数据点回归直线的理论确定方法及给定数据的回归直线求解

Got it, let's tackle this regression line problem from both theory and practice—super straightforward once you break it down!

At its core, a regression line is all about minimizing the sum of squared errors (SSE)—this is the "least squares" method, the gold standard for linear regression. Here's the step-by-step theory:

First, let's define our terms clearly:

• We have n data points: {(x₁,y₁), (x₂,y₂), ..., (xₙ,yₙ)}
• Our predicted line (hypothesis function) is h(x) = w₀ + w₁x, where w₀ is the y-intercept and w₁ is the slope (I think you might have had a typo with w + hx—this is the standard notation used in most regression contexts).

The squared error loss function we want to minimize is:

L(w₀, w₁) = Σᵢ₌₁ⁿ (yᵢ - h(xᵢ))² = Σᵢ₌₁ⁿ (yᵢ - w₀ - w₁xᵢ)²

To find the w₀ and w₁ that make this loss as small as possible, we take the partial derivatives of L with respect to each coefficient, set them equal to zero, and solve the resulting system of equations.

• Step 1: Solve for w₀
  Take the partial derivative of L with respect to w₀, set it to 0:

  ∂L/∂w₀ = -2Σᵢ₌₁ⁿ (yᵢ - w₀ - w₁xᵢ) = 0
  

  Simplify this, and you'll find:

  w₀ = ȳ - w₁x̄
  

  Where x̄ is the mean of all x values, and ȳ is the mean of all y values. A handy sanity check: this tells us the regression line always passes through the point (x̄, ȳ).

• Step 2: Solve for w₁
  Now take the partial derivative of L with respect to w₁, set it to 0, and substitute the w₀ we just found:

  ∂L/∂w₁ = -2Σᵢ₌₁ⁿ xᵢ(yᵢ - w₀ - w₁xᵢ) = 0
  

  After simplifying, the slope w₁ comes out to:

  w₁ = [Σᵢ₌₁ⁿ (xᵢ - x̄)(yᵢ - ȳ)] / [Σᵢ₌₁ⁿ (xᵢ - x̄)²]
  

  You might also recognize this as the ratio of the covariance between x and y to the variance of x (Cov(x,y)/Var(x)).

The key idea here: this line is the one where the total squared vertical distance from each data point to the line is smaller than any other possible line. If we assume our errors are normally distributed (a common regression assumption), this least squares solution is also the maximum likelihood estimate.

Let's walk through a concrete example to make this tangible—if your dataset is different, just swap in your numbers!

Suppose we have these data points: (1,2), (2,3), (3,5), (4,6), (5,8)

Step 1: Compute basic summary statistics #

• Number of points: n = 5
• Mean of x values: x̄ = (1+2+3+4+5)/5 = 3
• Mean of y values: ȳ = (2+3+5+6+8)/5 = 4.8
• Compute the numerator for w₁: Σ(xᵢ - x̄)(yᵢ - ȳ)

  (1-3)(2-4.8) + (2-3)(3-4.8) + (3-3)(5-4.8) + (4-3)(6-4.8) + (5-3)(8-4.8)
  = (-2)(-2.8) + (-1)(-1.8) + 0*0.2 + 1*1.2 + 2*3.2
  = 5.6 + 1.8 + 0 + 1.2 + 6.4 = 15
  

• Compute the denominator for w₁: Σ(xᵢ - x̄)²

  (1-3)² + (2-3)² + (3-3)² + (4-3)² + (5-3)² = 4 + 1 + 0 + 1 + 4 = 10
  

Step 2: Calculate coefficients #

• Slope: w₁ = 15/10 = 1.5
• Intercept: w₀ = 4.8 - (1.5 * 3) = 4.8 - 4.5 = 0.3

Step 3: Final regression function #

Our line that minimizes squared error is:

h(x) = 0.3 + 1.5x

A quick edge case to note: if all your x values are identical (variance of x is 0), the regression line is just h(x) = ȳ—a horizontal line, since there's no relationship between x and y to model.

内容的提问来源于stack exchange，提问作者Deepak Garg