Noise, mean, variance & bias

A sensor with two kinds of error

probe.read(name) measures a ball's position in metres. Every reading $m$ differs from the ball's true position $p$ in two ways:

$$m = p + b + v$$

• $v$ is noise: a fresh random error in every reading, about 12 mm per coordinate, as likely to be too big as too small.
• $b$ is bias: the same error in every reading, a few centimetres, set by how the probe is mounted. It changes every run.

Mean and standard deviation

For $N$ readings $m_1, \dots, m_N$ of one coordinate, the mean $\bar m$ is their average and the standard deviation $\sigma$ measures how spread out they are:

$$\bar m = \frac{1}{N}\sum_{i=1}^{N} m_i, \qquad \sigma = \sqrt{\frac{1}{N}\sum_{i=1}^{N} \left(m_i - \bar m\right)^2}$$

$\sigma^2$ is called the variance. Some books divide by $N - 1$ instead of $N$; either is fine here. A histogram of many readings makes a bell curve, the Gaussian (normal) distribution, centred on $\bar m$ and about $\sigma$ wide on each side.

Noise averages away, bias doesn't

The noise in each reading is different, so in an average much of it cancels. The mean's own error is about $\sigma/\sqrt N$: 400 readings pin it down to 12/20 = 0.6 mm. But the bias is the same every time, so the mean still sits at $p + b$. A few centimetres off is enough to miss the ball on most runs.

To measure a bias, measure something whose true position you know. The blue ball sits in the ring at world.targets()[0], so

$$b \approx \bar m_{\text{blue}} - p_{\text{blue}}$$

Then correct the red ball's mean: $\hat p_{\text{red}} = \bar m_{\text{red}} - b$, where the hat marks an estimate.

Noisy Sensors & Kalman Filters, step 1 averages noisy readings of a ball you can't see directly.

Your task

1. Write mean_and_std(samples). samples is an $N \times 3$ array (one reading per row); return (mean, std), each 3 numbers, one per coordinate.
2. Take 400 readings of the blue ball (the starter plots a histogram of their x-values) and store the bias in a global BIAS (3 numbers, in metres). It must be within 3 mm of the truth.
3. Average 400 readings of the red ball, subtract BIAS and put the red ball in the cup.