Part 1 · Probability & Statistics

1.4 Expected value and variance

“Average” and “spread” — the same average does not mean the same safety

9-minute readPrerequisite: 1.3 Probability distributions▶ Video (about 4 min)

What you will learn

Calculate the expected value as “an average weighted by probability”
Express “how spread out” something is with the variance and standard deviation
Explain why the same average can mean different safety when the spread differs

Build intuition

Two automated vehicles, A and B, were each tested six times on stopping before a stop line. The stopping positions (how many metres before the line) were as follows. A negative value means the vehicle crossed the line.

	Run 1	Run 2	Run 3	Run 4	Run 5	Run 6	Mean
Vehicle A	1.1	0.9	1.0	1.0	1.1	0.9	1.0 m
Vehicle B	1.8	0.2	1.5	−0.3	1.2	1.6	1.0 m

Both vehicles stop 1.0 m before the line on average. But B’s positions are widely scattered, and in one run out of six it crossed the line.

Looking only at the average, A and B have the same result: “1.0 m before the stop line”. Yet B crossed the line in one run out of six. To talk about safety, we need to express the spread in numbers as well as the average.

Check with math

① Expected value: an average weighted by probability

The “average value” of a quantity $X$ that follows a probability distribution (1.3) is called its expected value, $E[X]$. You multiply each possible value by its probability and add everything up.

$$E[X] = \sum_{i} x_i \, P(X = x_i)$$

Example: if the number of driver interventions $X$ on a certain route is 0 (50%), 1 (30%) or 2 (20%), then

$$E[X] = 0 \times 0.5 + 1 \times 0.3 + 2 \times 0.2 = 0.7$$

The data average $\bar{x}$ (0.2) is an estimate of this expected value from data. The 1.0 m averages in the table above follow the same idea.

② Variance and standard deviation: the size of the spread

Spread is measured as “deviation from the mean”. But positive and negative deviations cancel out, so we square them before averaging. The result is the variance. Because the variance is in squared units (m²), we take its square root to return to the original units; this is the standard deviation $\sigma$.

$$\mathrm{Var}(X) = E\big[(X - \mu)^2\big], \qquad \sigma = \sqrt{\mathrm{Var}(X)}$$

Here $\mu = E[X]$ (the mean). Let us calculate with the data in the table. A’s deviations are 0.1, −0.1, 0, 0, 0.1 and −0.1, so

$$\sigma_A = \sqrt{\frac{0.1^2 + 0.1^2 + 0^2 + 0^2 + 0.1^2 + 0.1^2}{6}} \approx 0.08\ \mathrm{m}$$

B’s deviations are 0.8, −0.8, 0.5, −1.3, 0.2 and 0.6, so

$$\sigma_B = \sqrt{\frac{0.8^2 + 0.8^2 + 0.5^2 + 1.3^2 + 0.2^2 + 0.6^2}{6}} \approx 0.78\ \mathrm{m}$$

B’s spread is about ten times A’s. Note that when estimating the “true” spread from sample data, we sometimes divide by 6 − 1 = 5 instead of 6. The reason is covered in 1.6.

③ Expected value and variance of the distributions in 1.3

Distribution	Expected value	Variance	In the 1.3 examples
Binomial (n, p)	$np$	$np(1-p)$	n = 10, p = 0.9 → expected 9, σ ≈ 0.95
Poisson (m)	$m$	$m$	m = 2 → expected 2, σ ≈ 1.41
Normal (μ, σ)	$\mu$	$\sigma^2$	μ and σ are directly the mean and spread

④ Setting a margin that accounts for the spread

If we assume the stopping positions follow a normal distribution, we can use the rule of thumb from 1.3 (about 99.7% within the mean ± 3σ).

Vehicle A: mean − 3σ = 1.0 − 3 × 0.08 ≈ 0.76 m. It almost certainly stops before the line
Vehicle B: mean − 3σ = 1.0 − 3 × 0.78 ≈ −1.3 m. The probability of crossing the line is estimated at about 10%

To bring B’s probability of crossing the line down to 0.1% or less, you would need to aim for a mean at least 3.1σ ≈ 2.4 m before the line — or reduce the spread σ itself. In safety design, this calculation is how you compare the two options: increase the margin or reduce the spread. That said, six runs are really too few to pin down σ; how reliable such estimates are is covered in 1.6.

Advanced: weighting per-condition rates by exposure

The expected value in ① was a sum of “value × probability”. The same form appears again and again in safety estimates. When the accident rate differs by condition (weather, road type, time of day and so on), the rate over the whole ODD is the expected value of the per-condition rates $\lambda_i$, weighted by the share of driving spent in each condition, $w_i$ (the exposure).

$$\lambda_{\text{overall}} = \sum_i w_i\, \lambda_i$$

Example: 90% of driving is in clear weather and 10% in rain, and the accident rate in rain is 10 times that in clear weather. Taking the clear-weather rate as 1 (a relative value), the overall rate is 0.9 × 1 + 0.1 × 10 = 1.9 — about twice what you would measure in clear weather alone. Moreover, more than half of all accidents (1.0 ÷ 1.9 ≈ 53%) happen in rain, which is only a tenth of the driving. If the share of rain rises to 20% the rate becomes 2.8; if rain is excluded from the ODD (the function is not used in rain) it returns to 1.

The formula teaches three things. (1) An average over everything is dragged by a few harsh conditions. (2) If the mix of test conditions differs from the real exposure (testing only on clear days), the estimate is off. (3) Both the per-condition rates $\lambda_i$ and the exposures $w_i$ have to be measured. Tabulating the ODD conditions to check test coverage (5.6, 5.9) and adding up the estimates of unknown hazards condition by condition (5.2) are both calculations of this form. How the ODD is defined is covered in 4.1.

Summary

The expected value is the sum of “value × probability”; the data average is an estimate of it
The variance is “the average of the squared deviations from the mean”, and the standard deviation is its square root (back in the original units)
Even with the same average, a large spread makes it easier to fall on the dangerous side. Safety margins are set with σ in mind

Quick quiz

Q1. If the number of interventions is 0 (60%), 1 (30%) or 3 (10%), what is the expected value?

0.6. 0 × 0.6 + 1 × 0.3 + 3 × 0.1 = 0.6.

Q2. What are the variance and standard deviation of the data 2, 4, 6? (Divide by the count, 3.)

Variance about 2.67, standard deviation about 1.63. The mean is 4 and the deviations are −2, 0 and 2, so the average of the squares is (4 + 0 + 4) ÷ 3 ≈ 2.67; its square root is about 1.63.

Q3. If the stopping position averages 2.0 m before the line with a standard deviation of 1.0 m (assume a normal distribution), roughly what is the probability of crossing the line?

About 2.5% (more precisely about 2.3%). The stop line lies 2σ below the mean. About 5% falls outside ±2σ, and only half of that is on the crossing side, so about 2.5%.

Distribution	Expected value	Variance	In the 1.3 examples
Binomial (n, p)	\(np\)	\(np(1-p)\)	n = 10, p = 0.9 → expected 9, σ ≈ 0.95
Poisson (m)	\(m\)	\(m\)	m = 2 → expected 2, σ ≈ 1.41
Normal (μ, σ)	\(\mu\)	\(\sigma^2\)	μ and σ are directly the mean and spread