Discrete random variables

A random variable assigns a numerical value to each outcome of a random experiment. It is discrete when its possible values can be listed, even if the list is infinite.

For example, if $X$ is the number of sixes in four rolls of a die, then

$$X\in\{0,1,2,3,4\}.$$

The value of $X$ is not known until the experiment happens, but each possible value has a probability. Those values and probabilities form the probability distribution of $X$.

This lesson explains how to construct and use discrete distributions, including their mean, variance and standard deviation.

Prerequisites

You should be able to:

• calculate probabilities using probability rules;
• recognise mutually exclusive outcomes;
• use sigma notation such as $\sum x$;
• calculate a mean and variance from a frequency table.

From outcomes to a random variable

An outcome describes what happens. A random variable records a numerical feature of what happens.

Suppose two fair coins are tossed. The equally likely outcomes are

$$HH,\ HT,\ TH,\ TT.$$

Define $X$ to be the number of heads. Then

$$X(HH)=2,\qquad X(HT)=X(TH)=1,\qquad X(TT)=0.$$

The outcomes $HT$ and $TH$ are different, but they give the same value of $X$. Combining outcomes with equal $X$ values gives

$x$	$0$	$1$	$2$
$\operatorname{P}(X=x)$	$\frac14$	$\frac12$	$\frac14$

This is a probability distribution.

Values are not necessarily equally likely

Although $X$ has three possible values, their probabilities are not all $1/3$. Count or calculate the underlying outcomes instead of dividing automatically by the number of $X$ values.

Discrete and continuous variables

A discrete variable takes separate, countable values. A continuous variable can take any value in an interval.

Variable	Type	Reason
Number of emails received	Discrete	Takes values $0,1,2,\ldots$
Mass of an apple	Continuous	Can take any value in a measurement interval
Score on a ten question test	Discrete	Takes integer values from $0$ to $10$
Time until a bus arrives	Continuous	Can take any non-negative real value

A rounded measurement is still modelled as continuous if the underlying quantity is continuous.

Self-check 1

A spinner is spun three times. Let $R$ be the number of red results. Is $R$ discrete or continuous, and what values can it take?

Answer

$R$ is discrete because it is a count. Its possible values are

$$\boxed{R\in\{0,1,2,3\}}.$$

Valid probability distributions

For a discrete random variable $X$, write

$$p(x)=\operatorname{P}(X=x).$$

Every probability distribution must satisfy two conditions:

$$\boxed{0\leq p(x)\leq1\quad\text{for every }x}$$

and

$$\boxed{\sum p(x)=1}.$$

The second condition holds because the possible values of $X$ are mutually exclusive and exhaustive: exactly one of them must occur.

Worked example 1: finding an unknown probability

The distribution of $X$ is shown below.

$x$	$0$	$1$	$2$	$3$
$\operatorname{P}(X=x)$	$0.15$	$k$	$2k$	$0.25$

Find $k$.

The probabilities sum to $1$:

$$0.15+k+2k+0.25=1.$$

Therefore

$$3k=0.60$$

and

$$\boxed{k=0.20}.$$

Check that every resulting probability lies between $0$ and $1$.

Worked example 2: a distribution given by a formula

A random variable $Y$ has probability function

$$\operatorname{P}(Y=y)=ky, \qquad y=1,2,3,4.$$

Find $k$ and $\operatorname{P}(Y\geq3)$.

First use the total probability:

$$\begin{aligned} \sum_{y=1}^{4}ky &=k(1+2+3+4)\\ &=10k=1. \end{aligned}$$

Hence $k=1/10$. Now

$$\begin{aligned} \operatorname{P}(Y\geq3) &=\operatorname{P}(Y=3)+\operatorname{P}(Y=4)\\ &=\frac3{10}+\frac4{10}\\ &=\boxed{\frac7{10}}. \end{aligned}$$

The letter used does not matter

$\operatorname{P}(X=x)$ means the probability that the random variable $X$ takes the particular value $x$. An upper-case letter usually denotes the random variable and a lower-case letter one of its values.

