GATE 2026 EE Q.64: solution and answer

GATE 2026 · EEQ.64 · 2 marks · NAT

Let X and Y be two real-valued random variables with $E (X) = 1$ , $E (Y) = 2$ , $E (X^{2}) = 4$ , $E (Y^{2}) = 9$ , and $E (X Y) = 0.9$ , where $E$ denotes the expectation operator.

The value of $α$ that minimizes $E ((X - α Y)^{2})$ is ______ (Round off to one decimal place)

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Answer and solution

ANSWER

$α = 0.1$ minimises $E ((X - α Y)^{2})$ . The minimum value is $3.91$ .

Official answer: 0.1Source: Official answer key, GATE 2026 EE (IIT Guwahati)

Expand the square. Because $α$ is a constant, linearity of expectation gives $E ((X - α Y)^{2}) = E (X^{2}) - 2 α E (X Y) + α^{2} E (Y^{2})$
Substitute the given second moments: $f (α) = 4 - 2 (0.9) α + 9 α^{2} = 9 α^{2} - 1.8 α + 4$
Set the first derivative to zero: $f^{'} (α) = 18 α - 1.8 = 0 \Rightarrow α = \frac{1.8}{18} = 0.1$ In general the minimiser is $α^{*} = \frac{E ( X Y )}{E ( Y ^{2} )} = \frac{0.9}{9} = 0.1$ .
Check it is a minimum: $f^{''} (α) = 18 = 2 E (Y^{2}) > 0$ , so $f$ is an upward-opening parabola and this stationary point is its global minimum. The minimum value is $f (0.1) = 4 - 0.18 + 0.09 = 3.91$ .
The means $E (X) = 1$ and $E (Y) = 2$ are not needed. The expression contains only raw (non-central) second moments, so it does not depend on the means. Hence $α = 0.1$ .

KEY IDEA

$E ((X - α Y)^{2}) = E (X^{2}) - 2 α E (X Y) + α^{2} E (Y^{2})$ is a quadratic in $α$ with a positive leading coefficient. Its vertex is at $α^{*} = E (X Y) / E (Y^{2})$ . This is the least-mean-square (projection) coefficient of $X$ onto $Y$ , and only raw second moments enter it.

WATCH OUT FOR

Using central moments by mistake, for example computing $Cov (X, Y) / Var (Y) = (0.9 - 2) / (9 - 4) = - 0.22$ . That ratio minimises $E ((X - μ_{X} - α (Y - μ_{Y}))^{2})$ , which is a different expression. Other slips are dropping the factor 2 on the cross term, which gives $0.2$ , and writing $E (X) E (Y)$ where the question needs $E (X Y)$ .

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GATE 2026 EE Q.64: solution and answer

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