GATE 2025 DA Q.11: solution and answer

GATE 2025 · DAQ.11 · 1 mark · MCQ

Suppose $X$ and $Y$ are random variables. The conditional expectation of $X$ given $Y$ is denoted by $E [X ∣ Y]$ . Then $E [E [X ∣ Y]]$ equals

(A)
$E [X ∣ Y]$
(B)
$\frac{E [ X ]}{E [ Y ]}$
(C)
$E [X]$
(D)
$E [Y]$

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Answer and solution

ANSWER

$E [E [X ∣ Y]] = E [X]$ , which is option (C).

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

$E [X ∣ Y]$ is itself a random variable: it is a function $g (Y)$ of $Y$ , where $g (y) = E [X ∣ Y = y]$ .
Take its expectation over the distribution of $Y$ : $E [E [X ∣ Y]] = \sum_{y} E [X ∣ Y = y] P (Y = y)$ , or $\int g (y) f_{Y} (y) d y$ for continuous $Y$ .
Write $E [X ∣ Y = y] = \sum_{x} x P (X = x ∣ Y = y)$ and substitute: $\sum_{y} \sum_{x} x P (X = x ∣ Y = y) P (Y = y)$ .
Since $P (X = x ∣ Y = y) P (Y = y) = P (X = x, Y = y)$ , this equals $\sum_{x} x \sum_{y} P (X = x, Y = y)$ .
Summing the joint probability over $y$ gives the marginal: $\sum_{y} P (X = x, Y = y) = P (X = x)$ . So the total is $\sum_{x} x P (X = x) = E [X]$ .
Option (A) is still a random variable, not a number. Options (B) and (D) involve $Y$ 's mean, which has no reason to appear. The answer is (C).

KEY IDEA

Law of iterated expectations (tower property): averaging the conditional mean $E [X ∣ Y]$ over the distribution of $Y$ gives back the unconditional mean $E [X]$ .

WATCH OUT FOR

Treating $E [X ∣ Y]$ as if it were already a number and picking (A), or confusing the outer expectation with a ratio or with $E [Y]$ . The inner expectation is a function of $Y$ , and the outer one averages it over $Y$ .

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.11: solution and answer

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