# Constant 1D Kalman Filter Is Exponential Or Cumulative Average

Understand the Kalman filter with a simple one-dimensional example. It converges to cumulative or exponential average in case of constant measurement uncertainty and process noise.

The Kalman filter (Kalman 1960 paper) allows combination of two estimates for a variable: the transition model (e.g. momentum of physical particle) and estimated variance of sensor measurement (outside-the-model measurement uncertainty) to achieve better precision. In other words, we use the knowledge of the “physics” of the process to predict the future position and then combine that with actual sensor measurement. This gives us higher precision. Additionally, the model can incorporate “control-input”, but that will be omitted from below. The Kalman filter, also known as linear quadratic estimation (LQE), is an iterative algorithm, that uses both knowledge of the state motion and noisy measurements to estimate values and variance of unknown variables.

### Kalman Filter’s State, Observation, and State Estimate

At the time $$k$$, let:

• the true hidden state be $$x_k$$,
• the observed measurement be $$z_k$$,
• the hidden state estimate be $$m_k$$.

The model has the following components:

### Kalman Filter’s State Transition Model

• the state-transition model (motion model), which maps from the previous step to the current estimate: $$F_k$$
• the process noise normal variance, which adds the noise of the transition model $$Q_k$$

These two give us the motion estimate: $$m_k = F_k x_{k-1} + \mathcal{N}(0, Q_k)$$

### Kalman Filter’s Observation Model

• the observation model, which maps from the current sensor outputs to the estimate: $$H_k$$
• the observation noise normal variance, which adds the noise of the sensor outputs: $$R_k$$

These two give us the observation estimate: $$z_k = H_k x_{k} + \mathcal{N}(0, R_k)$$

### Kalman Filter’s Estimate

Read other sources for details, but in short:

• the state estimate: $$m_k = F_k m_{k-1} + K_k (z_k - H_k F_k m_{k-1})$$
• with a covariance: $$P_k = (1 - K_k H_k) (F_k P_{k-1} F_k^\intercal + Q_k)$$
• and Kalman gain: $$K_k := (F_k P_{k-1} F_k^\intercal + Q_k) H_k^\intercal$$ $$\bigl( H_k (F_k P_{k-1} F^\intercal_k + Q_k) H_k^\intercal + R_k \bigr)^{-1}$$

### Kalman Filter Simplified

In some applications, we can often set measure the state directly $$H_k = 1$$, and the transition model, noise, and measurement error are nearly constant. Then we get simpler equations:

• the state estimate: $$m_k = F m_{k-1} + K_k (z_k - F m_{k-1})$$ $$= (1 - K_k) F m_{k-1} + K_k z_k$$
• with a covariance: $$P_k = (1 - K_k) (F P_{k-1} F^\intercal + Q)$$
• and Kalman gain: $$K_k := (F P_{k-1} F^\intercal + Q)$$ $$\bigl( (F P_{k-1} F^\intercal + Q)+ R \bigr)^{-1}$$

In above, we can see that the Kalman gain balances between measurement and process estimates based on comparison of the variances of the two sources.

## One-dimensional Kalman Filter

We can see that the Kalman filter is fairly complex. So it is interesting to understand, what the Kalman filter reduces to in trivial cases.

Find below proof that in 1D ($$F_k = 1, H_k = 1$$) with constant measurement uncertainty $$R_k = R$$ and process noise $$Q_k = Q$$ asymptotically behaves as:

• cumulative average in case of zero process noise $$Q = 0$$
• exponential average in case of non zero process noise $$Q > 0$$

Additionally, since the Kalman filter equations are differentiable, it is reasonable to expect that above could be generalized to nearly-constant uncertainty and process noise. The proofs rely on the fact that the Kalman filter asymptotically doesn’t depend on initial state.

### 1D Kalman Filter vs Cumulative Average

For the case where $$Q = 0$$, the proof relies on a choice of initial value of the Kalman variance $$P_1 = R$$.

