SR1–SR3 - Discrete distributions, mean and variance

SR1–SR3 - Discrete distributions, mean and variance

Describe a finite random variable by its possible values and their probabilities. SR1 turns tables, formulae and graphs into probability statements; SR2 and SR3 use the same information to measure the distribution's centre and spread.

Build a valid probability function

A game awards only 00, 22 or 55 points. Can its expected score be 2.52.5, even though nobody can score that in one play? We will distinguish an average from a possible outcome.

You already know

Mutually exclusive probabilities add, the probabilities of all possible outcomes sum to 11, and P(Ac)=1−P(A)P(A^{\mathrm{c}})=1-P(A). A random variable assigns a numerical value to each outcome of a random experiment.

For now the list of possible values is finite. A probability function can be a table, or an algebraic rule accompanied by the values on which it applies. The rule alone is incomplete: kxkx gives different distributions on different sets of values.

Discrete probability distribution

A discrete random variable XX takes separate possible values xix_i, with probabilities pi=P(X=xi)p_i=P(X=x_i). Its probability function must satisfy pi≥0p_i\ge0 for every value and ∑ipi=1\sum_i p_i=1. Probabilities at all other values are zero.

Use the total probability to find an unknown constant, then check that none of the resulting probabilities is negative. A list summing to 11 can still be invalid if it contains a negative entry.

Worked example 1: normalise an algebraic rule

Question. The probability function is P(X=x)=kxP(X=x)=kx for x∈{1,2,3}x\in\{1,2,3\}, and zero otherwise. Determine kk and P(X≥2)P(X\ge2).

Plan. Make the three probabilities sum to one, then add the probabilities belonging to the event.

k+2k+3k=1all possible values6k=1⇒k=16normaliseP(X≥2)=2k+3k=56values 2 and 3\begin{aligned} k+2k+3k&=1 &&\text{all possible values}\\ 6k&=1\quad\Rightarrow\quad k=\frac16 &&\text{normalise}\\ P(X\ge2)&=2k+3k=\frac56 &&\text{values }2\text{ and }3 \end{aligned}

Why this step? The probability rule weights the values differently; there are three possible values, but they are not equally likely.

Answer. k=1/6k=1/6 and P(X≥2)=5/6P(X\ge2)=5/6.

Check. The probabilities are 1/6,1/3,1/21/6,1/3,1/2: all are non-negative and sum to 11.

Change only the greatest possible value from 33 to 44. The normalising constant must change as well.

Worked example 2: extend the support

Question. Now P(X=x)=kxP(X=x)=kx for x∈{1,2,3,4}x\in\{1,2,3,4\}, and zero otherwise. Determine kk and P(X≥2)P(X\ge2).

Plan. Include the additional outcome when normalising and when finding the event probability.

k(1+2+3+4)=1new complete set10k=1⇒k=110solveP(X≥2)=1−P(X=1)=1−110=910complement\begin{aligned} k(1+2+3+4)&=1 &&\text{new complete set}\\ 10k&=1\quad\Rightarrow\quad k=\frac1{10} &&\text{solve}\\ P(X\ge2)&=1-P(X=1)=1-\frac1{10}=\frac9{10} &&\text{complement} \end{aligned}

Why this step? Keeping the old constant would make the total probability exceed one once the extra outcome was included.

Answer. k=1/10k=1/10 and P(X≥2)=9/10P(X\ge2)=9/10.

Check. 0.1+0.2+0.3+0.4=10.1+0.2+0.3+0.4=1.

The normalisation conditions are not listed as a separate formula in the booklet: learn them. They are also your first check on any table you construct.

Watch out

Wrong: k=1/6k=1/6 still works after adding the value 44. The probabilities would sum to (1+2+3+4)/6=5/3(1+2+3+4)/6=5/3, not one. Correct: normalise the complete new support, giving 10k=110k=1 and k=1/10k=1/10.

Check the total before using the next probability rule.

Read tables and probability graphs

A probability graph places a separate bar or vertical line at each possible value. Its height is P(X=x)P(X=x). Do not join the tops with a curve: values between the listed outcomes may be impossible.

Consider this distribution. Negative values can represent a loss or a signed change; probabilities themselves must still be non-negative.

xx−1-10022
P(X=x)P(X=x)1/41/41/21/21/41/4

[DIAGRAM: asset_name: A finite probability graph; asset_slug: drv_probability_graph; caption: Each height is a probability at one separate value, with no probability between the listed outcomes; description: A discrete probability graph has vertical lines at x minus one, zero and two with respective heights one quarter, one half and one quarter. The event X greater than zero highlights only the line at two. Both axes and all three probabilities are labelled.]
Each height is a probability at one separate value, with no probability between the listed outcomes

Worked example 3: a strict inequality

Question. For the table above, determine P(X>0)P(X>0).

Plan. List the possible values that satisfy the inequality, then read their probabilities.

X>0⟺X=2only positive listed valueP(X>0)=P(X=2)=14read the table\begin{aligned} X>0&\quad\Longleftrightarrow\quad X=2 &&\text{only positive listed value}\\ P(X>0)&=P(X=2)=\frac14 &&\text{read the table} \end{aligned}

Why this step? Zero is a possible outcome with substantial probability, but a strict inequality excludes it.

Answer. P(X>0)=1/4P(X>0)=1/4.

Check. The excluded outcomes have total probability 1/4+1/2=3/41/4+1/2=3/4, leaving 1/41/4.

Change only the inequality to include its boundary. That adds the entire probability at zero.

Worked example 4: include the boundary

Question. For the same distribution, determine P(X≥0)P(X\ge0).

Plan. Include both zero and the positive outcome.

X≥0⟺X∈{0,2}include equalityP(X≥0)=12+14=34add disjoint outcomes\begin{aligned} X\ge0&\quad\Longleftrightarrow\quad X\in\{0,2\} &&\text{include equality}\\ P(X\ge0)&=\frac12+\frac14=\frac34 &&\text{add disjoint outcomes} \end{aligned}

Why this step? In a discrete distribution, equality can carry positive probability, so changing the boundary can change the answer sharply.

Answer. P(X≥0)=3/4P(X\ge0)=3/4.

Check. P(X≥0)−P(X>0)=1/2=P(X=0)P(X\ge0)-P(X>0)=1/2=P(X=0).

For discrete probability graphs, the heights sum to one. The width of a decorative bar has no probability meaning; the heights at the separate values are what you add.

Watch out

Wrong: P(X≥0)=P(X>0)=1/4P(X\ge0)=P(X>0)=1/4. Equality includes the probability at zero, which is 1/21/2 in this table. Correct: P(X≥0)=1/2+1/4=3/4P(X\ge0)=1/2+1/4=3/4.

List the included values before drawing or adding the next probabilities.

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