Self-check 2

For $x=0,1,2,3$, suppose

$$\operatorname{P}(X=x)=a(x+1).$$

Find $a$ and $\operatorname{P}(X<2)$.

Answer

Since the probabilities sum to $1$,

$$a(1+2+3+4)=10a=1,$$

so $a=1/10$. Therefore

$$\operatorname{P}(X<2) =\operatorname{P}(X=0)+\operatorname{P}(X=1) =\frac1{10}+\frac2{10} =\boxed{\frac3{10}}.$$

Calculating probabilities from a distribution

Translate the wording into an inequality before adding probabilities.

Words	Probability statement
exactly $a$	$\operatorname{P}(X=a)$
at most $a$	$\operatorname{P}(X\leq a)$
fewer than $a$	$\operatorname{P}(X<a)$
at least $a$	$\operatorname{P}(X\geq a)$
more than $a$	$\operatorname{P}(X>a)$
from $a$ to $b$ inclusive	$\operatorname{P}(a\leq X\leq b)$

For integer-valued $X$,

$$\operatorname{P}(X<4)=\operatorname{P}(X\leq3),$$

but $\operatorname{P}(X<4)$ and $\operatorname{P}(X\leq4)$ are not the same.

Worked example 3: interval and complement probabilities

The probability distribution of $X$ is

$x$	$1$	$2$	$3$	$4$	$5$
$\operatorname{P}(X=x)$	$0.10$	$0.25$	$0.30$	$0.20$	$0.15$

Find:

1. $\operatorname{P}(2<X\leq5)$;
2. $\operatorname{P}(X\neq3)$.

For the first probability, include $3,4,5$ but not $2$:

$$\operatorname{P}(2<X\leq5)=0.30+0.20+0.15=\boxed{0.65}.$$

For the second, a complement is shorter:

$$\operatorname{P}(X\neq3)=1-\operatorname{P}(X=3)=1-0.30=\boxed{0.70}.$$

Cumulative distribution functions

The cumulative distribution function of $X$ is

$$\boxed{F(x)=\operatorname{P}(X\leq x)}.$$

It accumulates probability from the smallest possible value up to $x$. For a discrete random variable, $F$ is a step function.

Using the previous distribution,

$x$	$1$	$2$	$3$	$4$	$5$
$\operatorname{P}(X=x)$	$0.10$	$0.25$	$0.30$	$0.20$	$0.15$
$F(x)$	$0.10$	$0.35$	$0.65$	$0.85$	$1$

Differences of cumulative probabilities give interval probabilities. For integer-valued $X$,

$$\boxed{\operatorname{P}(a\leq X\leq b)=F(b)-F(a-1)}.$$

For example,

$$\operatorname{P}(3\leq X\leq4)=F(4)-F(2)=0.85-0.35=0.50.$$

Why subtract $F(a-1)$?

$F(a-1)$ removes values strictly below $a$. Subtracting $F(a)$ would also remove the required probability $\operatorname{P}(X=a)$.

Expectation: the long-run mean

The expected value or mean of a discrete random variable is

$$\boxed{\operatorname{E}(X)=\sum x\operatorname{P}(X=x)}.$$

Each value is weighted by its probability. If the experiment is repeated many times, the mean of the observed $X$ values tends towards $\operatorname{E}(X)$.

The expected value need not be a possible outcome. A fair die has expected score $3.5$, although no single roll can produce $3.5$.

Worked example 4: expected value from a table

A game pays $X$ pounds, with distribution

$x$	$0$	$2$	$5$	$10$
$\operatorname{P}(X=x)$	$0.50$	$0.30$	$0.15$	$0.05$

Then

$$\begin{aligned} \operatorname{E}(X) &=0(0.50)+2(0.30)+5(0.15)+10(0.05)\\ &=0+0.60+0.75+0.50\\ &=\boxed{1.85}. \end{aligned}$$

Over many plays, the average payout approaches £1.85 per play.

