FS1 1.1 - Mean and variance of discrete distributions | Edexcel A-Level Further Mathematics

A probability distribution tells you how often each value occurs; expectation and variance turn that information into a centre and a measure of spread. You will calculate both, find expectations of transformed values, work with linear coding, and use the results to judge whether a model describes the situation sensibly.

Expectation as a probability-weighted mean

A game pays the square of your score. Can you predict the average prize just by squaring the average score?

Start with a score $X$ taking values $-1,1,3$ , with probabilities $1/4,1/2,1/4$ . In a long run of independent plays under this model, the middle score occurs about twice as often as either extreme. A simple average of the three distinct values would ignore that unequal weighting.

You already know

From A-level Mathematics, a discrete random variable has separate possible values and an associated probability distribution. Each probability is between $0$ and $1$ , inclusive, and the probabilities over all possible values sum to $1$ . Check these conditions before treating a table as a distribution.

Multiply each outcome by its probability, then add. This is the probability version of dividing a frequency-weighted total by the total frequency.

Expectation

The expectation, or mean, of $X$ is its probability-weighted average, denoted $\mathrm E(X)=\mu$ . It describes the long-run average under the model, not a guaranteed result on the next trial. It need not be one of the possible values of $X$ .

The probabilities already add to one, so do not divide the resulting weighted sum by the number of outcomes again.

Mean of a discrete distribution

If $X$ takes values $x$ with probabilities $p(x)=\mathrm P(X=x)$ , then

\mathrm E(X)=\mu=\sum_x x\mathrm P(X=x)

Sum over every possible value, using a valid probability distribution. For infinitely many possible values, the expectation must exist; all tables in this lesson are finite. In the formula booklet.

Negative outcomes contribute negative terms, while an outcome of zero contributes zero.

Worked example 1: unequal probability weights

Question. Find $\mathrm E(X)$ for the distribution below.

$x$	$-1$	$1$	$3$
$\mathrm P(X=x)$	$1/4$	$1/2$	$1/4$

Plan. Check the probabilities, then add the outcome-probability products.

\begin{aligned} \frac14+\frac12+\frac14&=1 && \text{valid non-negative probabilities} \\ \mathrm E(X)&=(-1)\frac14+1\frac12+3\frac14 && \text{weight each outcome} \\ &=\frac{-1+2+3}{4}=1 && \text{add exactly} \end{aligned}

Why this step? Each probability represents the proportion of trials contributing that score to the long-run total.

Answer. $\mathrm E(X)=1$ .

Check. The distribution is symmetric about $1$ : scores two units either side have equal probability, so $1$ is a sensible balance point.

Worked example 2 changes only the highest score, from $3$ to $5$ . Its probability stays fixed, so you can predict which way the mean moves.

Worked example 2: the mean need not be attainable

Question. Find the mean for scores $-1,1,5$ with respective probabilities $1/4,1/2,1/4$ .

Plan. Keep the same weights and update the final weighted term.

\begin{aligned} \mathrm E(X)&=-\frac14+\frac12+\frac54 && \text{changed highest score} \\ &=\frac64=\frac32 && \text{collect fractions} \end{aligned}

Why this step? Increasing one score by $2$ , with probability $1/4$ , increases the mean by $2(1/4)=1/2$ .

Answer. $\mathrm E(X)=3/2$ , even though the score can never equal $3/2$ .

Check. The mean lies between the smallest and largest possible values, $-1$ and $5$ , and is half a unit above the previous mean.

Expectation is a balance point, not another outcome to add to the table.

Watch out

Wrong: divide $\sum x\mathrm P(X=x)$ by the number of listed values. Right: the probabilities already supply normalised weights, so the sum itself is the mean.

Change the highest score once more in your turn.

Variance measures squared spread

The mean alone cannot tell you whether scores cluster tightly or vary widely, so measure their squared distances from the mean.

The average signed deviation is always zero: $\mathrm E(X-\mu)=0$ . Squaring before averaging prevents positive and negative deviations cancelling. The result is the variance $\sigma^2$ ; its non-negative square root $\sigma$ is the standard deviation.

Variance of a discrete distribution

For a valid distribution with finite second moment,

\operatorname{Var}(X)=\sigma^2 =\sum_x(x-\mu)^2\mathrm P(X=x) =\mathrm E(X^2)-\mu^2

where $\mu=\mathrm E(X)$ and $\mathrm E(X^2)=\sum_x x^2\mathrm P(X=x)$ . In the formula booklet.

The shortcut follows by expanding the square: $\mathrm E(X^2-2\mu X+\mu^2)=\mathrm E(X^2)-2\mu^2+\mu^2$ . Variance is non-negative, and is zero exactly when the variable is constant with probability one. Its units are the square of the variable's units; standard deviation returns to the original units. Use $\sigma=\sqrt{\operatorname{Var}(X)}$ ; this square-root relation is not printed as a separate formula in the booklet, so learn it.

