GATE 2025 DA Q.34: solution and answer

GATE 2025 · DAQ.34 · 1 mark · NAT

Given data ${(- 1, 1), (2, - 5), (3, 5)}$ of the form $(x, y)$ , we fit a model $y = w x$ using linear least-squares regression. The optimal value of $w$ is ____________. (Round off to three decimal places)

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Answer and solution

ANSWER

The optimal value is $w = \frac{2}{7} \approx 0.286$ .

Official answer: 0.285 to 0.287Source: Official answer key, GATE 2025 DA (IIT Roorkee)

The model $y = w x$ has no intercept, so the least-squares objective is $J (w) = \sum_{i} (y_{i} - w x_{i})^{2}$ , a function of the single parameter $w$ .
Setting $\frac{dJ}{d w} = - 2 \sum_{i} x_{i} (y_{i} - w x_{i}) = 0$ gives $w \sum_{i} x_{i}^{2} = \sum_{i} x_{i} y_{i}$ . Since $\frac{d ^{2} J}{d w ^{2}} = 2 \sum_{i} x_{i}^{2} > 0$ , this stationary point is the minimum. Hence $w = \frac{\sum _{i} x _{i} y _{i}}{\sum _{i} x _{i}^{2}}$ .
Numerator: $\sum x_{i} y_{i} = (- 1) (1) + (2) (- 5) + (3) (5) = - 1 - 10 + 15 = 4$ .
Denominator: $\sum x_{i}^{2} = (- 1)^{2} + 2^{2} + 3^{2} = 1 + 4 + 9 = 14$ .
So $w = \frac{4}{14} = \frac{2}{7} \approx 0.2857$ , which rounds to $0.286$ .

KEY IDEA

With no intercept, least squares in one parameter gives $w = \frac{\sum x _{i} y _{i}}{\sum x _{i}^{2}}$ , which is the projection of the $y$ vector onto the $x$ vector.

WATCH OUT FOR

Fitting a line with an intercept, $y = w_{0} + w_{1} x$ , or using the mean-centred formula $\frac{\sum ( x _{i} - x ˉ ) ( y _{i} - y ˉ )}{\sum ( x _{i} - x ˉ ) ^{2}}$ . The question fixes the model as $y = w x$ , so no centring applies.

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GATE 2025 DA Q.34: solution and answer

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