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I want to construct a random variable $X_i$ which can take one of three values: $-1,0,1$. I want to construct a probability function such that the variance of this random variable to be $1$. How can I do that?

Note that $\mu=0$, and $\sum (x_i-\mu)^2=(-1)^2+0^2+1^2=2$. Hence, one such probability function could be $p(-1)=\frac{1}{2}$ and $p(1)=\frac{1}{2}$. However, that would mean that there is no probability of the random variable taking the value $0$, which is weird.

matilda
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  • Are there any other solutions? – kimchi lover Oct 04 '23 at 18:43
  • It is not particularly weird. You cannot get a variance bigger than $1$ here, so a variance of $1$ extreme and you should expect an extreme distribution – Henry Oct 04 '23 at 18:50
  • @kimchilover- I don't there are any solutions where there is a non-zero probability of the variable taking the value $0$ – matilda Oct 04 '23 at 18:50

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You just need to start from the basic definitions. In your case, the probability (mass) function is defined by three numbers, say $a$, $b$ and $c$ such that \begin{align} \Pr[X=-1] &= a \ge 0 \\ \Pr[X=0] &= b \ge 0 \\ \Pr[X=1] &= c \ge 0 \\ a+b+c&=1. \end{align} You want $X$ to be zero-mean, then $$ \mu=-1\times \Pr[X=-1] + 0 \times \Pr[X=0]+1\times \Pr[X=1]=-a+c=0. $$ Also, you want the variance to be one, that is $$ \sigma^2 = (-1-\mu)^2\times \Pr[X=-1] + (0-\mu)^2 \times \Pr[X=0]+(1-\mu)^2\times \Pr[X=1]=a+c=1. $$ Putting everything together, the masses $a$, $b$ and $c$ must satisfy the following system of equations $$ \left\{ \begin{aligned} a+b+c &= 1 \\ -a+c &= 0 \\ a+c &= 1. \end{aligned} \right. $$ The only solution is the one you gave - but now you can ask yourself what would happen if $\mu \ne 0$ or $\sigma^2 \ne 1$ and play with the masses.

dvdgrgrtt
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    I suspect that in general that $|\mu| - \mu^2 \le \sigma^2 \le 1-\mu^2$ and if you have an equality then at least one of the points has probability $0$. – Henry Oct 04 '23 at 20:37