GATE EE 2024 Set 1 — Question 29

Let $S = \{-10, -9, \dots, 0, \dots, 9, 10\}$. The set $S$ contains $10 - (-10) + 1 = 21$ distinct values. Since $X$ is uniformly distributed over $S$, $P(X=x) = \frac{1}{21}$ for each $x \in S$.A random variable $Y$ is uniformly distributed if all its possible values have the same probability.(A) Consider $Y = X^2$.
Possible values for $X^2$ are $0^2=0$, $(\pm 1)^2=1$, $(\pm 2)^2=4$, ..., $(\pm 10)^2=100$.
For $Y=0$, $P(Y=0) = P(X=0) = \frac{1}{21}$.
For $Y=1$, $P(Y=1) = P(X=1) + P(X=-1) = \frac{1}{21} + \frac{1}{21} = \frac{2}{21}$.
Since $P(Y=0) \neq P(Y=1)$, $X^2$ is not uniformly distributed.(B) Consider $Y = X^3$.
Possible values for $X^3$ are $(-10)^3=-1000$, $(-9)^3=-729$, ..., $0^3=0$, ..., $9^3=729$, $10^3=1000$.
For any $x_1, x_2 \in S$, if $x_1^3 = x_2^3$, then $x_1 = x_2$. This means that each value in $S$ maps to a unique value in the set of $X^3$ values.
Therefore, for each $y_j$ in the set of $X^3$ values, $P(Y=y_j) = P(X=x_i)$ where $y_j = x_i^3$. Since $P(X=x_i) = \frac{1}{21}$ for all $x_i$, $P(Y=y_j) = \frac{1}{21}$ for all $y_j$ in its range. Thus, $X^3$ is uniformly distributed.(C) Consider $Y = (X-5)^2$.
Let $Z = X-5$. The values for $Z$ are $\{-10-5, -9-5, \dots, 10-5\} = \{-15, -14, \dots, 5\}$.
$P(Z=z) = \frac{1}{21}$ for each $z$ in its range.
Now consider $Y = Z^2$.
For $Y=0$, $P(Y=0) = P(Z=0) = P(X=5) = \frac{1}{21}$.
For $Y=1$, $P(Y=1) = P(Z=1) + P(Z=-1) = P(X=6) + P(X=4) = \frac{1}{21} + \frac{1}{21} = \frac{2}{21}$.
Since $P(Y=0) \neq P(Y=1)$, $(X-5)^2$ is not uniformly distributed.(D) Consider $Y = (X+10)^2$.
Let $Z = X+10$. The values for $Z$ are $\{-10+10, -9+10, \dots, 10+10\} = \{0, 1, \dots, 20\}$.
$P(Z=z) = \frac{1}{21}$ for each $z$ in its range. So $Z$ is uniformly distributed over $\{0, 1, \dots, 20\}$.
Now consider $Y = Z^2$.
Possible values for $Y$ are $0^2=0$, $1^2=1$, $2^2=4$, ..., $20^2=400$.
For any $z_1, z_2 \in \{0, 1, \dots, 20\}$, if $z_1^2 = z_2^2$, then $z_1 = z_2$ (since $z$ is non-negative).
Therefore, each value in the set $\{0, 1, \dots, 20\}$ maps to a unique value in the set of $Z^2$ values.
For each $y_j$ in the set of $Y$ values, $P(Y=y_j) = P(Z=z_i)$ where $y_j = z_i^2$. Since $P(Z=z_i) = \frac{1}{21}$ for all $z_i$, $P(Y=y_j) = \frac{1}{21}$ for all $y_j$ in its range. Thus, $(X+10)^2$ is uniformly distributed.Both (B) and (D) are uniformly distributed.The final answer is $\boxed{\text{B, D}}$.