S1 5.4 - The discrete uniform model

S1 5.4 - The discrete uniform model

Recognise a discrete uniform distribution by its equally likely values, then calculate its probabilities, mean and variance. Distinguish a fair underlying process from a random variable whose values are themselves equally likely.

Equal probabilities on finite outcomes

Can a fair process give unequal probabilities? Two fair coin tosses have four equally likely ordered results, but the number of heads has only three possible values. We will return to that distinction after learning exactly what a uniform model says.

Discrete uniform distribution

A random variable has a discrete uniform distribution on nn distinct possible values if each has probability 1/n1/n. Its support is the set of values it can take.

The support must be stated. A fair six-sided die has uniform scores on {1,2,3,4,5,6}\{1,2,3,4,5,6\}. A fair die labelled 1,3,5,7,9,111,3,5,7,9,11 also gives a discrete uniform score, although those values are not consecutive integers. The gaps do not decide whether the distribution is uniform: equal probabilities do.

For an event, count the support values that satisfy it.

Uniform probabilities

If XX is uniform on a finite support with nn distinct values, and kk of them satisfy event AA,

P(A)=kn.\mathrm P(A)=\frac{k}{n}.

This follows from adding kk probabilities of 1/n1/n; it is not a separately printed formula in the S1 booklet.

An inequality may include a value, exclude it or fall between support values. List the eligible values before counting if there is any doubt.

Worked example 1: a consecutive support

XX is the score on a fair die labelled 1,2,3,4,5,61,2,3,4,5,6. Find P(X≥5)\mathrm P(X\ge5).

Plan. Count the labels at least 55, then divide by the six equally likely labels.

X≥5⟺X∈{5,6},P(X≥5)=26=13.X\ge5\quad\Longleftrightarrow\quad X\in\{5,6\},\qquad \mathrm P(X\ge5)=\frac26=\frac13.

Answer. 13\frac13.

Check. The complementary event has labels 1,2,3,41,2,3,4, with probability 4/64/6. The two probabilities add to 11.

Change only the die's labels to odd numbers, keeping six equally likely faces. The same inequality now selects a different number of values.

Worked example 2: gaps in the support

YY is the score on a fair die labelled 1,3,5,7,9,111,3,5,7,9,11. Find P(Y≥5)\mathrm P(Y\ge5).

Plan. Use the actual support, rather than counting every integer between its endpoints.

Y≥5⟺Y∈{5,7,9,11},P(Y≥5)=46=23.Y\ge5\quad\Longleftrightarrow\quad Y\in\{5,7,9,11\},\qquad \mathrm P(Y\ge5)=\frac46=\frac23.

Answer. 23\frac23.

Check. Only 11 and 33 fail the condition. Their probability is 2/6=1/32/6=1/3, leaving 2/32/3.

For consecutive integers l,l+1,…,ul,l+1,\ldots,u, the number of values is u−l+1u-l+1, because both endpoints are included. This count is different from the length u−lu-l of a continuous interval.

Watch out

A discrete variable has no probability at values outside its support. For the odd-labelled die, P(Y=6)=0\mathrm P(Y=6)=0, even though 66 lies between the smallest and largest labels.

Count the supported values in the requested interval.

Mean and variance

With equal probabilities, expectation is the ordinary arithmetic mean of the distinct support values. For equally spaced values, pairing the smallest with the largest shows why the mean is the midpoint.

For 1,2,3,4,5,61,2,3,4,5,6, the pairs (1,6)(1,6), (2,5)(2,5), (3,4)(3,4) each total 77. Their common midpoint is 3.53.5. Variance still measures squared distances from this midpoint; it is not determined by the midpoint alone.

Moments of a finite uniform distribution

For nn distinct equally likely values x1,…,xnx_1,\ldots,x_n,

E(X)=1n∑i=1nxi,Var⁡(X)=1n∑i=1nxi2−[E(X)]2.\mathrm E(X)=\frac1n\sum_{i=1}^{n}x_i,\qquad \operatorname{Var}(X)=\frac1n\sum_{i=1}^{n}x_i^2-[\mathrm E(X)]^2.

These are the general discrete-moment formulae with probability 1/n1/n, using the formulae provided in the booklet.

For the particular support 1,2,…,n1,2,\ldots,n, the finite sums simplify to

E(X)=n+12,Var⁡(X)=n2−112.\mathrm E(X)=\frac{n+1}{2},\qquad \operatorname{Var}(X)=\frac{n^2-1}{12}.

These shortcuts are not separately printed in the S1 booklet. You can always calculate the two moments directly. A shift to consecutive values l,…,ul,\ldots,u gives mean (l+u)/2(l+u)/2 and the same variance formula with n=u−l+1n=u-l+1. If the spacing is dd instead of 11, variance is multiplied by d2d^2.

Worked example 3: the moments of a die score

UU is uniform on 1,2,3,4,5,61,2,3,4,5,6. Find its mean and variance.

Plan. Average the values and their squares, then subtract the squared mean.

E(U)=1+2+3+4+5+66=72E(U2)=1+4+9+16+25+366=916Var⁡(U)=916−(72)2=3512.\begin{aligned} \mathrm E(U)&=\frac{1+2+3+4+5+6}{6}=\frac72\\ \mathrm E(U^2)&=\frac{1+4+9+16+25+36}{6}=\frac{91}{6}\\ \operatorname{Var}(U)&=\frac{91}{6}-\left(\frac72\right)^2=\frac{35}{12}. \end{aligned}

Answer. Mean 7/27/2, variance 35/1235/12.

Check. The mean is the midpoint (1+6)/2(1+6)/2. The variance shortcut gives (62−1)/12=35/12(6^2-1)/12=35/12, agreeing with the moments.

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