## Derivation of Expressions for Multiple Regression Coefficients

Supose we suspect that an observable
$Y$
and random variables
$X+- , \: i=1,2,3,...n$
are linearly related. We can try to find an relationship of the form
$y_i =\beta_0+\beta_1(x_{i1}- \bar{x}_1)+ \beta_2(x_{i2}-\bar{x}_2))+...+\beta_n(x_{in}- \bar{x}_n)$
.
The statistically most desirable way to find the coefficients
$\beta_k, \: k=1,2,...,n$
is to minimise the sum of the squares of the errors, so that if the observable
$Y$
and the random variables
$X_i, \: 1,2,3,...,n$
to have paired values, then we want to minimise
$E=\sum^n_{i=1} e^2_i= \sum^n_{i=1} (y_i-(\beta_0+\beta_1(x_{i1}- \bar{x}_1)+ \beta_2(x_{i2}-\bar{x}_2))+...+\beta_n(x_{in}- \bar{x}_n))^2$
.
Find the partial derivatives and set each equal to zero.
$\frac{\partial E}{\partial \beta_0}=\sum^n_{i=1} 2 (\beta_0 - y_i) =0 \rightarrow n \beta_0 = \sum^n_{i=1} y_i \rightarrow \beta_0 = \frac{\sum^n_{i=1} y_i}{n}$

$\frac{\partial E}{\partial \beta_k}=\sum^n_{i=1} (-2(x_{ik}-\bar{x}_k) (y_i- \bar{y})+ 2(x_{ik}- \bar{x}_k)^2 =0 \rightarrow n \beta_k = \frac{\sum^n_{i=1} (x_{ik}-\bar{x}_k) (y_i- \bar{y})}{\sum^n_{i=1} (x_{ik}- \bar{x}_k)^2}$

Refresh