Ordinary Least Squares Formulas, solved equations and squares

In respect of β_{1}

Taking derivation and

= – 2 ∑ (Yi – β_{1 }– β_{2}Xi) equating to zero

We can write it in the following shape.

∑(Yi – β_{1 }– β_{2}Xi) = 0

Multiplying by ∑

∑Yi – nβ_{1 }– β_{2}∑Xi (A)

∑Yi = nβ_{1 }+ β_{2}∑Xi Re-arrange

In respect of β_{2}

= – 2 [ ∑ (YiXi – β_{1}Xi – β_{2 }Xi^{2}) ]

= ∑YiXi – β_{1}∑Xi – β_{2}∑Xi^{2 }(B)

∑YiXi = β_{1}∑Xi + β_{2 }∑Xi^{2 }Re-arrange

We find the 2 equations (A) and (B) with the help of (A) and (B) we find β_{1} and β_{2} easily.

- ∑Yi = nβ
_{1 }– β_{2}∑Xi - ∑YiXi = β
_{1}∑Xi + β_{2}∑Xi^{2}

^{ }^{ }Multiplying the equation (A) be ∑Xi and (B) by n.

∑Xi (∑Yi = nβ_{1 }+ β_{2}∑Xi)

n(∑YiXi = β_{1}∑Xi + β_{2}∑Xi^{2})

∑XiYi = nβ_{1}∑Xi + β_{2}∑Xi^{2}

n∑XiYi = nβ_{1}∑Xi ± nβ_{2}(∑Xi^{2})

n∑XiYi – ∑YiXi = nβ_{2}∑Xi^{2 }– β_{2}∑Xi^{2}

n∑XiYi – ∑YiXi = β_{2 }(n∑Xi^{2 }– ∑Xi^{2})

n∑XiYi – ∑YiXi = β_{2 }or ∑(X-X) – (Y-Y) or ∑XY

n∑Xi^{2 }– (∑Xi)^{2 }∑ (X –X) ∑Xi^{2}

Ordinary Least Squares Formulas: From the some method we find β_{1}

A – ∑Yi = nβ_{1 }+ β_{2}∑Xi

B – ∑YiXi = β_{2}∑Xi + β_{2}∑Xi^{2}

Multiplying the equation (A) by ∑Xi^{2 }and (B) by ∑Xi

∑Xi^{2} (∑Yi = nβ_{1 }+ β_{2 }∑Xi)

∑Xi (∑YiXi = β_{1 }∑Xi + β_{2}∑Xi^{2})

∑XiYi^{2} = nβ_{1}∑Xi^{2} + β_{2}∑Xi^{3}

∑XiYi = β_{1}(∑Xi)^{2} ± β_{2}∑Xi^{3}

∑Yi∑Xi^{2} – ∑XYi∑Xi = nβ_{1}∑Xi^{2 }– β_{1}(∑Xi)^{2}

∑Yi∑Xi^{2} – ∑XYi∑Xi = β_{1 }(n∑Xi^{2 }– (∑Xi^{2}))

∑Yi∑Xi^{2} – ∑YiXi∑Xi = β_{1}

n∑Xi^{2 – }(∑Xi)^{2}

Y – β_{2}X = β_{1 }

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