如何从 ODR 结果计算标准误差? [英] How to compute standard error from ODR results?

问题描述

我使用 scipy.odr 以适应 x 和 y 上的不确定性，并遵循这个问题 使用 scipy curve_fit 进行正确拟合，包括 x 中的误差?

拟合后，我想计算参数的不确定性.因此，我查看协方差矩阵的对角元素的平方根.我明白了:

<预><代码>>>>打印(np.sqrt(np.diag(output.cov_beta)))[ 0.17516591 0.33020487 0.27856021]

但是在 Output 中还有 output.sd_beta，根据 odr 上的文档

<块引用>

估计参数的标准误差，形状为 (p,).

但是，它并没有给我相同的结果:

<预><代码>>>>打印(输出.sd_beta)[ 0.19705029 0.37145907 0.31336217]

编辑

这是笔记本上的示例:https://nbviewer.jupyter.org/github/gvallverdu/cookbook/blob/master/fit_odr.ipynb

最小二乘

停止原因:['平方和收敛']参数:[-1.94792946 11.03369235 -5.43265555]信息:1sd_beta:[0.26176284 0.49877962 0.35510071]sqrt(diag(cov): [ 0.25066236 0.47762805 0.34004208]

有 ODR

停止原因:['平方和收敛']参数:[-1.93538595 6.141885 -3.80784384]信息:1sd_beta:[0.6941821 0.88909997 0.17292514]sqrt(diag(cov): [ 0.01093697 0.01400794 0.00272447]

解决方案

产生差异的原因是 sd_beta 是由残差方差缩放的，而 cov_beta 不是'

scipy.odr 是 ODRPACK FORTRAN 库的接口，它被薄薄地包裹在 __odrpack.c 中.sd_beta 和 cov_beta 通过索引到 FORTRAN 例程内部使用的 work 向量来恢复.work 中第一个元素的索引是名为 sd 和 vcv 的变量(参见 这里).

来自 ODRPACK 文档(第 85 页):

<块引用>

WORK(SDI) 是 p × 1 数组 SD 的第一个元素，包含函数参数β的标准差̂σβK，即协方差矩阵对角线项的平方根，其中

WORK(SDI-1+K) = SD(K) = ̂V 1/2 β (K, K) = ̂σβK

对于 K = 1,... ,p.

WORK(VCVI) 是 p × p 数组 VCV 的第一个元素，包含参数β 之前的协方差矩阵的值通过残差进行缩放，其中

WORK(VCVI-1+I+(J-1)*(NP)) = VCV(I,J) = ̂σ⁻²V β(I, J)

对于 I = 1,... ,p 和 J = 1,... ,p.

换句话说，np.sqrt(np.diag(output.cov_beta * output.res_var)) 会给你与 output.sd_beta 相同的结果.

我在此处打开了错误报告.

I use scipy.odr in order to make a fit with uncertainties on both x and y following this question Correct fitting with scipy curve_fit including errors in x?

After the fit I would like to compute the uncertainties on the parameters. Thus I look at the square root of the diagonal elements of the covariance matrix. I get :

>>> print(np.sqrt(np.diag(output.cov_beta)))
[ 0.17516591  0.33020487  0.27856021]

But in the Output there is also output.sd_beta which is, according to the doc on odr

Standard errors of the estimated parameters, of shape (p,).

But, it does not give me the same results :

>>> print(output.sd_beta)
[ 0.19705029  0.37145907  0.31336217]

EDIT

This is an example on a notebook : https://nbviewer.jupyter.org/github/gvallverdu/cookbook/blob/master/fit_odr.ipynb

With least square

stop reason: ['Sum of squares convergence']
        params: [ -1.94792946  11.03369235  -5.43265555]
          info: 1
       sd_beta: [ 0.26176284  0.49877962  0.35510071]
sqrt(diag(cov): [ 0.25066236  0.47762805  0.34004208]

With ODR

stop reason: ['Sum of squares convergence']
        params: [-1.93538595  6.141885   -3.80784384]
          info: 1
       sd_beta: [ 0.6941821   0.88909997  0.17292514]
sqrt(diag(cov): [ 0.01093697  0.01400794  0.00272447]

解决方案

The reason for the discrepancy is that sd_beta is scaled by the residual variance, whereas cov_beta isn't.

scipy.odr is an interface for the ODRPACK FORTRAN library, which is thinly wrapped in __odrpack.c. sd_beta and cov_beta are recovered by indexing into the work vector that's used internally by the FORTRAN routine. The indices of their first elements in work are variables named sd and vcv (see here).

From the ODRPACK documentation (p.85):

WORK(SDI) is the first element of a p × 1 array SD containing the standard deviations ̂σβK of the function parameters β, i.e., the square roots of the diagonal entries of the covariance matrix, where

WORK(SDI-1+K) = SD(K) = ̂V 1/2 β (K, K) = ̂σβK

for K = 1,... ,p.

WORK(VCVI) is the first element of a p × p array VCV containing the values of the covariance matrix of the parameters β prior to scaling by the residual variance, where

WORK(VCVI-1+I+(J-1)*(NP)) = VCV(I,J) = ̂σ⁻²V β(I, J)

for I = 1,... ,p and J = 1,... ,p.

In other words, np.sqrt(np.diag(output.cov_beta * output.res_var)) will give you the same result as output.sd_beta.

I've opened a bug report here.

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