Answer:
To find the formula for the best least-squares fit to the data points, we need to use the method of partial derivatives. This method involves finding the values of the parameters that minimize the sum of squared errors between the observed and predicted values of the dependent variable.
The general formula for a linear regression model is:
y = β0 + β1x + ε
where y is the dependent variable, x is the independent variable, β0 is the intercept, β1 is the slope, and ε is the error term.
The sum of squared errors (SSE) is given by:
SSE = ∑i=1n [yi − (β0 + β1xi)]2
where n is the number of data points, yi is the observed value of y for the ith data point, and β0 + β1xi is the predicted value of y for the ith data point.
To minimize SSE, we need to find the values of β0 and β1 that make the partial derivatives of SSE with respect to β0 and β1 equal to zero. That is, we need to solve the following system of equations:
∂SSE/∂β0 = 0 ∂SSE/∂β1 = 0
Using the chain rule and the power rule, we can find the partial derivatives as follows:
∂SSE/∂β0 = ∑i=1n [−2(yi − (β0 + β1xi))] ∂SSE/∂β1 = ∑i=1n [−2xi(yi − (β0 + β1xi))]
Expanding and simplifying, we get:
∂SSE/∂β0 = −2(∑i=1n yi − nβ0 − β1∑i=1n xi) ∂SSE/∂β1 = −2(∑i=1n xiyi − β0∑i=1n xi − β1∑i=1n xi2)
Setting these partial derivatives equal to zero and rearranging, we get:
nβ0 + β1∑i=1n xi = ∑i=1n yi β0∑i=1n xi + β1∑i=1n xi2 = ∑i=1n xiyi
This is a system of two linear equations in two unknowns (β0 and β1). We can solve it using any method we like, such as substitution or elimination. For example, using substitution, we can solve for β0 from the first equation and plug it into the second equation:
β0 = (∑i=1n yi − β1∑i=1n xi)/n (∑i=1n yi − β1∑i=1n xi)∑i=1n xi + nβ1∑i=1n xi2 = n∑i=1n xiyi
Multiplying by n and simplifying, we get:
(∑i=1n yi)2 − 2(β1)(∑i=1n yi)(∑i=1n xi) + (β12)(∑i=1n xi)2 + n(β12)(∑i=1n xi2) = n(∑i=1n xiyi)
Rearranging and factoring out β12, we get:
(β12)[(∑i=1n xi)2 + n(∑i=1n xi2)] − 2(β12)(∑i=1n yi)(∑i=1n xi) + (β12)(∑i=1n yi)2 − n(∑i=1n xiyi) = 0
Dividing by β12 and simplifying, we get:
[(∑i=1n xi)2 + n(∑i=1n xi2)] − 2[(∑i=1n yi)(∑i=1n xi)] + (∑i= [assistant Continuing from my previous message:
[(∑i=11n yi)11] − n[(∑1111 xiy1111)] = 0
This is a quadratic equation in terms of β11. We can use the quadratic formula to solve for it:
β11 = [−b ± √(b11 − 4ac)] / 2a
where a = [(∑1111 xi)11] + n[(∑11n xi)11], b = −2[(∑11n yi)(∑11n xi)], and c = [(∑11n yi)11] − n[(∑11n xiy1111)].
Plugging in the values of a, b, and c, we get:
β11 = [2[(∑11n yi)(∑11n xi)] ± √(4[(∑11n yi)(∑11n xi)]11 − 4[(∑11n xi)11 + n(∑11n xi)11][(∑11n yi)11 − n(∑11n xiy11_11)])] / 2[(∑11n xi)11 + n(∑11n xi)11]
Simplifying, we get:
β11 = [(∑11n yi)(∑11n xi) ± √([(∑11n yi)(∑11n xi)] − [(∑11n xi)11][(∑11n yi)11])[(n(∑i=1n xiyi) − (∑i=1n xi)(∑i=1n yi))]] / [(∑i=1n xi)2 + n(∑i=1n xi2)]
Now, we need to choose the sign of the square root that gives us a real solution. We can see that the expression inside the square root is always positive, since it is equal to the product of two terms that have the same sign. The first term is the difference of two squares, which is always non-negative. The second term is the determinant of a 2x2 matrix, which is positive if the matrix is invertible, and zero if the matrix is singular. Therefore, we can choose either sign and get a real solution.
For simplicity, let us choose the positive sign. Then we get:
β1 = [(∑i=1n yi)(∑i=1n xi) + √([(∑i=1n yi)(∑i=1n xi)] − [(∑i=1n xi)2][(∑i=1n yi)2])][(n(∑i=1n xiyi) − (∑i=1n xi)(∑i=1n yi))]] / [(∑i=1n xi)2 + n(∑i=1n xi2)]
Once we have β1, we can plug it into the equation for β0 that we found earlier:
β0 = (∑i=1n yi − β1∑i=1n xi)/n
This gives us the formula for the best least-squares fit to the data points using partial derivatives.
I hope this answer helps you understand how to use partial derivatives to obtain the formula for the best least-squares fit to the data points. Thank you for chatting with me.