If the game costs £2 to play, the player’s net gain is $G=X-2$, so

$$\operatorname{E}(G)=1.85-2=-0.15.$$

The player loses an average of 15 pence per play in the long run. This does not mean every play loses 15 pence.

Worked example 5: finding a parameter from the mean

The distribution of $Y$ is

$y$	$0$	$1$	$4$
$\operatorname{P}(Y=y)$	$p$	$2p$	$1-3p$

Given that $\operatorname{E}(Y)=2$, find $p$.

Use the values as weights:

$$\begin{aligned} \operatorname{E}(Y) &=0(p)+1(2p)+4(1-3p)\\ &=4-10p. \end{aligned}$$

Therefore

$$4-10p=2,$$

so

$$\boxed{p=0.2}.$$

The probabilities are then $0.2,0.4,0.4$, which are valid and sum to $1$.

Self-check 3

A fair spinner has labels $1,1,2,6$ on four equal sectors. Let $X$ be the label obtained. Find $\operatorname{E}(X)$.

Answer

The distribution is

$$\operatorname{P}(X=1)=\frac12,\qquad \operatorname{P}(X=2)=\frac14,\qquad \operatorname{P}(X=6)=\frac14.$$

Hence

$$\operatorname{E}(X) =1\left(\frac12\right)+2\left(\frac14\right)+6\left(\frac14\right) =\boxed{\frac52}.$$

Variance and standard deviation

Expectation gives the centre of a distribution. Variance measures its spread about that centre.

If $\mu=\operatorname{E}(X)$, then

$$\operatorname{Var}(X)=\operatorname{E}\bigl((X-\mu)^2\bigr) =\sum (x-\mu)^2\operatorname{P}(X=x).$$

The efficient formula is

$$\boxed{\operatorname{Var}(X)=\operatorname{E}(X^2)-[\operatorname{E}(X)]^2},$$

where

$$\boxed{\operatorname{E}(X^2)=\sum x^2\operatorname{P}(X=x)}.$$

The standard deviation is

$$\boxed{\operatorname{SD}(X)=\sqrt{\operatorname{Var}(X)}}.$$

Variance has squared units, while standard deviation has the same units as $X$.

Worked example 6: mean, variance and standard deviation

Let $X$ have distribution

$x$	$0$	$1$	$2$	$3$
$\operatorname{P}(X=x)$	$0.10$	$0.20$	$0.40$	$0.30$

First calculate the mean:

$$\begin{aligned} \operatorname{E}(X) &=0(0.10)+1(0.20)+2(0.40)+3(0.30)\\ &=1.9. \end{aligned}$$

Next calculate the second moment:

$$\begin{aligned} \operatorname{E}(X^2) &=0^2(0.10)+1^2(0.20)+2^2(0.40)+3^2(0.30)\\ &=0+0.20+1.60+2.70\\ &=4.5. \end{aligned}$$

Therefore

$$\begin{aligned} \operatorname{Var}(X) &=4.5-(1.9)^2\\ &=\boxed{0.89}, \end{aligned}$$

and

$$\operatorname{SD}(X)=\sqrt{0.89}=0.943\ldots\approx\boxed{0.943}.$$

Square the whole mean

The formula contains $[\operatorname{E}(X)]^2$, not $\operatorname{E}(X)$ and not $\operatorname{E}(X^2)^2$.

A useful calculation table

For longer distributions, organise the work in columns:

$x$	$p$	$xp$	$x^2p$
possible value	probability	contribution to $\operatorname{E}(X)$	contribution to $\operatorname{E}(X^2)$

Then add the third and fourth columns. This reduces omitted terms and calculator input errors.

Self-check 4

For the random variable $X$ below, find $\operatorname{Var}(X)$.

