GATE 2018 IN Q.3: solution and answer

GATE 2018 · INQ.3 · 1 mark · MCQ

$X$ and $Y$ are two independent random variables with variances 1 and 2, respectively. Let $Z = X - Y$ . The variance of $Z$ is

(A)
0
(B)
1
(C)
2
(D)
3

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Answer and solution

ANSWER

The variance of $Z = X - Y$ is $1 + 2 = 3$ , which is option (D).

Official answer: DSource: Official answer key, GATE 2018 IN (IIT Guwahati)

Given: $Var (X) = 1$ and $Var (Y) = 2$ . Because $X$ and $Y$ are independent, $Cov (X, Y) = 0$ .
The variance of a linear combination is $Var (a X + bY) = a^{2} Var (X) + b^{2} Var (Y) + 2 ab Cov (X, Y)$ .
For $Z = X - Y$ , take $a = 1$ and $b = - 1$ . Then $b^{2} = 1$ , and the covariance term is zero because $X$ and $Y$ are independent.
$Var (Z) = (1)^{2} (1) + (- 1)^{2} (2) + 0 = 1 + 2 = 3$
The variance of $Z$ is 3, which is option (D).

KEY IDEA

When two random variables are independent, their variances add, whether you add or subtract the variables: $Var (X \pm Y) = Var (X) + Var (Y)$ . The coefficient $- 1$ is squared, so the minus sign disappears.

WATCH OUT FOR

Subtracting the variances ( $1 - 2 = - 1$ , or $∣1 - 2∣ = 1$ , which gives option B) because $Z$ is a difference. Variance cannot be negative. The coefficient of $Y$ is squared ( $(- 1)^{2} = 1$ ), so the variances add. Only the means subtract: $E [Z] = E [X] - E [Y]$ .

From the GATE 2018 IN paper. Every question of this paper · Download the full paper · All 33 Probability & statistics questions, with notes

GATE 2018 IN Q.3: solution and answer

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