SA3, SA4 - Averages and spread

SA3, SA4 - Averages and spread

The most likely outcome and the long-run average need not be the same. SA3 introduces mean, mode, median, variance and standard deviation for a discrete random variable; SA4 calculates them using expectation and second moments. You will also derive the shortcut that makes variance practical to calculate.

Average and spread

A game pays £0 with probability 0.6 and £5 with probability 0.4. How can its average payment be £2 when no player ever receives £2? An average describes a distribution; it need not be a possible individual outcome.

Imagine many repetitions with proportions near the stated probabilities. In 100 plays, about 60 payments of £0 and 40 of £5 give an average near (60(0)+40(5))/100=2(60(0)+40(5))/100=2 pounds. This is the model's mean, or expectation, E(X)E(X). It is a probability-weighted average, not the unweighted average of the two listed values. It describes a long-run average under repeated comparable play, not a guaranteed average in a small sample.

Measures of a distribution

The mode is a value with greatest probability. A median mm satisfies both P(X≤m)≥1/2P(X\leq m)\geq1/2 and P(X≥m)≥1/2P(X\geq m)\geq1/2. The variance is the mean squared distance from the mean; the standard deviation is the nonnegative square root of the variance.

For the game, the mode is 0, not its probability 0.6. Its unique median is also 0: at least half the probability is at or below 0 and at least half is at or above 0. Mean, mode and median answer different questions. Variance and standard deviation describe spread, rather than another central value. If XX is measured in pounds, variance has units of pounds squared and standard deviation has units of pounds.

To find a median, put the values in increasing order and accumulate their probabilities. A jump from below 1/21/2 to above 1/21/2 identifies a unique median. If the total is exactly 1/21/2 in a gap between adjacent values, every point in that gap, including both endpoints, satisfies the definition. The midpoint is a conventional single choice when one is requested. There can also be several modes if the greatest probability is shared.

Worked example 1: locating mode and median

Question. XX has values 0,2,50,2,5 with probabilities 0.2,0.5,0.30.2,0.5,0.3. Find its mode and median.

Plan. Find the largest probability, then locate the cumulative probability crossing 1/21/2.

x025P(X≤x)0.20.71\begin{array}{c|ccc} x&0&2&5\\\hline P(X\leq x)&0.2&0.7&1 \end{array}

The largest mass, 0.5, is attached to 2. The cumulative probability crosses 1/21/2 at 2.

Answer. Mode =2=2, median =2=2.

Check. P(X≤2)=0.7P(X\leq2)=0.7 and P(X≥2)=0.8P(X\geq2)=0.8 both exceed 1/21/2. A number just below or above 2 fails one of these tests.

Keep the values fixed and move 0.3 of probability from 2 to 5. Watch how the central values change.

Worked example 2: changing the probabilities

Question. XX now has values 0,2,50,2,5 with probabilities 0.2,0.2,0.60.2,0.2,0.6. Find its mode and median.

Plan. Recalculate using the new masses, even though the possible values are unchanged.

x025P(X≤x)0.20.41\begin{array}{c|ccc} x&0&2&5\\\hline P(X\leq x)&0.2&0.4&1 \end{array}

The greatest mass is 0.6, at 5; the cumulative probability crosses 1/21/2 at 5.

Answer. Mode =5=5, median =5=5.

Check. P(X≥5)=0.6P(X\geq5)=0.6; for any m<5m<5, P(X≤m)≤0.4P(X\leq m)\leq0.4, so no smaller median is possible.

The probabilities, as well as the values, determine an average.

Watch out

The mode is the value at the highest mass, not the mass itself. For a median, accumulate probabilities in numerical order; do not take the middle column of an unsorted table.

Apply these distinctions to new values.

Expectation and second moments

Expectation weights each value by its probability. Recall the arithmetic and frequency means of data, xˉ=∑xi/n\bar{x}=\sum x_i/n and xˉ=∑fixi/∑fi\bar{x}=\sum f_i x_i/\sum f_i. Replacing observed relative frequencies by the model probabilities gives the expectation formula.

First and second moments

For values xix_i with probabilities pip_i,

E(X)=∑ixipi,E(X2)=∑ixi2piE(X)=\sum_i x_i p_i,\qquad E(X^2)=\sum_i x_i^2p_i

These are the first and second moments about zero. For finite support these sums always exist. For countably infinite support, use them only when the respective moment is finite; an absolutely convergent first-moment sum gives a finite expectation. These general discrete moment formulae are required knowledge and are not printed in the formula booklet. The data mean formulae are required recall too.

To calculate E(X2)E(X^2), square each value before weighting. Do not square each probability, and do not square the final mean. If XX has negative values, its expectation can be negative; E(X2)E(X^2) is always nonnegative.

Worked example 3: two weighted sums

Question. XX takes values 0,2,40,2,4 with probabilities 1/4,1/2,1/41/4,1/2,1/4. Find E(X)E(X) and E(X2)E(X^2).

Plan. Form separate rows for the value and its square, then weight each by its original probability.

E(X)=0(14)+2(12)+4(14)=2E(X2)=02(14)+22(12)+42(14)=6\begin{aligned} E(X)&=0\left(\frac14\right)+2\left(\frac12\right)+4\left(\frac14\right)=2\\ E(X^2)&=0^2\left(\frac14\right)+2^2\left(\frac12\right)+4^2\left(\frac14\right)=6 \end{aligned}

Answer. E(X)=2E(X)=2, E(X2)=6E(X^2)=6; in particular, E(X2)≠(E(X))2=4E(X^2)\ne(E(X))^2=4.

Check. Symmetry of the equally weighted outer values about 2 gives mean 2. The squared values 0,4,160,4,16 have weighted average 6, between their minimum and maximum.

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