Variance of a discrete random variable

8 minAS Level5.4.1

AI-drafted, independently checked · not yet teacher-reviewed

By the end of this lesson

By the end you can calculate $E(X^2)$ , use it to find $Var(X) = E(X^2) - (E(X))^2$ , and take the square root to obtain the standard deviation.

Why the mean is not the whole story

Two random variables can share the same expectation and still behave completely differently. One might always give a value near its mean; the other might swing wildly.

Variance measures that difference. It is the average squared distance of $X$ from its mean, so a small variance means the values cluster tightly around $E(X)$ , and a large variance means they spread out.

Squaring does two jobs: it removes the sign, so values above and below the mean both count as distance, and it punishes large deviations more heavily than small ones.

Definition

Variance and standard deviation

For a discrete random variable $X$ ,

$Var(X) = E(X^2) - \left(E(X)\right)^2$

where $E(X^2) = \sum x^2 P(X = x)$ , computed by squaring each value before multiplying by its probability.

The standard deviation is $\sigma = \sqrt{Var(X)}$ , which returns to the original units of $X$ .

Variance is never negative, because a squared distance cannot be negative. If you get a negative answer, the arithmetic is wrong.

Why this formula measures spread

The definition of variance is $E\left((X - \mu)^2\right)$ : find the distance from the mean, square it, and average. That is exactly "average squared distance".

Here $\mu$ ("mu") is the mean, $E(X)$ . The next step uses two facts about expectation (you do not need to prove them): the expectation of a sum is the sum of the expectations, and a constant multiplier can be taken outside, so $E(2\mu X) = 2\mu E(X)$ and $E(\mu^2) = \mu^2$ .

Expanding the bracket gives $E(X^2 - 2\mu X + \mu^2) = E(X^2) - 2\mu E(X) + \mu^2$ . Since $E(X) = \mu$ , this becomes $E(X^2) - 2\mu^2 + \mu^2 = E(X^2) - \mu^2$ .

So the two forms are the same quantity. The calculation form is preferred because it needs only two sums, rather than working out $(x - \mu)^2$ separately for every value. Keep $E(X)$ exact when you square it, or the answer can lose accuracy.

Worked example: Example 1: variance from a table

$X$ has the distribution below. Find $E(X)$ , $E(X^2)$ , $Var(X)$ and the standard deviation.

$x$	$0$	$1$	$2$	$3$
$P(X = x)$	$0.1$	$0.2$	$0.3$	$0.4$

Have a go first?

Check 1 — the calculation

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