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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 $\{-1,0,1 \}$ ?

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    $\begingroup$ What do you mean by "Find the uniform distribution"? Do you mean you want to find a distribution function? $\endgroup$
    – Arthur
    Apr 4, 2021 at 6:41
  • $\begingroup$ @Arthur Yes you are right $\endgroup$ Apr 4, 2021 at 6:42
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    $\begingroup$ 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. $\endgroup$ Apr 4, 2021 at 6:49
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    $\begingroup$ 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 $\endgroup$
    – Henry
    Apr 4, 2021 at 7:27

1 Answer 1

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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 $$ F(x) \begin{cases} 0 & x < -1 \\ 1/3 & -1 \le x < 0 \\ 2/3 & 0 \le x < 1 \\ 1 & x \ge 1 \end{cases} $$

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  • $\begingroup$ why is it 1/3?? sorry for asking $\endgroup$ Apr 4, 2021 at 7:04
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    $\begingroup$ Is it because we divide the elements -1, 0, 1 = which is 3? $\endgroup$ Apr 4, 2021 at 7:12
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    $\begingroup$ @헬창공돌이: Yes, we have three elements which can occur with the same probability, hence occuring of one has probability 1/3. This is similar to probability of some outcome on a dice which is 1/6. This is because the dice has six sides. $\endgroup$ Apr 4, 2021 at 7:15

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