If $X$ is a uniformly distributed discrete random variable, what is the condition that $Y=\phi (X)$ is too?

Question

More specifically I was solving the following problem:

Let 𝑋 be a discrete random variable that is uniformly distributed over the set $S=\{−10, −9, ⋯ , 0, ⋯ , 9, 10\}$. Which of the following random variables is/are uniformly distributed?
(A) $𝑋^2$
(B) $𝑋^3$
(C) $(𝑋 − 5)^2$
(D) $(𝑋 + 10)^2$

The official answer is (B) and (D). The only plausible reason I find is that if $S= \{s_i \in \mathbb Z: -10\leq s_i\leq 10 \}$, then since there is a one-one correspondence between the sets $(S+10)^2 = \{(s_i+10)^2 : s_i\in S\}$ and $S^3 = \{s_i^3 : s_i\in S\}$, and $S$, therefore the variables $(X+10)^2$ and $X^3$ are uniformly distributed, with PMF $= \frac{1}{21}$. More generally, I postulate that if $X$ is a discrete random variable $X$ is a uniformly distributed over some set $S= \{s_i \in \mathbb Z: m\leq s_i\leq n \text{ for integers } m,n\}$, then the variable $Y=\phi(X)$ is also uniformly distributed if there exists a one-one correspondence between the sets $S$ and $\phi(S):=\{\phi(s_i):s_i\in S\}$ and $\text{PMF}_X=\text{PMF}_Y = \frac{1}{|S|}.$ Is it right or I am missing something?

Answer 1

"Uniformly distributed" is slightly counter-intuitive here, but seems to be intended to mean that the probabilities for each possible value are either $0$ or some constant $k$ for each possible value in the support, no matter what the pattern of the values are.

(B) is uniformly distributed because $f(S)\to \mathbb R:f_B(x)=x^3$ is an injective function so, since $|S|=21$, each possible value in the support of $X^3$ here has probability $\frac1{21}$.

(D) is uniformly distributed because $f(S)\to \mathbb R:f_D(x)=(x+10)^2$ is also an injective function for this particular $S$ so each possible value in the support of $X^3$ again has probability $\frac1{21}$.

(A) and (C) are not uniformly distributed. For example $\mathbb P(X^2=0)=\frac1{21}$ while $\mathbb P(X^2=1)=\frac2{21}$ and similarly $\mathbb P((X-5)^2=0)=\frac1{21}$ while $\mathbb P((X-5)^2=1)=\frac2{21}$.

Your final part of your question is correct: if you start with a uniformly distributed discrete distribution and then apply an injective function, you get another uniformly distributed discrete distribution. But

• You do not always need an injective function: if $\operatorname{rem}(a, b)$ is the remainder when dividing $a$ by $b$, i.e. the amount in $[0,b)$ by which $a$ exceeds a multiple of $b$, then here $\operatorname{rem}(X, 3)$ would be supported on $\{0,1,2\}$ each with probability $\frac13$ and so would be uniformly distributed.

• An injective function is insufficient for a uniformly distributed continuous distribution. If you had started with $X\sim U[-10,10]$ uniformly distributed on the real interval, then $X^3$ would not be uniformly distributed despite $f(x)=x^3$ being an injective function.