GATE 2024 EE Q.29: solution and answer

GATE 2024 · EEQ.29 · 1 mark · MSQ

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

(A)
$X^{2}$
(B)
$X^{3}$
(C)
$(X - 5)^{2}$
(D)
$(X + 10)^{2}$

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Answer and solution

ANSWER

(B) $X^{3}$ and (D) $(X + 10)^{2}$ are uniformly distributed. $X^{2}$ and $(X - 5)^{2}$ are not.

Official answer: B, DSource: Official answer key, GATE 2024 EE (IISc Bengaluru)

$X$ takes 21 equally likely values, so $P (X = k) = \frac{1}{21}$ for $k = - 10, - 9, \dots, 10$ .
For a function $g$ of a discrete random variable, $P (g (X) = y) = \sum_{x : g (x) = y} P (X = x)$ . Every support point carries $\frac{1}{21}$ , so $P (g (X) = y) = \frac{n ( y )}{21}$ , where $n (y)$ is the number of support points with $g (x) = y$ . So $g (X)$ is uniform on its range exactly when every value $y$ in the range has the same count $n (y)$ . In particular, if $g$ is one-to-one on the support then every $n (y) = 1$ and $g (X)$ is uniform. If some values have $n (y) = 2$ and others have $n (y) = 1$ , it is not uniform.
(A) $X^{2}$ : the value $0$ comes only from $X = 0$ , so $P (X^{2} = 0) = \frac{1}{21}$ . But $X = k$ and $X = - k$ give the same $k^{2}$ , so $P (X^{2} = k^{2}) = \frac{2}{21}$ for $k = 1, \dots, 10$ . Not uniform.
(B) $X^{3}$ : the cube is strictly increasing, so it is one-to-one. $X^{3}$ takes the 21 distinct values $- 1000, - 729, \dots, 729, 1000$ , each with probability $\frac{1}{21}$ . Uniform.
(C) $(X - 5)^{2}$ : $X - 5$ runs over $- 15, - 14, \dots, 5$ . For $k = 1, \dots, 5$ , both $X - 5 = k$ and $X - 5 = - k$ lie in this range, so $P ((X - 5)^{2} = k^{2}) = \frac{2}{21}$ . The value $0$ comes only from $X = 5$ , and for $k = 6, \dots, 15$ only $X - 5 = - k$ is possible, so $0$ and $6^{2}, \dots, 1 5^{2}$ each have probability $\frac{1}{21}$ . (Count check: $1 + 5 \cdot 2 + 10 = 21$ support points.) For example, $P ((X - 5)^{2} = 0) = \frac{1}{21} \neq = \frac{2}{21} = P ((X - 5)^{2} = 1)$ . Not uniform.
(D) $(X + 10)^{2}$ : $X + 10$ runs over $0, 1, \dots, 20$ , and all of these are non-negative. Squaring is strictly increasing on non-negative numbers, so it is one-to-one there. $(X + 10)^{2}$ therefore takes the 21 distinct values $0, 1, 4, \dots, 400$ , each with probability $\frac{1}{21}$ . Uniform.
Correct options: (B) and (D).

KEY IDEA

When $X$ is uniform on $N$ points, $P (g (X) = y)$ equals (the number of support points that $g$ sends to $y$ ) divided by $N$ . So $g (X)$ is uniform exactly when every value in its range has the same number of preimages. Any one-to-one map keeps the distribution uniform: a strictly monotone function such as $x^{3}$ , or squaring on a set whose values all have the same sign, as with $X + 10 \geq 0$ . A square of a set that contains both $k$ and $- k$ for some $k$ , next to values that have no partner, is not uniform.

WATCH OUT FOR

(1) Rejecting (D) because "a square is never one-to-one". Squaring is one-to-one on ${0, 1, \dots, 20}$ because every value is non-negative. (2) Thinking the shift in (C) removes the $\pm$ pairing. Since $X - 5$ still runs from $- 15$ up to $+ 5$ , the values $\pm 1, \dots, \pm 5$ still pair up. (3) Treating "uniform" as meaning evenly spaced values and rejecting (B) or (D) because $- 1000, \dots, 1000$ or $0, 1, 4, \dots, 400$ are unequally spaced. Uniformity is about equal probabilities, not about spacing.

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GATE 2024 EE Q.29: solution and answer

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