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Levenberg-Marquardt covariance matrix

I see that the Levenberg Marquardt vi returns the 1/2 the inverse of the Hessian as the parameter covariance matrix.  Calling the data y_i and the parameters a_j, I thought the correct formula for the inverse of parameter covariance matrix was [sigma_a^2]^-1 = J^T [sigma_y^2]^{-1} J, where J_ij = d yfit_i/d a_j are the Jacobian elements and [sigma_y^2] is the covariance matrix for the data (typically sigma_i^2 along the diagonal for independent y_i).  This formula gives the correct units for [sigma_a^2]_jk, with elements having the units of a_ja_k.  Can someone give me the formula for the covariance matrix from this vi, both when a weighting matrix is and is not supplied?

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OK sorry about the post.  Assuming the data point weights supplied to the Levenberg-Marquardt vi are 1/sigma_i^2, the returned covariance matrix is the inverse of J^T [sigma_y^2]^{-1} J.  This is the correct formula.

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