I-have-a-question-I-am-unsure-how-this-is-done-because-I-have-never-learnt-it-How-do-you-determine-the-line-of-best-fit-

Question Number 5117 by FilupSmith last updated on 14/Apr/16

I have a question . I am unsure how this

is done because I have never learnt it .

How do you determine the line of best fit ?

Commented by Yozzii last updated on 14/Apr/16

L e a s t s q u a r e s m e t h o d i s o n e w a y .

G i v e n a s e t o f n p o i n t s (x_{i}, y_{i}), s u p p o s e

t h a t t h e l i n e o f b e s t f i t h a s t h e f o r m

y = m x + c . T h e n, f o r e a c h x_{i}, w e g e t

t h e r e s u l t y_{j} = {m x}_{i} + c . T h e l e a s t s q u a r e s

m e t h o d r e q u i r e s t h a t f o r t h e r e g r e s s i o n l i n e

y o n x, t h e q u a n t i t y Q m u s t b e m i n i m i s e d

w h e r e Q = \underset{i = 1}{\sum^{n}} {(y_{i} - y_{j})}^{2} = \underset{i = 1}{\sum^{n}} {(y_{i} - {m x}_{i} - c)}^{2}

T h i s m e a n s t h a t w e a i m t o m i n i m i s e

t h e s q u a r e d d i s t a n c e s b e t w e e n a g i v e n y_{i} a n d

t h e p o i n t o n t h e l i n e y_{j} c o r r e s p o n d i n g t o

x_{i} . Q i s a f u n c t i o n o f t w o v a r i a b l e s m

a n d c s i n c e w e a s s u m e t h a t (x_{i}, y_{i}) a r e

k n o w n . W e c a n t h e n e m p l o y t e c h n i q u e s

o f m u l t i v a r i a b l e c a l c u l u s t o f i n d m a n d

c . S i n c e Q i s o f a q u a d r a t i c f o r m i t s

s t a t i o n a r y v a l u e i s a m i n i m u m o n e;

i n 3 D s p a c e, t h e l o c u s o f p o i n t s (m, c, Q) i s a b o w l s u r f a c e

w h e r e Q ⩾ 0 .

\Rightarrow \frac{\partial Q}{\partial m} = \underset{i = 1}{\sum^{n}} (- 2 x_{i}) (y_{i} - {m x}_{i} - c)

\frac{\partial Q}{\partial m} = 2 m (\underset{i = 1}{\sum^{n}} x_{i}^{2}) + 2 c (\underset{i = 1}{\sum^{n}} x_{i}) - 2 (\underset{i = 1}{\sum^{n}} x_{i} y_{i})

a n d \frac{\partial Q}{\partial c} = \underset{i = 1}{\sum^{n}} (- 2) (y_{i} - {m x}_{i} - c)

\frac{\partial Q}{\partial c} = 2 m (\underset{i = 1}{\sum^{n}} x_{i}) + 2 c (\underset{i = 1}{\sum^{n}} 1) - 2 \underset{i = 1}{\sum^{n}} y_{i}

o r \frac{\partial Q}{\partial c} = 2 m (\underset{i = 1}{\sum^{n}} x_{i}) + 2 c n - 2 (\underset{i = 1}{\sum^{n}} y_{i})

A t t h e s t a t i o n a r y p o i n t, \frac{\partial Q}{\partial m} = 0 a n d \frac{\partial Q}{\partial c} = 0 .

Y o u g e t t h e f o l l o w i n g r e s u l t f o r m

f r o m t h e s e t w o e q u a t i o n s b y e l i m i n a t i n g c .

m = \frac{n \underset{i = 1}{\sum^{n}} x_{i} y_{i} - (\underset{i = 1}{\sum^{n}} x_{i}) (\underset{i = 1}{\sum^{n}} y_{i})}{n \underset{i = 1}{\sum^{n}} x_{i}^{2} - {(\underset{i = 1}{\sum^{n}} x_{i})}^{2}}

I t c a n s h o w n t h a t (\bar{x}, \bar{y}) = (\frac{\underset{i = 1}{\sum^{n}} x_{i}}{n}, \frac{\underset{i = 1}{\sum^{n}} y_{i}}{n})

l i e s o n t h e l i n e o f b e s t f i t, s o c c a n b e

f o u n d f r o m c = \bar{y} - m \bar{x} . T h e r e s u l t

o f y = m x + c i s t h e l e a s t s q u a r e s l i n e o f

b e s t f i t .

(\bar{x}, \bar{y}) i s t h e c e n t r o i d o f a l l t h e p o i n t s

(x_{i}, y_{i}) . m h a s a n o t h e r f o r m .

m = \frac{\underset{i = 1}{\sum^{n}} x_{i} y_{i} - n \bar{x} \bar{y}}{\underset{i = 1}{\sum^{n}} x_{i}^{2} - n {(\bar{x})}^{2}} .

Facebook Tweet Pin