GATE 2025 DA Q.20: solution and answer

GATE 2025 · DAQ.20 · 1 mark · MCQ

Let $X = a Z + b$ , where $Z$ is a standard normal random variable, and $a$ , $b$ are two unknown constants. It is given that $E [X] = 1$ , $E [(X - E [X]) Z] = - 2$ , $E [(X - E [X])^{2}] = 4$ , where $E [X]$ denotes the expectation of random variable $X$ . The values of $a, b$ are:

(A)
$a = - 2, b = 1$
(B)
$a = 2, b = - 1$
(C)
$a = - 2, b = - 1$
(D)
$a = 1, b = 1$

See the question in the paper

Tap to open full size

Answer and solution

ANSWER

The values are $a = - 2$ and $b = 1$ , which is option (A).

Official answer: ASource: Official answer key, GATE 2025 DA (IIT Roorkee)

Since $Z$ is standard normal, $E [Z] = 0$ and $E [Z^{2}] = Var (Z) + (E [Z])^{2} = 1$ .
By linearity, $E [X] = a E [Z] + b = b$ . Given $E [X] = 1$ , we get $b = 1$ .
Then $X - E [X] = a Z + b - b = a Z$ .
$E [(X - E [X]) Z] = E [a Z^{2}] = a E [Z^{2}] = a$ . Given this equals $- 2$ , we get $a = - 2$ .
Check with the variance condition: $E [(X - E [X])^{2}] = a^{2} E [Z^{2}] = a^{2} = (- 2)^{2} = 4$ , which matches the given value. This condition alone only gives $a = \pm 2$ , so the sign comes from the covariance condition in the previous step.
So $a = - 2$ , $b = 1$ , which is option (A).

KEY IDEA

Write $X - E [X] = a Z$ . Then $E [(X - E [X]) Z] = a E [Z^{2}] = a$ because $E [Z^{2}] = 1$ for a standard normal. The mean gives $b$ and the covariance condition gives $a$ with its sign.

WATCH OUT FOR

Using only $E [(X - E [X])^{2}] = a^{2} = 4$ gives $a = \pm 2$ , so option (A) and option (B) both look possible. Option (B) has $b = - 1$ , which contradicts $E [X] = 1$ , and $a = 2$ contradicts the covariance condition. Another slip is to write $E [Z^{2}] = 0$ by confusing it with $E [Z] = 0$ .

From the GATE 2025 DA paper. Every question of this paper · Download the full paper · All 8 Random variables questions, with notes

GATE 2025 DA Q.20: solution and answer

Answer and solution

More Random variables questions