The linear regression model is a fairly simple model, nevertheless it is used in many applications.

What follows is a fairly mathematical derivation of the linear regression parameters.

Let us denote the observations as $y[n]$ and $x[n]$ for $n=0,1,..,N-1$ for the model

$y[n] = ax[n]+b$

We assume that we can measure both $x[n]$ and $y[n]$ - the unknowns are $a$ and $b$, the gradient and intercept respectively. We can use the least squares error criterion to find estimates for $a$ and $b$, from data $x[n]$ and $y[n]$.

Let $e[n] = y[n] - ax[n]-b$ be the error. We can represent the samples $e[n]$, $x[n]$ and $y[n]$ as length $N$ column vectors $\underline{e}$,$\underline{x}$ and $\underline{y}$ respectively.

Thus the error sum of squares (which can be regarded as a cost function $C$ to minimise over) will be

$\underline{e}^T\underline{e}=(\underline{y}-a\underline{x}-b\underline{1})^T(\underline{y}-a\underline{x}-b\underline{1}$).

where $\underline{1}$ is a column vector with all elements set to 1.

Expanding this cost function results in:-

$C=\underline{y}^T\underline{y}-2a\underline{x}^T\underline{y}-2b\underline{y}^T\underline{1}+a^2(\underline{x}^T\underline{x})+2ab(\underline{x}^T\underline{1})+b^2(\underline{1}^T\underline{1})$

Differentiating $C$ w.r.t. to $a$ and $b$:-

$\frac{\partial C}{\partial a}=2a(\underline{x}^T\underline{x})+2b(\underline{x}^T\underline{1})-2\underline{x}^T\underline{y}$

$\frac{\partial C}{\partial b}=2a(\underline{x}^T\underline{1})+2b(\underline{1}^T\underline{1})-2\underline{y}^T\underline{1}$

In order to estimate $\hat{a}$ and $\hat{b}$ (the least squares estimates of $a$ and $a$ respectively) we need to set the above two partial differential equations to 0, and solve for $a$ and $b$. We need to solve the following matrix equation

$\left[\begin{array}{cc}\underline{x}^T\underline{x}&\underline{x}^T\underline{1}\\ \underline{x}^T\underline{1}&\underline{1}^T\underline{1}\end{array}\right]$$\left[\begin{array}{c}\hat{a}\\ \hat{b}\end{array}\right]$ = $\left[\begin{array}{c}\underline{x}^T\underline{y} \\ \underline{y}^T\underline{1} \end{array}\right]$

Inverting the matrix and pre multiplying both sides of the above matrix equation by the inverse results in:

$\left[\begin{array}{c}\hat{a} \\ \hat{b} \end{array}\right]$ = $\frac{1}{N(\underline{x}^T\underline{x})-(\underline{x}^T\underline{1})^2}$$\left[\begin{array}{cc}\underline{1}^T\underline{1}&-\underline{x}^T\underline{1}\\ -\underline{x}^T\underline{1}&\underline{x}^T\underline{x}\end{array}\right]$$\left[\begin{array}{c}\underline{x}^T\underline{y} \\ \underline{y}^T\underline{1} \end{array}\right]$

If we revert to a more explicit notation in terms of the samples, we have

$\hat{a}=\frac{1}{N(\sum_{n=0}^{N-1}(x[n])^2)-(\sum_{n=0}^{N-1}x[n])^2}[N(\sum_{n=0}^{N-1}x[n]y[n])-(\sum_{n=0}^{N-1}y[n])(\sum_{n=0}^{N-1}x[n])]$

$\hat{b}=\frac{1}{N(\sum_{n=0}^{N-1}(x[n])^2)-(\sum_{n=0}^{N-1}x[n])^2}[(\sum_{n=0}^{N-1}(x[n])^2)(\sum_{n=0}^{N-1}y[n])-(\sum_{n=0}^{N-1}x[n])(\sum_{n=0}^{N-1}x[n]y[n])]$

where the denominator of the fraction in the above two equations is the matrix determinant.

