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2.3 Discrete Random Variables

Figure 2.1 (Link) shows the CDF for a particular random variable, R (a Poisson random variable with λ=2). It is horizontal, except at r∈{0,1,2,3,4,…,}, at which point it jumps (is discontinuous).

Figure 2.1: Link, Caption: The function 𝖯⁡(R≤x) with R a discrete (Poisson⁡(2)) random variable.

Random variables with CDFs which are horizontal except at jump points are called discrete random variables.

The probability of outcome r (r an integer) for a discrete random variable R is given by the probability mass function (pmf) defined by

pR⁢(r)=𝖯⁡(R=r)=FR⁢(r)-limi→∞⁡FR⁢(r-1/i).

The probability 𝖯⁡(R=r) is, therefore, only non-zero where there are jumps. The cdf in Figure 2.1 (Link) corresponds to non-zero probabilities for r∈{0,1,2,3,4,…,}. For r∉{0,1,2,3,4,…,} we have 𝖯⁡(R=r)=0.

A discrete random variable R has a countable sample space (set of possible values), often only the integers or the non-negative integers. For simplicity we will take the sample space to be the integers in the following presentation.

Discrete random variables arise in a variety of ways:

  1. 1.

    from experiments with a natural integer valued outcome (e.g. rolling a die),

  2. 2.

    from experiments with outcomes to which integer values are assigned.

To write the cdf in terms of the pmf as a function of r for any r we need the function int⁡(r), which denotes the largest integer smaller than or equal to r, e.g. int⁡(3.9)=3, int⁡(2)=2, int⁡(-1.5)=-2.

FR⁢(r)=𝖯⁡(R≤r)=∑i=-∞int⁡(r)pR⁢(i). (2.1)

If p⁢(r) is a probability mass function then

  1. 1.

    0≤p⁢(r)≤1 for all r,

  2. 2.

    ∑r=-∞∞p⁢(r)=1.

Example 2.3.1.

Why are the following not valid probability mass functions? (p⁢(r)=0 unless otherwise specified)

  1. (a)

    p⁢(r)=15⁢(4-r), r=1,2,3,4,5;

  2. (b)

    p⁢(r)=r/2, r=1,2,3,4;

  3. (c)

    p⁢(r)=r/20, r=1,2,3,4.

Solution. 

  1. (a)

    p⁢(5)=-1/5;

  2. (b)

    p⁢(4)=2;

  3. (c)

    p⁢(1)+p⁢(2)+p⁢(3)+p⁢(4)=1/2.

For any event A defined as a set of possible values of the random variable R, then

𝖯⁡(A)=𝖯⁡(R∈A)=∑i∈Ap⁢(i)=∑i∈A∑ω:R⁢(ω)=i𝖯⁡(ω)=∑ω:R⁢(ω)∈A𝖯⁡(ω).

E.g. if A={i:0≤i≤m} then 𝖯⁡(A)=∑i=0mp⁢(i).