卡尔曼滤波的置信区间估计
我一直在努力实现一个卡尔曼滤波器，以搜索异常的二维数据集。非常类似于我在这里找到的优秀帖子。作为下一步，我想预测置信区间(例如，95%对下限和上限值的置信度)，我预测的下一个值将下降。因此，除了下面这条线之外，我还想生成两条额外的线，它们代表95%的信心，认为下一个值将在地板或天花板以下。
我假设我将使用由Kalman滤波器生成的每个预测返回的不确定性协方差矩阵(P)，但我不确定它是否正确。任何指导或参考如何做到这一点，将不胜感激！
上述文章中的代码会随着时间的推移生成一组测量值，并使用卡尔曼滤波来平滑结果。
import numpy as np
import matplotlib.pyplot as plt
def kalman_xy(x, P, measurement, R,
motion = np.matrix('0. 0. 0. 0.').T,
Q = np.matrix(np.eye(4))):
"""
Parameters:
x: initial state 4-tuple of location and velocity: (x0, x1, x0_dot, x1_dot)
P: initial uncertainty convariance matrix
measurement: observed position
R: measurement noise
motion: external motion added to state vector x
Q: motion noise (same shape as P)
"""
return kalman(x, P, measurement, R, motion, Q,
F = np.matrix('''
1. 0. 1. 0.;
0. 1. 0. 1.;
0. 0. 1. 0.;
0. 0. 0. 1.
'''),
H = np.matrix('''
1. 0. 0. 0.;
0. 1. 0. 0.'''))
def kalman(x, P, measurement, R, motion, Q, F, H):
'''
Parameters:
x: initial state
P: initial uncertainty convariance matrix
measurement: observed position (same shape as H*x)
R: measurement noise (same shape as H)
motion: external motion added to state vector x
Q: motion noise (same shape as P)
F: next state function: x_prime = F*x
H: measurement function: position = H*x
Return: the updated and predicted new values for (x, P)
See also http://en.wikipedia.org/wiki/Kalman_filter
This version of kalman can be applied to many different situations by
appropriately defining F and H
'''
# UPDATE x, P based on measurement m
# distance between measured and current position-belief
y = np.matrix(measurement).T - H * x
S = H * P * H.T + R # residual convariance
K = P * H.T * S.I # Kalman gain
x = x + K*y
I = np.matrix(np.eye(F.shape[0])) # identity matrix
P = (I - K*H)*P
# PREDICT x, P based on motion
x = F*x + motion
P = F*P*F.T + Q
return x, P
def demo_kalman_xy():
x = np.matrix('0. 0. 0. 0.').T
P = np.matrix(np.eye(4))*1000 # initial uncertainty
N = 20
true_x = np.linspace(0.0, 10.0, N)
true_y = true_x**2
observed_x = true_x + 0.05*np.random.random(N)*true_x
observed_y = true_y + 0.05*np.random.random(N)*true_y
plt.plot(observed_x, observed_y, 'ro')
result = []
R = 0.01**2
for meas in zip(observed_x, observed_y):
x, P = kalman_xy(x, P, meas, R)
result.append((x[:2]).tolist())
kalman_x, kalman_y = zip(*result)
plt.plot(kalman_x, kalman_y, 'g-')
plt.show()
demo_kalman_xy()回答 3
Stack Overflow用户
发布于 2014-06-28 14:07:35
1西格玛间期的二维推广是以方程(x-mx).T P^{-1}.(x-mx)==1为特征的置信椭圆，其中x为参数2D -向量，mx为2D均值或椭圆中心，P^{-1}为逆协方差矩阵。关于如何绘制一个图，请参阅此回答。与西格玛间隔一样，椭圆面积对应于真实值在此范围内的固定概率。通过使用因子n (缩放间隔长度或椭圆半径)进行缩放，可以获得更高的置信度。请注意，因子n在一维和二维上有不同的概率：
|`n` | 1D-Intverval | 2D Ellipse |
==================================
1 | 68.27% | 39.35%
2 | 95.5% | 86.47%
3 | 99.73% | 98.89%在2D中计算这些值有点复杂，不幸的是，我没有公开引用它。
Stack Overflow用户
发布于 2014-07-21 06:53:03
Stack Overflow用户
发布于 2015-01-17 15:37:38
因为你的统计数据当然是从一个样本中得到的，所以人口统计量大于2西格玛标准差的概率是0.5。因此，如果你没有应用2x标准偏差的上置信度因子，那么我会考虑考虑是否有一个很好的预测值的重要性，如果你没有应用2x标准偏差的上置信度因子，那么下一个测度的概率将低于0.95。这一因素的大小将取决于得出0.5人口概率所使用的样本大小。求协方差矩阵的样本规模越小，求出0.95概率的因子越大，总体0.95统计量就越小。
https://stackoverflow.com/questions/24398811
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