * (N/4) = 1/N.

Given this, I can't understand why the entries on the diagonal are 1/sqrt(N) and not 1/N. Can someone explain?

Edited 1 time(s). Last edit at 06-Nov-2011 18:23 by jmclaugh.

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