The next procedure accepts positive and/or negative values of $y$ (as well mixed values, some positive, some negative). The $x$-values can also be scattered.
[A typo was corrected. Thanks to ccorn for pointing it out. ]
For information, see : https://fr.scribd.com/doc/14674814/Regressions-et-equations-integrales
The integral equation involved is very simple : $\quad y(x)=c\int y(x)dx\:+$constant.
As an example, the graph published by Hagtar was scanned in order to get the scattered data "Series2". The graphical scanning can cause additional deviation, but certainly of low importance compared to the scatter of the data.
The approximate values of the parameters, computed thanks to the above procedure, are given on the figure, on which the fitted curve is drawn in red.
IN ADDITION, a brief appraisal of the effect of scatter :
The figure below shows a series of results depending on the scatter level.
The theoretical function is, for example: $y=c\:e^{b\:x}$ with $b=1.$ and $c=-0.1$
On a first graph "Noise amplitude =$0.$ " The very low discrepancy is entirely due to the calculus process (essentially the numerical integration).
The following graphs show the increase of RMSE as the scatter is increased.
Of course, the last graphs with extremely high scatter are not realistic. They are shown only to appraise the robustness of this special method of regression.
If someone knows a simpler method with such a robustness, I would be very grateful to make me know about.
There certainly is a drawback : The criteria of fitting isn't exactly the least mean square error. It is not far, but not exactly. Nevertheless, outside the scope of theoretical studies where the exactness is a rule, the criteria of least mean square error is not to take so strictly into account. Sometimes this criteria isn't even the best.
THE QUETION OF HAGTAR'S DATA :
Until Hagtar posts his Data (corresponding to the orange points on his graph) we have to use rough data imported from graphical scan of the Hagtar's graph.
I am reluctant to publish this data because it is certainly not correct, due to the graphical treatment. This is obvious, looking at the values of $x$ which should be integers.
Nevertheless I post the data record below, in order to answer to the demands of several people. This is the data directly issued from scanning software, without rounding : of course, many digits are without signifiance. While using this data, be aware that it is not the true Hagtar's data.
1.00536193029491 , 0.908450704225352
2.01072386058981 , 0.832394366197183
3.01608579088472 , 0.853521126760563
4.02144772117962 , 0.752112676056338
5.02680965147453 , 0.659154929577465
5.96514745308311 , 0.616901408450704
6.97050938337802 , 0.477464788732394
7.97587131367292 , 0.388732394366197
8.98123324396783 , 0.354929577464789
9.98659517426274 , 0.452112676056338
10.9919571045576 , 0.43943661971831
11.9973190348525 , 0.329577464788732
13.0026809651475 , 0.266197183098592
14.0080428954424 , 0.270422535211268
14.9463806970509 , 0.147887323943662
15.9517426273458 , 0.164788732394366
17.0241286863271 , 0.236619718309859
18.029490616622 , 0.185915492957746
19.0348525469169 , 0.147887323943662
19.9731903485255 , 0.0591549295774648
21.0455764075067 , 0.190140845070423
21.9839142091153 , 0.109859154929577
22.9892761394102 , 0.0464788732394366
23.9946380697051 , 0.156338028169014
25 , 0.0633802816901408
26.0053619302949 , 0.156338028169014
27.0107238605898 , 0
28.0160857908847 , 0.0845070422535211
29.0214477211796 , 0.0591549295774648
29.9597855227882 , 0.109859154929577
31.0321715817694 , 0.130985915492958
31.970509383378 , 0.101408450704225
32.9758713136729 , 0.0464788732394366
33.9812332439678 , -0.00422535211267606
35.0536193029491 , 0.0929577464788732
35.9919571045576 , -0.0380281690140845
37.0643431635389 , 0.0802816901408451
38.0026809651475 , 0.0338028169014084
39.0080428954424 , 0.0211267605633803
40.0134048257373 , -0.0380281690140845
41.0187667560322 , -0.0802816901408451
42.0241286863271 , 0.071830985915493
43.029490616622 , -0.0845070422535211
44.0348525469169 , -0.0422535211267606
45.0402144772118 , 0.0295774647887324
46.0455764075067 , -0.0126760563380282
47.0509383378016 , 0.0591549295774648
47.9892761394102 , 0.0802816901408451
48.9946380697051 , 0.105633802816901
Best Answer
HINT :
Before loosing one pair, the linear regression to fit the function $y=ax+b$ (with 5 pairs) leads to the values of $a=2$ and $b=-0.1$ : $$b=\frac{\sum y\sum x^2-\sum x\sum xy}{5\sum x^2-\left(\sum x\right)^2}$$ $$a=\frac{5\sum xy-\sum x\sum y}{5\sum x^2-\left(\sum x\right)^2}$$ Let $(X,Y)$ be the missing pair : $$\sum x=1+X\quad;\quad \sum x^2=0.3+X^2\quad;\quad \sum xy=0.47+XY\quad;\quad \sum y=1.5+Y$$ The $\sum$ continues to be for the 5 pairs. Putting them into the above equations : $$\begin{cases} -0.1=\frac{(1.5+Y)(0.3+X^2)-(1+X)(0.47+XY)}{5(0.3+X^2)-\left(1+X\right)^2}\\ 2=\frac{5(0.47+XY)-(1+X)(1.5+Y)}{5(0.3+X^2)-\left(1+X\right)^2} \end{cases}$$ Solve this system of two equations for the two unknowns $X,Y$.
LATTER ADDIION :
The result of solving is $X=0.3$ , $Y=0.6$
A short way :
One observe that the pairs $(0.1,0.1)$ , $(0.2,0.3)$ , $(0.4,0.7)$ are exactly on the regression line $y=2x-0.1$
On the other hand, the pair $(0.3,0.4)$ is not on the regression line since the point should be $(0.3,0.5)$ to be on the regression line, because $2*0.3-0.1=0.5$
The missing pair $(X,Y)$ must compensate so that $(0.4+Y)/2=0.5 \quad\to\quad Y=0.6$
So, the result is $X=0.3$ , $Y=0.6$