Variance and standard deviation

Two games can share the same expected value and have nothing in common: one calm, the other risky. What the expected value ignores, the variance measures.

The problem

Game	Rule	Expectation
A	win €1 with probability $\dfrac{1}{2}$, lose €1 otherwise	$E = 0$
B	win €1000 with probability $\dfrac{1}{2}$, lose €1000 otherwise	$E = 0$

Same expectation, same fairness — and yet nobody confuses them. A measure of the spread of values around the mean is missing.

Variance

$$V(X) = E\left[\,(X - E(X))^2\,\right]$$

We measure each value's distance to the mean, square it — so that negative deviations do not cancel positive ones — and take the weighted average:

$$V(X) = \sum (x_i - E(X))^2 \cdot P(X = x_i)$$

The variance is always positive or zero. It is zero only if $X$ is constant.

The computational formula

Expanding the square gives a much faster formula, the König-Huygens one:

$$V(X) = E(X^2) - \left[E(X)\right]^2$$

« The mean of the squares minus the square of the mean. » This is the one used in practice. For a fair die:

$$E(X) = 3.5 \qquad E(X^2) = \frac{1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2}{6} = \frac{91}{6} \approx 15.17$$

$$V(X) = 15.17 - 3.5^2 = 15.17 - 12.25 = 2.92$$

Standard deviation

The variance is expressed in the square of the unit: euros² for a gain, which means nothing concrete. So we take its square root:

$$\sigma(X) = \sqrt{V(X)}$$

This is the standard deviation, in the unit of $X$. For the die, $\sigma = \sqrt{2.92} \approx 1.71$: results deviate on average by about $1.7$ from the mean of $3.5$.

Back to the two games above:

Game	Variance	Standard deviation
A	$V = 1$	$\sigma = 1$ €
B	$V = 10^6$	$\sigma = 1000$ €

Game B is a thousand times riskier. The standard deviation puts an exact figure on what intuition said.

The properties

$$V(aX + b) = a^2 \cdot V(X) \qquad \sigma(aX + b) = |a| \cdot \sigma(X)$$

Two lessons:

• adding a constant translates the whole distribution without changing its spread — dispersion is insensitive to shifts;
• multiplying by $a$ multiplies the variance by $a^2$, hence the standard deviation by $|a|$.

             μ                        μ + 5
     ___/\___                     ___/\___
    /        \        + 5        /        \          same shape,
   /          \     ------>     /          \         same standard deviation
  ------------------          ------------------

Careful though: $V(X + Y) = V(X) + V(Y)$ holds only if $X$ and $Y$ are independent. That is the big difference from the expected value, which always adds.

Summary

• The variance $V(X) = E[(X - E(X))^2]$ measures the spread around the mean.
• Computational formula: $V(X) = E(X^2) - [E(X)]^2$.
• The standard deviation $\sigma = \sqrt{V}$ is in the unit of $X$: that is what gets interpreted.
• $V(aX + b) = a^2 V(X)$: the shift $b$ has no effect on dispersion.
• $V(X + Y) = V(X) + V(Y)$ requires independence — unlike the expected value.