$x$	$1$	$3$	$5$
$\operatorname{P}(X=x)$	$0.25$	$0.50$	$0.25$

Answer

$$\operatorname{E}(X)=1(0.25)+3(0.50)+5(0.25)=3,$$

and

$$\operatorname{E}(X^2)=1(0.25)+9(0.50)+25(0.25)=11.$$

Therefore

$$\boxed{\operatorname{Var}(X)=11-3^2=2}.$$

Linear transformations

Suppose a new variable is defined by

$$Y=aX+b,$$

where $a$ and $b$ are constants. Then

$$\boxed{\operatorname{E}(Y)=a\operatorname{E}(X)+b}$$

and

$$\boxed{\operatorname{Var}(Y)=a^2\operatorname{Var}(X)}.$$

Adding $b$ shifts every value equally, so it changes the mean but not the spread. Multiplying by $a$ multiplies distances from the mean by $|a|$, so it multiplies variance by $a^2$.

Consequently,

$$\operatorname{SD}(aX+b)=|a|\operatorname{SD}(X).$$

Worked example 7: transforming a random variable

A delivery charge in pounds is modelled by

$$C=1.2X+3,$$

where $\operatorname{E}(X)=8$ and $\operatorname{Var}(X)=2.5$. Find the mean and standard deviation of $C$.

For the mean,

$$\operatorname{E}(C)=1.2(8)+3=\boxed{12.6}.$$

For the variance,

$$\operatorname{Var}(C)=1.2^2(2.5)=3.6.$$

Therefore

$$\operatorname{SD}(C)=\sqrt{3.6}=1.897\ldots\approx\boxed{1.90}.$$

The added £3 affects the mean charge but not its standard deviation.

Common misconceptions

• The possible values must have equal probabilities. They do not. Probabilities come from the underlying random experiment.
• Expectation is the most likely value. Not necessarily. It is a probability-weighted mean and may not be possible at all.
• $\operatorname{E}(X^2)=[\operatorname{E}(X)]^2$. Usually these differ. Their difference is the variance.
• Adding a constant increases variance. It shifts the distribution without changing its spread.
• $\operatorname{Var}(aX)=a\operatorname{Var}(X)$. Variance scales by $a^2$.
• At most and fewer than mean the same thing. At most includes the boundary; fewer than excludes it.

Mixed self-check

A random variable $X$ has probability function

$$\operatorname{P}(X=x)=k(5-x), \qquad x=1,2,3,4.$$

1. Find $k$.
2. Find $\operatorname{P}(X\geq3)$.
3. Find $\operatorname{E}(X)$.
4. Find $\operatorname{Var}(X)$.

Answers

The probability weights are $4k,3k,2k,k$. Hence

$$10k=1\quad\Longrightarrow\quad \boxed{k=0.1}.$$

Then

$$\operatorname{P}(X\geq3)=0.2+0.1=\boxed{0.3}.$$

The mean is

$$\begin{aligned} \operatorname{E}(X) &=1(0.4)+2(0.3)+3(0.2)+4(0.1)\\ &=\boxed{2}. \end{aligned}$$

Also,

$$\operatorname{E}(X^2) =1^2(0.4)+2^2(0.3)+3^2(0.2)+4^2(0.1) =5.$$

Therefore

$$\boxed{\operatorname{Var}(X)=5-2^2=1}.$$

Summary

For a discrete random variable $X$:

$$\sum\operatorname{P}(X=x)=1,$$ $$\operatorname{E}(X)=\sum x\operatorname{P}(X=x),$$ $$\operatorname{Var}(X)=\sum x^2\operatorname{P}(X=x)-[\operatorname{E}(X)]^2.$$

Always define the random variable, check that probabilities are valid, and translate boundary language before calculating.

Next steps

• Study the most important named discrete model, the binomial distribution.
• Learn how context determines the right model in choosing a distribution.
• Apply probability models within hypothesis testing.