Once we have calculated $\hat{a}$ and $\hat{b}$, we can see how well the linear model accounts for the data. This is achieved by finding the Coefficient of Determination ($r^2$). This is the ratio of the residual sum of squares ($rss$) to the total sum of squares ($tss$),

$\Large r^2=\frac{rss}{tss}$

and has a maximum value of 1, which is attained when the model is exactly linear. The larger $r^2$ is, the better the linear model fits the data. We can think of $r^2$ as the proportion of variation in data explained by the linear model, or the "goodness of fit" of the linear model.

The total sum of squares is given by

$tss=\sum_{n=0}^{N-1}(y[n]-\bar{y})^2$

where

$\bar{y}=\frac{1}{N}\sum_{n=0}^{N-1}y[n]$ is the mean of $y$.

The residual sum of squares ($rss$) is the total sum of squares ($tss$) minus the sum of squared errors ($sse$)

where

$sse=\sum_{n=0}^{N-1}(y[n]-\hat{a}x[n]-\hat{b})^2$

If the data perfectly fits the linear model, $sse$ will equal 0, so that $tss$ will equal $rss$ - the coefficient of determination $r^2$ will be $1$.

Summarising, we have

Now, let us consider the following example:-

$x=[1,2,3,4]$, $y=[1.5,6,7,8]$

The matrix determinant will be $(4\times (1 + 2^2+3^2+4^2))-(1+2+3+4)^2=20$.

The least squares estimate of $a$ is

$\hat{a}=[4\times((1\times 1.5) + (2\times 6) + (3\times 7) + (4\times 8)) \\ -( (1.5+6+7+8)\times(1+2+3+4))]/20 \\ =2.05$

The least squares estimate of $b$ is

$\hat{b}=[(1+2^2+3^2+4^2)\times(1.5+6+7+8)-((1+2+3+4)\times((1\times 1.5) \\+ (2\times 6) + (3\times 7) +(4\times 8)))]/20 \\ = (675 - 665)/20 \\ =0.5$

The $tss$ is

$tss=(1.5-5.625)^2+(6-5.625)^2+(7-5.625)^2+(8-5.625)^2=24.688$

where the average of $y$ is $5.625$.

The $sse$ is

$(1.5 - (2.05\times 1) - 0.5)^2+(6 - (2.05\times 2) - 0.5)^2\\+(7 - (2.05\times 3) - 0.5)^2+(8 - (2.05\times 4) - 0.5)^2=3.675$

So that the $rss$ is $24.688-3.675=21.013$.

The coefficient of determination is

$r^2=21.013/24.688=0.8511$

This result is quite high, and is fairly indicative of a linear model.

There is a Linear Regression Calculator in this blog, which can be found here

Let $e[n] = y[n] - ax[n]-b$ be the error. We can represent the samples $e[n]$, $x[n]$ and $y[n]$ as length $N$ column vectors $\underline{e}$,$\underline{x}$ and $\underline{y}$ respectively.

Thus the error sum of squares (which can be regarded as a cost function $C$ to minimise over) will be

$\underline{e}^T\underline{e}=(\underline{y}-a\underline{x}-b\underline{1})^T(\underline{y}-a\underline{x}-b\underline{1}$).

where $\underline{1}$ is a column vector with all elements set to 1.

Expanding this cost function results in:-

$C=\underline{y}^T\underline{y}-2a\underline{x}^T\underline{y}-2b\underline{y}^T\underline{1}+a^2(\underline{x}^T\underline{x})+2ab(\underline{x}^T\underline{1})+b^2(\underline{1}^T\underline{1})$

Differentiating $C$ w.r.t. to $a$ and $b$:-

$\frac{\partial C}{\partial a}=2a(\underline{x}^T\underline{x})+2b(\underline{x}^T\underline{1})-2\underline{x}^T\underline{y}$

$\frac{\partial C}{\partial b}=2a(\underline{x}^T\underline{1})+2b(\underline{1}^T\underline{1})-2\underline{y}^T\underline{1}$

In order to estimate $\hat{a}$ and $\hat{b}$ (the least squares estimates of $a$ and $a$ respectively) we need to set the above two partial differential equations to 0, and solve for $a$ and $b$. We need to solve the following matrix equation

