# How do I find the distribution function of uniform distribution on $\{-1,0,1\}$?

I only know that if it is uniformly distributed the formula for probability function will be $$\frac{1}{b-a}$$ but in my question I have a uniform distribution on $$\{-1,0,1\}$$.

For example if $$X$$ has uniform distribution on $$(a,b)$$ then it will be

$$F(x) \begin{cases} 0 & x < a \\ \frac{x-a}{b-a} & a \le x \le b \\ 1 & x > b \end{cases}$$

How do I find uniform distribution on $$\{-1,0,1 \}$$ ?

• What do you mean by "Find the uniform distribution"? Do you mean you want to find a distribution function? Apr 4, 2021 at 6:41
• @Arthur Yes you are right Apr 4, 2021 at 6:42
• Just note that the distribution function of the uniform distribution on interval $(a,b)$ is $\frac{x-a}{b-a}$ for $x \in (a,b)$, zero for $x<a$ and $1$ for $x > b$. I edited your question in accordingly. Apr 4, 2021 at 6:49
• The (continuous) uniform distribution on the interval $[a,b]$ has density $\frac{1}{b-a}$ in that interval, while the (discrete) uniform distribution on the integers $\{a,a+1,\ldots,b\}$ has probability mass function $\frac1{b-a+1}$ on those integers Apr 4, 2021 at 7:27

Isn't $$\{-1,0,1\}$$ a set with three elements? In that case we have $$P[-1]=P[0]=P[1]=1/3$$ and zero otherwise. The distribution function will be $$F(x) \begin{cases} 0 & x < -1 \\ 1/3 & -1 \le x < 0 \\ 2/3 & 0 \le x < 1 \\ 1 & x \ge 1 \end{cases}$$