Multiple Linear Regression – Problem & Solution

Multiple Linear Regression explains the relationship between one dependent variable (Y) and two or more independent variables (X₁, X₂, …).

The general form:

$$Y = a + b_1 X_1 + b_2 X_2$$

Where:

• a = Intercept

• b₁, b₂ = Partial regression coefficients

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🔢 Example Problem (Two Independent Variables)

🔹 Question

Fit the regression equation:

$$Y = a + b_1 X_1 + b_2 X_2$$

Given data:

Obs	X₁	X₂	Y
1	1	2	4
2	2	1	3
3	3	4	8
4	4	3	7

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✅ Step 1: Model

$$Y = a + b_1 X_1 + b_2 X_2$$

We use Normal Equations:

$$\sum Y = na + b_1\sum X_1 + b_2\sum X_2$$ $$\sum X_1Y = a\sum X_1 + b_1\sum X_1^2 + b_2\sum X_1X_2$$ $$\sum X_2Y = a\sum X_2 + b_1\sum X_1X_2 + b_2\sum X_2^2$$

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✅ Step 2: Calculate Required Sums

First prepare working table:

X₁	X₂	Y	X₁²	X₂²	X₁X₂	X₁Y	X₂Y
1	2	4	1	4	2	4	8
2	1	3	4	1	2	6	3
3	4	8	9	16	12	24	32
4	3	7	16	9	12	28	21

Now totals:

$$\sum X_1 = 10$$ $$\sum X_2 = 10$$ $$\sum Y = 22$$ $$\sum X_1^2 = 30$$ $$\sum X_2^2 = 30$$ $$\sum X_1X_2 = 28$$ $$\sum X_1Y = 62$$ $$\sum X_2Y = 64$$ $$n = 4$$

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✅ Step 3: Form Normal Equations

$$22 = 4a + 10b_1 + 10b_2$$

$$62 = 10a + 30b_1 + 28b_2$$

$$64 = 10a + 28b_1 + 30b_2$$

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✅ Step 4: Solve Equations

From (2) − (3):

$$62 - 64 = (30b_1 - 28b_1) + (28b_2 - 30b_2)$$ $$-2 = 2b_1 - 2b_2$$ $$b_1 - b_2 = -1$$ $$b_1 = b_2 - 1$$

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Substitute into equation (1):

$$22 = 4a + 10(b_2 - 1) + 10b_2$$ $$22 = 4a + 10b_2 - 10 + 10b_2$$ $$22 = 4a + 20b_2 - 10$$ $$32 = 4a + 20b_2$$ $$8 = a + 5b_2$$ $$a = 8 - 5b_2$$

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Substitute into equation (2):

$$62 = 10(8 - 5b_2) + 30(b_2 -1) + 28b_2$$ $$62 = 80 - 50b_2 + 30b_2 - 30 + 28b_2$$ $$62 = 50 + 8b_2$$ $$12 = 8b_2$$ $$b_2 = 1.5$$

Then:

$$b_1 = b_2 - 1 = 0.5$$ $$a = 8 - 5(1.5)$$ $$a = 8 - 7.5 = 0.5$$

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✅ Final Regression Equation

$$Y = 0.5 + 0.5X_1 + 1.5X_2$$

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🔍 Interpretation

• If X₁ increases by 1, Y increases by 0.5 (keeping X₂ constant).

• If X₂ increases by 1, Y increases by 1.5 (keeping X₁ constant)

• Correlation and Regression