如果参数完全拟合，为什么“curve_fit"不能估计参数的协方差? [英] Why isn't `curve_fit` able to estimate the covariance of the parameter if the parameter fits exactly?

问题描述

我不明白 curve_fit 无法估计参数的协方差，因此引发了下面的 OptimizeWarning.以下 MCVE 解释了我的问题:

I don't understand curve_fit isn't able to estimate the covariance of the parameter, thus raising the OptimizeWarning below. The following MCVE explains my problem:

MCVE python 代码段

from scipy.optimize import curve_fit
func = lambda x, a: a * x
popt, pcov = curve_fit(f = func, xdata = [1], ydata = [1])
print(popt, pcov)

输出

python-3.4.4libsite-packagesscipyoptimizeminpack.py:715:
OptimizeWarning: Covariance of the parameters could not be estimated
category=OptimizeWarning)

[ 1.] [[ inf]]

对于 a = 1，该函数完全适合 xdata 和 ydata.为什么错误/方差不是 0 或接近 0 的东西，而是 inf ?

For a = 1 the function fits xdata and ydata exactly. Why isn't the error/variance 0, or something close to 0, but inf instead?

引用自 curve_fit SciPy 参考指南:

There is this quote from the curve_fit SciPy Reference Guide:

如果解中的雅可比矩阵没有满秩，则 'lm' 方法返回一个填充了 np.inf 的矩阵，另一方面，'trf' 和 'dogbox' 方法使用 Moore-Penrose 伪逆来计算计算协方差矩阵.

If the Jacobian matrix at the solution doesn’t have a full rank, then ‘lm’ method returns a matrix filled with np.inf, on the other hand ‘trf’ and ‘dogbox’ methods use Moore-Penrose pseudoinverse to compute the covariance matrix.

那么，潜在的问题是什么?为什么解处的雅可比矩阵没有满秩?

So, what's the underlying problem? Why doesn't the Jacobian matrix at the solution have a full rank?

推荐答案

参数协方差的公式 (维基百科) 在分母中有自由度数.自由度计算为(数据点数)-(参数数)，在您的示例中为 1 - 1 = 0.而这个是SciPy检查在除以它之前的自由度数.

The formula for the covariance of the parameters (Wikipedia) has the number of degrees of freedom in the denominator. The degrees of freedoms are computed as (number of data points) - (number of parameters), which is 1 - 1 = 0 in your example. And this is where SciPy checks the number of degrees of freedom before dividing by it.

使用 xdata = [1, 2], ydata = [1, 2] 你会得到零协方差(注意模型仍然完全拟合:精确拟合不是问题).

With xdata = [1, 2], ydata = [1, 2] you would get zero covariance (note that the model still fits exactly: exact fit is not the problem).

如果样本大小 N 为 1(样本方差公式的分母为 (N-1))，则这与未定义样本方差属于同一类问题.如果我们只从总体中抽取 size=1 的样本，我们不会将方差估计为零，我们对方差一无所知.

This is the same sort of issue as sample variance being undefined if the sample size N is 1 (the formula for sample variance has (N-1) in the denominator). If we only took size=1 sample out of the population, we don't estimate the variance by zero, we know nothing about the variance.

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