Discrete random variables and their probability distributions, given as a table, a formula p(x) or built from an experiment

Using ΣP(X=x)=1 to find an unknown constant

The cumulative distribution function F(x)=P(X≤x), and moving between F(x) and p(x)

E(X), E(X²), E(g(X)) and Var(X), with the standard deviation

E(aX+b)=aE(X)+b and Var(aX+b)=a²Var(X)

Finding unknown probabilities from conditions such as a given mean

The discrete uniform distribution: its mean and variance, and uniform distributions that do not start at 1

1. A discrete random variable X takes separate values x with probabilities P(X=x); these probabilities are all between 0 and 1 and add to 1, and everything on this page is a sum built from that list.

2. The cumulative distribution function is F(x)=P(X≤x). For whole-number values, P(X=x)=F(x)-F(x-1).

3. Mean: E(X)=Σx P(X=x). More generally E(g(X))=Σg(x) P(X=x).

4. Variance: Var(X)=E(X²)-[E(X)]²; the standard deviation is its square root.

5. Linear change: E(aX+b)=aE(X)+b and Var(aX+b)=a²Var(X): adding a constant shifts the mean but not the spread.

6. Discrete uniform on 1,2,…,n: each value has probability 1/n, mean (n+1)/2, variance (n²-1)/12.

Capital letters name the variable (X, Y); lower-case letters name a value it can take (x, y). P(X=3) means "the probability that X takes the value 3". A distribution can be a table, a formula p(x)=P(X=x), or a vertical line diagram. Counts (heads, calls, a score on a dice) give discrete variables; measurements such as a time or a mass are continuous and are treated differently in S2.

To build a distribution from an experiment, list the equally likely outcomes (a sample-space grid or a tree), work out the value of X for each, and collect the probabilities.

WORKED EXAMPLE · QUESTION

Two fair spinners, each numbered 1, 2, 3, 4, are spun. The random variable X is the larger of the two scores (if the scores are equal, X is that score). (a) Show that P(X=3)=(5/16). (b) Find the probability distribution of X, and show that it can be written as P(X=x)=(2x-1)/16. (c) Find P(X≥3).

SOLUTION

(a) (1, 3), (2, 3), (3, 3), (3, 2), (3, 1): 5 of the 16 equally likely outcomes, so P(X=3)=(5/16) as required.

(b) Counting cells for each value:

1, 3, 5, 7 are 2x-1 for x=1,2,3,4, so P(X=x)=(2x-1)/16; check: (1+3+5+7)/16=1.

(c) P(X≥3)=(5/16)+(7/16)= ¾.

A probability function gives P(X=x) as a formula, often with an unknown constant k and sometimes in pieces for different ranges of x. Outside the listed values, p(x)=0. The method is always the same: write out every probability in terms of k first, then use Σp(x)=1 to find k.

The mode is the value with the largest probability, visible at a glance on a line diagram. Probabilities of events are sums: P(X is odd) adds the probabilities of the odd values. Conditional probability works as in S1.3: P(A|B)=P(A∩B)/P(B).

WORKED EXAMPLE · QUESTION

The discrete random variable X has probability function p(x)=kx for x=1,2,3 and p(x)=k(8-x) for x=4,5, where k is a constant; p(x)=0 otherwise. (a) Find the value of k. (b) Find P(X is odd). (c) Find P(X=4|X≥3).

SOLUTION

(a) k+2k+3k+4k+3k=13k=1⇒ k=(1/13).

(b) Odd values 1, 3, 5: (1+3+3)/13= (7/13).

(c) X=4 is inside X≥3, so the intersection is just X=4:

(P(X=4))/(P(X≥3))=(4/13)/((3+4+3)/13)=(4/10)= (2/5).

F is a running total of the probabilities. It never decreases and reaches 1 at the largest value of X. Plotted, it is a step graph: flat between values, jumping up at each value by exactly P(X=x). That picture gives every rule you need:

For whole-number values, P(X=x)=F(x)-F(x-1); the first probability is just F at the smallest value.

Ranges come straight from F: P(a<X≤b)=F(b)-F(a). Convert other inequalities first: P(X<4)=P(X≤3)=F(3) and P(X≥2)=1-F(1).

If F is given as a formula with a constant, use F(largest value)=1 to find it.

