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My probability book defines a discrete uniform random variable as a variable X such that P(X=x) = \frac{1}{b-a+1}, for all x=a,a+1...b. My doubt is, in a discrete uniform distribution must the numbers that the random variable may take on always be intenger and sequential? And if so, why? I mean, could there be a uniform distribution of the following form?

$$P(X=x) = \begin{cases}\frac{1}{4} &,& \text{if }x \in\{ 3,5,6.4,9\} \\ 0 &, & \text{otherwise} \end{cases}$$ Athough it does make sense to me (since there are 4 possible outcomes with equal probability, so each individual probability must = 1/4), it doesn't seem to fit that definition: $\frac{1}{b-a+1} = \frac{1}{9-3+1}$ is different from $\frac{1}{4}$.

What am I getting wrong? Thanks in advance.

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    The sum of the probabilities must be 1 – 2017-01-29

2 Answers 2

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Your probability mass function is indeed that of a uniform discrete random variable; although not a standard one so the usual formula does not apply.

It is just, because the support is not a set of consecutive integer values, that the count of elements in the sample space is not conveniently counted by subtracting its minimum from its maximum and adding 1.

$$X\sim\mathcal{U}\{ 3,5,6.4,9\} ~\iff~ \mathsf P(X=x) = \begin{cases}\frac{1}{4} &,& \text{if }x \in\{ 3,5,6.4,9\} \\0 &, & \text{otherwise} \end{cases}$$

We have three unequally-spaced, non-unit-length steps between the four massive points.   However, those points do have a uniform probability distribution.

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They definitely don't need to be sequential. The only point of a uniform distribution is to have a constant probability mass function (or pdf in the continuous case). Just take the number  $K$ of distinct values $X$ can take, and assign $1/K$ probability to each.

When they use the usual definition for the pmf, i.e. $$ \frac{1}{b-a+1} $$ the interval $a, a+1, \dots, b$ contains $b-a+1$ elements and hence is a special case of simply using the number of elements as the denominator.

When you wrote $\frac{1}{9-3+1}$ you treat your $X$ as if it can take values in the interval $3$ to $9$, whereas it actually only takes values in the set $\lbrace 3, 5, 6.4, 9 \rbrace$