SR7–SR9 - The discrete uniform distribution

SR7–SR9 - The discrete uniform distribution

Use equally likely values to build a discrete uniform model. SR7 and SR8 identify the model and count event probabilities; SR9 calculates its mean and variance, including derivations and distributions whose labels do not start at one.

Recognise a discrete uniform model

A fair spinner has seven equal sections labelled 4,5,6,7,8,9,104,5,6,7,8,9,10. Is the probability of a score above its mean exactly one half? Its symmetry is tempting, but the middle value has probability too.

You already know

A finite probability function assigns non-negative probabilities that sum to one. Find its mean from ∑xp\sum xp and its variance from ∑x2p−E(X)2\sum x^2p-E(X)^2. A constant shift does not change variance.

The spinner's equal sections suggest equal probabilities for its labels. Fairness is a modelling assumption about the mechanism, not a consequence of having written down a list of numbers.

Discrete uniform distribution

A random variable is discrete uniform over a specified finite set when every value in that set is equally likely. If there are nn distinct values, each has probability 1/n1/n. State the set as well as the distribution's name.

The values need not be consecutive, but the familiar booklet mean and variance formulae apply specifically to the set {1,2,…,n}\{1,2,\ldots,n\}. Other sets can be handled by a transformation or by the general weighted sums.

A fair die's single score is uniform. The sum of two independent fair dice is not: there are more ways to make middle totals than extreme totals. Random does not mean equally likely.

Key idea

Uniform describes equal probabilities of the listed values, not equal spacing on its own and not the word “random”.

Choose the mechanism that justifies those equal probabilities.

Count the values in an event

For a uniform variable, every included value contributes the same probability. Translate an inequality into the actual integer values first; only then count them.

Uniform probabilities

For XX uniform over {1,2,…,n}\{1,2,\ldots,n\},

P(X=r)=1n,r=1,2,…,nP(X=r)=\frac1n,\qquad r=1,2,\ldots,n

In the formula booklet.

Adding the equal probabilities gives P(X∈A)=number of included values/nP(X\in A)=\text{number of included values}/n. For consecutive integers from aa to bb, inclusive, the count is b−a+1b-a+1, not b−ab-a.

Worked example 1: include the boundary

Question. A fair ten-sided spinner is labelled 1,2,…,101,2,\ldots,10. Find the probability that its score is at least 77.

Plan. List the qualifying labels and divide their count by ten.

X≥7⟺X∈{7,8,9,10}include 7P(X≥7)=410=25four equal probabilities\begin{aligned} X\ge7&\quad\Longleftrightarrow\quad X\in\{7,8,9,10\} &&\text{include }7\\ P(X\ge7)&=\frac4{10}=\frac25 &&\text{four equal probabilities} \end{aligned}

Why this step? There are four qualifying integers even though the difference between the endpoints is three.

Answer. 2/52/5.

Check. The six excluded labels have probability 6/106/10, leaving 4/104/10.

Change only “at least” to “greater than”. One label is removed from the event.

Worked example 2: remove the boundary

Question. For the same spinner, find P(X>7)P(X>7).

Plan. Exclude the boundary value before counting.

X>7⟺X∈{8,9,10}strict inequalityP(X>7)=310three of ten labels\begin{aligned} X>7&\quad\Longleftrightarrow\quad X\in\{8,9,10\} &&\text{strict inequality}\\ P(X>7)&=\frac3{10} &&\text{three of ten labels} \end{aligned}

Why this step? The change is exactly the probability at 77, namely 1/101/10.

Answer. 3/103/10.

Check. 2/5−1/10=3/102/5-1/10=3/10.

When an inequality involves an expression such as ∣X−μ∣|X-\mu|, solve the inequality first and identify the permitted integer values. A sketch of the integers prevents losing one side of an absolute-value condition.

Watch out

Wrong: P(X≥7)=(10−7)/10P(X\ge7)=(10-7)/10 for the ten-sided spinner. Subtracting endpoints misses one included integer. Correct: P(X≥7)=(10−7+1)/10=4/10P(X\ge7)=(10-7+1)/10=4/10; the strict event X>7X>7 has probability 3/103/10.

Decide whether the boundary is included before counting the next event.

Derive mean and variance from the sums

The mean of 1,2,…,n1,2,\ldots,n is the midpoint, but a question asking for a derivation needs the expectation calculation. Equal probabilities make the sums especially simple.

Use ∑r=1nr=n(n+1)/2\sum_{r=1}^n r=n(n+1)/2 and ∑r=1nr2=n(n+1)(2n+1)/6\sum_{r=1}^n r^2=n(n+1)(2n+1)/6. The arithmetic-series and square-sum results are in the booklet.

E(X)=∑r=1nr1n=1nn(n+1)2=n+12E(X2)=∑r=1nr21n=(n+1)(2n+1)6Var⁡(X)=(n+1)(2n+1)6−(n+12)2=(n+1)(2(2n+1)−3(n+1))12=n2−112\begin{aligned} E(X)&=\sum_{r=1}^n r\frac1n=\frac1n\frac{n(n+1)}2=\frac{n+1}{2}\\ E(X^2)&=\sum_{r=1}^n r^2\frac1n=\frac{(n+1)(2n+1)}6\\ \operatorname{Var}(X)&=\frac{(n+1)(2n+1)}6-\left(\frac{n+1}2\right)^2\\ &=\frac{(n+1)\bigl(2(2n+1)-3(n+1)\bigr)}{12}=\frac{n^2-1}{12} \end{aligned}

The subtraction is essential: E(X2)E(X^2) alone is not the variance.

Moments on the standard set

For a discrete uniform variable over {1,2,…,n}\{1,2,\ldots,n\},

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

These formulae are supplied, but their derivations may be required.

In the formula booklet.

The numerical examples show what direct calculation looks like when detailed reasoning is requested.

Worked example 3: direct calculation on five values

Question. For XX uniform over {1,2,3,4,5}\{1,2,3,4,5\}, determine the mean and variance from the expectation sums.

Plan. Use probability 1/51/5 at each value and calculate both moments.

E(X)=1+2+3+4+55=3weighted first momentE(X2)=1+4+9+16+255=11weighted squaresVar⁡(X)=11−32=2subtract mean squared\begin{aligned} E(X)&=\frac{1+2+3+4+5}{5}=3 &&\text{weighted first moment}\\ E(X^2)&=\frac{1+4+9+16+25}{5}=11 &&\text{weighted squares}\\ \operatorname{Var}(X)&=11-3^2=2 &&\text{subtract mean squared} \end{aligned}

Why this step? Showing the sums establishes the result; quoting the supplied moment formulae would not be a derivation.

Answer. Mean 33; variance 22.

Check. The booklet formula gives (52−1)/12=2(5^2-1)/12=2.

Change only the upper endpoint to 66. The mean now lies between two possible outcomes.

Continue this lesson.

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