WORKED EXAMPLE · QUESTION

The discrete random variable X takes the values 1, 2, 3, 4 and has cumulative distribution function F(x)=(2ˣ+k)/20 for x=1,2,3,4, where k is a constant. (a) Find k. (b) Find P(X=3). (c) Find the probability distribution of X. (d) Find P(1<X≤3).

SOLUTION

(a) F at the largest value is 1: F(4)=(16+k)/20=1⇒ k=4.

(b) P(X=3)=F(3)-F(2)=(12/20)-(8/20)= ⅕.

(c) F(1)=(6/20), F(2)=(8/20), F(3)=(12/20), F(4)=1; take differences:

(d) P(1<X≤3)=F(3)-F(1)=(12/20)-(6/20)= 0.3 (check: 0.1+0.2).

E(X) is a long-run average over many observations; it need not be a value X can actually take. Think of the probabilities as weights on a beam: E(X) is the balance point of the line diagram. Lay the work out in rows: x, P(X=x), xP(X=x), x²P(X=x), and add the last two rows. For any function g, apply g to each value, not to the mean: in general E(g(X))≠g(E(X)).

WORKED EXAMPLE · QUESTION

The random variable X has the probability distribution below. Find (a) E(X), (b) E(X²), (c) E(1/(X+2)), and compare (c) with 1/(E(X)+2).

SOLUTION

(a) (-1)(0.2)+0(0.1)+1(0.3)+2(0.25)+4(0.15)=-0.2+0+0.3+0.5+0.6= 1.2

(b) 1(0.2)+0+1(0.3)+4(0.25)+16(0.15)=0.2+0.3+1+2.4= 3.9

(c) g(x)=1/(x+2) takes the values 1,½,⅓,¼,⅙: 0.2+0.05+0.1+0.0625+0.025= 0.4375. But 1/(1.2+2)=0.3125, which is not 0.4375: the function must be applied to each value before averaging.

Variance measures spread about the mean, in squared units; the standard deviation is back in the units of X. Find E(X²) and E(X) first, then subtract, keeping exact values until the last line. A variance can never be negative: a negative answer usually means you forgot to square the mean. With the standard deviation you can describe a range such as "within one standard deviation of the mean" and find its probability from the table.

WORKED EXAMPLE · QUESTION

For the random variable X in Worked example 4, (a) find Var(X), (b) find the standard deviation of X, (c) find P(μ-σ<X<μ+σ).

SOLUTION

(a) Var(X)=3.9-1.2²=3.9-1.44= 2.46

(b) σ=√2.46=1.5684…= 1.57 (3 s.f.)

(c) μ±σ=1.2±1.5684…, from −0.37 to 2.77. The values inside are 0, 1, 2: 0.1+0.3+0.25= 0.65.

Check by the second form: Σ(x-1.2)²P(X=x)=0.968+0.144+0.012+0.16+1.176=2.46.

Adding b moves every value by the same amount, so it changes the mean but not the spread. Multiplying by a stretches every gap by a, so the variance gains a factor a². The sign of a does not matter for the variance: Var(4-2X)=(-2)²Var(X)=4Var(X). The same rules work backwards (given E(Y) and Var(Y) for Y=aX+b, solve for the unknowns), and they match the coding rules for data in S1.2. A rearrangement that is often needed: E(X²)=Var(X)+[E(X)]².

WORKED EXAMPLE · QUESTION

In a game, the number of points X that a player scores has E(X)=4 and Var(X)=2.25. The prize, in pence, is Y=20X-30. (a) Find E(Y) and the standard deviation of Y. (b) Find E(X²). (c) A second scoring rule uses T=aX+b, where a>0 and b are constants, chosen so that E(T)=50 and Var(T)=9. Find a and b.

SOLUTION

(a) E(Y)=20×4-30= 50p; Var(Y)=20²×2.25=900, so s.d. = 30p.

(b) E(X²)=Var(X)+[E(X)]²=2.25+16= 18.25

(c) Variance first (it involves a only): a²×2.25=9⇒a²=4⇒a=2 (since a>0). Then 2×4+b=50⇒b=42. a=2, b=42.

Each piece of information gives an equation. "Probabilities add to 1" is always one; a given E(X), a given value of F or a stated relationship (such as P(X=2)=2P(X=1)) give others. Two unknowns need two equations: solve them simultaneously, then check that every probability lies between 0 and 1. Once the table is complete, carry on as usual.