$\left[\begin{array}{cc}\underline{x}^T\underline{x}&\underline{x}^T\underline{1}\\ \underline{x}^T\underline{1}&\underline{1}^T\underline{1}\end{array}\right]$$\left[\begin{array}{c}\hat{a}\\ \hat{b}\end{array}\right]$ = $\left[\begin{array}{c}\underline{x}^T\underline{y} \\ \underline{y}^T\underline{1} \end{array}\right]$

Inverting the matrix and pre multiplying both sides of the above matrix equation by the inverse results in:

$\left[\begin{array}{c}\hat{a} \\ \hat{b} \end{array}\right]$ = $\frac{1}{N(\underline{x}^T\underline{x})-(\underline{x}^T\underline{1})^2}$$\left[\begin{array}{cc}\underline{1}^T\underline{1}&-\underline{x}^T\underline{1}\\ -\underline{x}^T\underline{1}&\underline{x}^T\underline{x}\end{array}\right]$$\left[\begin{array}{c}\underline{x}^T\underline{y} \\ \underline{y}^T\underline{1} \end{array}\right]$

If we revert to a more explicit notation in terms of the samples, we have

$\hat{a}=\frac{1}{N(\sum_{n=0}^{N-1}(x[n])^2)-(\sum_{n=0}^{N-1}x[n])^2}[N(\sum_{n=0}^{N-1}x[n]y[n])-(\sum_{n=0}^{N-1}y[n])(\sum_{n=0}^{N-1}x[n])]$

$\hat{b}=\frac{1}{N(\sum_{n=0}^{N-1}(x[n])^2)-(\sum_{n=0}^{N-1}x[n])^2}[(\sum_{n=0}^{N-1}(x[n])^2)(\sum_{n=0}^{N-1}y[n])-(\sum_{n=0}^{N-1}x[n])(\sum_{n=0}^{N-1}x[n]y[n])]$

where the denominator of the fraction in the above two equations is the matrix determinant.

__Coefficient of Determination__Once we have calculated $\hat{a}$ and $\hat{b}$, we can see how well the linear model accounts for the data. This is achieved by finding the Coefficient of Determination ($r^2$). This is the ratio of the residual sum of squares ($rss$) to the total sum of squares ($tss$),

$\Large r^2=\frac{rss}{tss}$

and has a maximum value of 1, which is attained when the model is exactly linear. The larger $r^2$ is, the better the linear model fits the data. We can think of $r^2$ as the proportion of variation in data explained by the linear model, or the "goodness of fit" of the linear model.

The total sum of squares is given by

$tss=\sum_{n=0}^{N-1}(y[n]-\bar{y})^2$

where

$\bar{y}=\frac{1}{N}\sum_{n=0}^{N-1}y[n]$ is the mean of $y$.

The residual sum of squares ($rss$) is the total sum of squares ($tss$) minus the sum of squared errors ($sse$)

where

$sse=\sum_{n=0}^{N-1}(y[n]-\hat{a}x[n]-\hat{b})^2$

If the data perfectly fits the linear model, $sse$ will equal 0, so that $tss$ will equal $rss$ - the coefficient of determination $r^2$ will be $1$.

Summarising, we have

$\Large r^2=\frac{\sum_{n=0}^{N-1}(y[n]-\bar{y})^2-\sum_{n=0}^{N-1}(y[n]-\hat{a}x[n]-\hat{b})^2}{\sum_{n=0}^{N-1}(y[n]-\bar{y})^2}$

__A worked example__Now, let us consider the following example:-

$x=[1,2,3,4]$, $y=[1.5,6,7,8]$

The matrix determinant will be $(4\times (1 + 2^2+3^2+4^2))-(1+2+3+4)^2=20$.