Worked example 3: calculate the second moment first

Question. Find the variance of scores $-1,1,3$ , with probabilities $1/4,1/2,1/4$ .

Plan. Use the mean $1$ , calculate the weighted squares, then subtract the square of the mean.

\begin{aligned} \mathrm E(X)&=1 && \text{weighted mean} \\ \mathrm E(X^2)&=(-1)^2\frac14+1^2\frac12+3^2\frac14=3 && \text{square values, retain probabilities} \\ \operatorname{Var}(X)&=3-1^2=2 && \text{subtract the squared mean} \end{aligned}

Why this step? $\mathrm E(X^2)$ and $(\mathrm E(X))^2$ are different quantities: the square goes inside the weighted sum in the first one.

Answer. $\operatorname{Var}(X)=2$ , so the standard deviation is $\sqrt2$ .

Check. Using squared deviations directly gives $4(1/4)+0(1/2)+4(1/4)=2$ .

Worked example 4 changes only the highest score to $5$ , as in the earlier mean calculation. Recalculate both moments; the old mean no longer applies.

Worked example 4: both moments change

Question. Find the variance of scores $-1,1,5$ , with probabilities $1/4,1/2,1/4$ .

Plan. Use the new mean $3/2$ and calculate the new second moment exactly.

\begin{aligned} \mathrm E(X)&=\frac32 && \text{new weighted mean} \\ \mathrm E(X^2)&=\frac14+\frac12+\frac{25}4=7 && \text{weighted squared scores} \\ \operatorname{Var}(X)&=7-\left(\frac32\right)^2=\frac{19}4 && \text{subtract the new squared mean} \end{aligned}

Why this step? A wider upper tail changes the centre as well as the spread, so holding the old mean fixed would not calculate variance.

Answer. $\operatorname{Var}(X)=19/4$ .

Check. Direct squared deviations give $(25/4)(1/4)+(1/4)(1/2)+(49/4)(1/4)=19/4$ .

Keep exact fractions or full calculator precision until the final answer. A rounded mean gets squared, which can magnify an early rounding error.

Watch out

Wrong: $\operatorname{Var}(X)=\mathrm E(X^2)-\mathrm E(X)$ . Right: subtract $(\mathrm E(X))^2$ . A negative variance is a warning to check the arithmetic, probabilities and brackets.

Use a new set of scores for the next calculation.

Expectations and variances of transformed values

To average a function of a random value, apply the function to each possible outcome before taking the probability-weighted mean.

The outcome probabilities do not get squared when the outcomes do. If different original outcomes give the same transformed value, add their probabilities when writing the new distribution.

Expectation of a function

For a valid discrete distribution and a real-valued function $g$ , provided the expectation exists,

\mathrm E(g(X))=\sum_x g(x)\mathrm P(X=x)

If $Y=g(X)$ has a finite second moment, calculate its variance as $\operatorname{Var}(Y)=\mathrm E((g(X))^2)-(\mathrm E(g(X)))^2$ . This uses the supplied expectation and variance formulae. In the formula booklet.

For $Y=X^2$ , this means $\operatorname{Var}(X^2)=\mathrm E(X^4)-(\mathrm E(X^2))^2$ . The required fourth power comes from squaring the transformed variable, not from an extra rule to memorise.

Worked example 5: a square can merge outcomes

Question. For scores $-1,1,3$ with probabilities $1/4,1/2,1/4$ , let $Y=X^2$ . Find $\mathrm E(Y)$ and $\operatorname{Var}(Y)$ .

Plan. Form the transformed values, merge repeated outcomes, then apply the usual moment formulae.

$y$	$1$	$9$
$\mathrm P(Y=y)$	$3/4$	$1/4$

\begin{aligned} \mathrm E(Y)&=1\frac34+9\frac14=3 && \text{average squared scores} \\ \mathrm E(Y^2)&=1^2\frac34+9^2\frac14=21 && \text{average fourth powers of }X \\ \operatorname{Var}(Y)&=21-3^2=12 && \text{variance of the new variable} \end{aligned}

Why this step? Both $X=-1$ and $X=1$ produce $Y=1$ , so its probability is $1/4+1/2=3/4$ .

Answer. $\mathrm E(Y)=3$ and $\operatorname{Var}(Y)=12$ .

Check. Directly from $X$ , $\mathrm E(X^4)=(1+2+81)/4=21$ , matching $\mathrm E(Y^2)$ .

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FS1 1.1 - Mean and variance of discrete distributions

Expectation as a probability-weighted mean

Variance measures squared spread

Expectations and variances of transformed values

Continue this lesson.

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