WORKED EXAMPLE · QUESTION

The random variable X has the probability distribution below, where p and q are constants. Given that E(X)=3.25, (a) find p and q, (b) find Var(X), (c) find Var(4-2X), (d) find P(X≤3).

SOLUTION

(a) Sum = 1: 0.1+p+q+0.3+0.2=1⇒p+q=0.4. Mean: 0.1+2p+3q+1.2+1.0=3.25⇒2p+3q=0.95. Subtract 2 × the first: q=0.95-0.8=0.15, so p=0.25, q=0.15.

(b) E(X²)=0.1+4(0.25)+9(0.15)+16(0.3)+25(0.2)=12.25; Var(X)=12.25-3.25²=12.25-10.5625= 1.6875.

(c) a=-2, so the factor is (-2)²=4: Var(4-2X)=4×1.6875= 6.75.

(d) P(X≤3)=0.1+0.25+0.15= 0.5.

Where the formulae come from. The mean uses the arithmetic series sum from P2: E(X)=1/n(1+2+⋯+n)=1/n×(n(n+1))/2=(n+1)/2. For the variance, Σr²=⅙n(n+1)(2n+1) gives E(X²)=((n+1)(2n+1))/6, so Var(X)=((n+1)(2n+1))/6-((n+1)²)/4=((n+1)(n-1))/12. For a fair six-sided dice (n=6): mean (7/2)=3.5 and variance (35/12).

The formulae are for values 1 to n. For other equally spaced values, write X as aU+b with U uniform on 1,…,n. A whole number chosen at random from 5, 6, …, 14 is X=U+4 with U uniform on 1,…,10, so E(X)=5.5+4=9.5 and Var(X)=(10²-1)/12=8.25, unchanged by the shift.

WORKED EXAMPLE · QUESTION

A fair spinner has n equal sectors numbered 1,2,…,n. The score X on one spin has a discrete uniform distribution, and Var(X)=10. (a) Find the value of n. (b) Write down E(X) and find P(X>E(X)). (c) A player's points are Y=3X-2. Find E(Y) and Var(Y). (d) Find P(Y>20). (e) The spinner is spun twice. Find the probability that both scores are greater than E(X).

SOLUTION

(a) (n²-1)/12=10⇒n²=121⇒ n=11 (n>0).

(b) E(X)=(11+1)/2=6;

X>6 for 7, 8, 9, 10, 11, so P=(5/11).

(c) E(Y)=3×6-2= 16; Var(Y)=3²×10= 90.

(d) 3X-2>20⇒X>(22/3)=7.33…, so X≥8: values 8, 9, 10, 11, P=(4/11). X must be a whole number, so X>7.33… means X≥8, not X≥7.

(e) Independent spins: (5/11)²= (25/121).

S1.3 Probability and S1.2 Measures of location and dispersion

Building a distribution from an experiment uses sample spaces, trees, independence and conditional probability. E(X) and Var(X) are the mean and variance of a frequency distribution with probabilities in place of relative frequencies, and the rules for aX+b are the coding rules.

S2 The binomial and Poisson distributions

These are discrete distributions with formulae for p(x), for which E(X) and Var(X) reduce to np, np(1-p) and λ; continuous random variables replace sums with integrals.

P2 Series

The mean of the discrete uniform distribution uses the sum 1+2+⋯+n of an arithmetic series.

1. Can a discrete random variable have P(X=1)=0.6 and P(X=2)=0.6?

2. E(X)=3 and Var(X)=2. Find E(X²).

3. Var(X)=2. Find Var(5-3X).

4. For a whole-number random variable, F(3)=0.55 and F(4)=0.7. Find P(X=4).

5. Find the mean and variance of the score on one roll of a fair six-sided dice.

1. No. The probabilities of all the values must add to 1, and these two alone add to 1.2.

2. E(X²)=Var(X)+[E(X)]²=2+9=11.

3. (-3)²×2=18. The 5 does not affect the spread, and the minus sign disappears when squared.

4. F(4)-F(3)=0.7-0.55=0.15.

5. Discrete uniform with n=6: mean (7/2)=3.5, variance (6²-1)/12=(35/12).