GATE DA 2025 | Question 20

MCQ · 1 mark · −0.33

Let \(X = aZ + b\), where \(Z\) is a standard normal random variable, and \(a\), \(b\) are two unknown constants. It is given that

\[E[X] = 1, \quad E[(X - E[X])Z] = -2, \quad E[(X - E[X])^{2}] = 4,\]

where \(E[X]\) denotes the expectation of random variable \(X\). The values of \(a, b\) are:

  1. Option A: \(a = -2, b = 1\) (correct)
  2. Option B: \(a = 2, b = -1\)
  3. Option C: \(a = -2, b = -1\)
  4. Option D: \(a = 1, b = 1\)

Answer

(A)

Explanation

Use \(E[Z] = 0\) and \(E[Z^{2}] = \operatorname{Var}(Z) = 1\).

  • \(E[X] = aE[Z] + b = b\), so \(b = 1\).
  • \(X - E[X] = aZ\), so \(E[(X - E[X])Z] = aE[Z^{2}] = a\), which gives \(a = -2\).
  • Check: \(E[(X - E[X])^{2}] = a^{2}E[Z^{2}] = 4\) holds. This condition alone cannot fix the sign of \(a\); the second one does.

Option C has the right \(a\) but the wrong mean, while B and D fail the condition \(E[(X - E[X])Z] = -2\).

Answer: A