The least squares estimate of $a$ is

$\hat{a}=[4\times((1\times 1.5) + (2\times 6) + (3\times 7) + (4\times 8)) \\ -( (1.5+6+7+8)\times(1+2+3+4))]/20 \\ =2.05$

The least squares estimate of $b$ is

$\hat{b}=[(1+2^2+3^2+4^2)\times(1.5+6+7+8)-((1+2+3+4)\times((1\times 1.5) \\+ (2\times 6) + (3\times 7) +(4\times 8)))]/20 \\ = (675 - 665)/20 \\ =0.5$

The $tss$ is

$tss=(1.5-5.625)^2+(6-5.625)^2+(7-5.625)^2+(8-5.625)^2=24.688$

where the average of $y$ is $5.625$.

The $sse$ is

$(1.5 - (2.05\times 1) - 0.5)^2+(6 - (2.05\times 2) - 0.5)^2\\+(7 - (2.05\times 3) - 0.5)^2+(8 - (2.05\times 4) - 0.5)^2=3.675$

So that the $rss$ is $24.688-3.675=21.013$.

The coefficient of determination is

$r^2=21.013/24.688=0.8511$

This result is quite high, and is fairly indicative of a linear model.

There is a Linear Regression Calculator in this blog, which can be found here

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ReplyDeleteGreatest thanks to Dr Oyagu for his herbal medicine that he prepared for me and when I start using it in just 2weeks I was completely cured and that ended my HERPES SIMPLEX 1&2 DISEASE i am so happy and grateful to Dr Oyagu . after reading about him on a testimony of Jason Cash on a blogger. I knew suddenly Dr Oyagu was the right Doctor to cure my HERPES SIMPLEX 1&2 DISEASE. I discuss with Dr Oyagu and he prepared a herbal medicine for me and when it got sent to me in south Korean . I used the herbal medicine in 2weeks and i went for check up again. after 15years of suffering from HERPES SIMPLEX 1&2 at last i am smiling once again. Dr Oyagu also has remedy to others disease like COLD SORES,HIV/AIDS,DIABETES.CANCER,HIGH BLOOD PRESSURE AND MANY MORE. I oblige everyone to contact this powerful herbalist Dr Oyagu and be free from your suffering. contact his WhatsApp line: +2348101755322 or visit his Website https://oyaguspellcaster.wixsite.com/oyaguherbalhome Email:Oyaguherbalhome@gmail.com

ReplyDeleteI got diagnose of herpes virus last year, and i was taking some drugs prescribed for me by my family doc the drugs could not work and the herpes in my system was very terrifying that i was so depressed the good news is that i never gave up in searching for natural cure for herpes virus ,cause i believe so much in herbal medication.. One faithful morning i read a comment from a lady called Destiny Hudson on how she was cured with natural herbs made by DR.OYAGU from somewhere is West Africa, i immediately copied out his contact email via oyaguherbalhome@gmail.com and explain all my herpes problem to him via his email. the big truth is that it took just two week for his herbal medication to cure me completely without side affect. Till date am herpes virus free and all thanks to Dr OYAGU for his good deeds for me, Once again am very happy to share this great testimony of DR OYAGU cure hurry up now and contact him via his email address (oyaguherbalhome@gmail.com Call & WhatsApp him on (+2348101755322 ) and see what he can do

ReplyDeleteDear friends online My name is Destiny Hudson And I live in the USA, Dominica I am here today to appreciate the good work of Dr Oyagu, for helping me curing my HSV 1&2 Disease, I contacted the virus from my Partner, i found out he was cheating on me, so i broke up with him, 2 month later i went to test myself in the hospital when i was seeing sore on my mouth and Private areas, i was diagnosed of HSV 1&2. i tried using the acyclovir medicine but it didn't help in curing it, i went on a search and found out Dr Oyagu curing people illness, and so i got his email and i was told how i can get his herbal cure, 3 days later i received his medicine after using it for 2 weeks i was tested again and was confirm negative, today i am totally free. If you are over there looking for solution email: oyaguherbalhome@gmail.com or contact him on whatsapp +2348101755322 or visit his website https://oyaguspellcaster.wixsite.com/oyaguherbalhome today and share your own testimony. And He also has herbal cure for the Following DISEASES, eczema, urethra wart, chronic problems. Herpes, Cancer, Als, Hepatitis, Diabetes, HPV ,Infections, ulcer ETC

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