Let X be a discrete random variable that is uniformly distributed over the set {-10, -9, ..., 0, ..., 9, 10}. Which of the following random variables is/are uniformly distributed ?

The correct options are X³ and (X + 10)².

Let us analyze the discrete random variable $X$. It is uniformly distributed over the set:

$S=\{-10,-9,...,0,...,9,10\}$

The set $S$ contains exactly $21$ distinct elements. Since $X$ is uniformly distributed, each value in $S$ occurs with equal probability:
$P(X=k)=\frac{1}{21}$ for each $k\in S$.

For a transformed random variable $Y=g\left(X\right)$ to be uniformly distributed, the function $g\left(x\right)$ must map the elements of $S$ to its range in a one-to-one (injective) manner. If multiple distinct elements in $S$ map to the same value in the range, the resulting distribution will not be uniform because some values in the range will have higher probabilities than others.

Let us test each option step-by-step:

1. Testing the random variable $X^3$:
The function $g\left(x\right)=x^3$ is strictly increasing and therefore one-to-one (injective) for all real numbers. Let us look at the values produced for each element in $S$:
$(-10{)}^3=-1000$, $(-9{)}^3=-729$, ..., $0^3=0$, ..., $9^3=729$, ${10}^3=1000$.
Since every element in $S$ maps to a unique value, the set of values that $X^3$ can take is:
$\{-1000,-729,...,0,...,729,1000\}$ (containing $21$ elements).
The probability of $X^3$ taking any of these values is exactly $\frac{1}{21}$. Thus, $X^3$ is uniformly distributed.

2. Testing the random variable $(X+10{)}^2$:
Let us compute the values of $Y=(X+10{)}^2$ as $X$ ranges from $-10$ to $10$:
For $X=-10\Rightarrow Y=(-10+10{)}^2=0$
For $X=-9\Rightarrow Y=(-9+10{)}^2=1$
For $X=-8\Rightarrow Y=(-8+10{)}^2=4$
...
For $X=0\Rightarrow Y=(0+10{)}^2=100$
...
For $X=9\Rightarrow Y=(9+10{)}^2=361$
For $X=10\Rightarrow Y=(10+10{)}^2=400$.
Since $X\ge -10$, the term $X+10$ is always non-negative ($X+10\ge 0$). The function $f\left(t\right)=t^2$ is strictly increasing for $t\ge 0$. Therefore, the transformation $(X+10{)}^2$ is one-to-one on the set $S$. Every element in $S$ maps to a unique perfect square in the set $\{0,1,4,...,400\}$.
Since the mapping is one-to-one, each of the $21$ possible outcomes in the range has a probability of exactly $\frac{1}{21}$. Thus, $(X+10{)}^2$ is uniformly distributed.

3. Why the other options are not uniformly distributed:

Let us look at $X^2$:
Here, both $X=-1$ and $X=1$ map to the same value $1$. This means:
$P(X^2=1)=P(X=1)+P(X=-1)=\frac{2}{21}$
However, for $X=0$, $P(X^2=0)=P(X=0)=\frac{1}{21}$.
Since the probabilities are not equal for all outcomes in the range, $X^2$ is not uniformly distributed.

Let us look at $(X-5{)}^2$:
Similarly, values symmetric around $5$ will map to the same square. For example, $X=4$ and $X=6$ both yield $(4-5{)}^2=1$ and $(6-5{)}^2=1$.
Thus, the mapping is not one-to-one, and $(X-5{)}^2$ is not uniformly distributed.