• $$m_{k+1} = m_k + K_{k+1} (z_{k+1} - m_k)$$
• $$P_{k+1} = (1-K_{k+1}) P_k$$
• $$K_{k+1} = \frac{P_k}{P_k + R}$$

$$P_{k+1} = (1 - K_{k+1}) P_k = \frac{R P_k}{P_k + R}$$

$$P_1 := R$$, $$P_2 = R / 2$$, $$P_3 = R / 3$$, …, $$P_k = R / k$$

$$K_{k+1} = \frac{R}{k (R / k + R)} = \frac{1}{k+1}$$

Which gives us recursive equation that match cumulative average equation: $$m_{k+1} = m_k + \frac{z_{k+1} - m_k}{k + 1}$$

We can double-check above proof by plotting the convergence.

### 1D Kalman Filter vs Exponential Average

Now we approach the case where $$Q > 0$$. Below is the proof relies on setting initial value of Kalman variance $$P_0$$ such that $$P_k$$ becomes constant for recursive equation to match exponential moving average equation.

• $$m_{k+1} = m_k + K_{k+1} (z_{k+1} - m_k)$$
• $$P_{k+1} = (1-K_{k+1}) (P_k + Q)$$
• $$K_{k+1} = \frac{P_k + Q}{P_k + Q + R}$$

$$P_{k+1} = (1 - \frac{P_k + Q}{P_k + Q + R}) (P_k + Q)$$ $$= \frac{P_k + Q + R - P_k - Q}{P_k + Q + R} (P_k + Q) = \frac{R (P_k + Q)}{P_k + Q + R}$$

Let’s select $$P_0$$, such that following holds true:

$$P_0 = \frac{R (P_k + Q)}{P_k + Q + R}$$ $$P_0^2 + (Q+R) P_0 - RP_0 - QR = 0$$

Since $$P_0 > 0$$ and $$Q > 0$$, then

$$\implies P_0 = \frac{-Q + \sqrt{Q^2 + 4 QR}}{2}$$

Which also leads to constant covariance and Kalman gain:

$$P_k = P_0$$, $$K_{k+1} = K_1$$

So we get recursive equation for exponential moving average:

$$m_{k+1} = K_1 z_{k+1} + (1-K_1) m_k$$

Plot of the convergence to exponential moving average:

## Kalman Filter Applications

The Kalman filter can be used in to keep a system in a state of control. Read more about application of Kalman filter in PID Controller.

### Kalman Filter Python Implementation in 1D

Below is simplistic implementation of Kalman filter in one dimension in Python used to generate plots presented above.

import matplotlib.pyplot as plt
import random
from statistics import mean
import pandas as pd

def current_k(p: float) -> float:
return (p + q) / (p + q + r)

def next_p(p: float, k: float) -> float:
return (1 - k) * (p + q)

def next_m(m, z, k: float) -> float:
return m + k * (z - m)

# configuration of the Kalman filter
r = 1
q = 0
p = 1
m = 1

# zs is the measured with noise
zs = []

# variables of the Kalman filter
# variance estimate
ps = []
# kalman gain
ks = []

# m is the smoothed output
ms = []
cumulative_avg = []
exponential_avg = []
count = 50

for i in range(count):
zs.append(random.gauss(0, 1))

m = zs[0]

for i in range(count):
k = current_k(p)
ks.append(k)
p = next_p(p, k)
ps.append(p)
m = next_m(m, zs[i], k)
ms.append(m)
cumulative_avg.append(mean(zs[:i + 1]))

exponential_avg = pd.Series(zs).ewm(alpha=ks[-1]).mean()

plt.plot(ks, label='ks')
plt.plot(ps, label='ps')
plt.legend()
plt.show()

plt.plot(ms, label='kalman filter', linewidth=1)
plt.plot(cumulative_avg, label='cumulative avg', linestyle='dashed')
plt.plot(exponential_avg, label='exponential moving avg', linewidth=3, linestyle='dashed')
plt.legend()

plt.savefig('../images/test.png')
plt.show()


Created on 28 Aug 2019. Updated on: 27 Aug 2022.