I'm writing my master thesis and have a problem with a multiple linear regression analysis:

I have a non-linear relationship between the independent and dependent variables, as you can see in the picture of the partial regression plot (showing the relationship between one independent and the dependent variable, the others look quite similar). So the condition for a regression is not met. My question: What can I do?

I have tried the transformation of the variables, but that didn't work. I have read that there are also non-linear and non-parametric methods. But most of them are not supported by SPSS. Thats all I know with my very basic and limited knowledge in statistics.

For my thesis I would like to know, if there is a "easy" way to analyse my data. If not, I would need a good explanation, I guess.

Ah and by the way, I have two samples (same problem) I want to compare. Maybe I could use the regression to compare both samples nevertheless?

I hope you understand my issue and can help me. Thanks :)

Partial Regression Plot

  • $\begingroup$ Where does this data come from? $\endgroup$
    – J.G.
    Commented Aug 15, 2019 at 5:53

2 Answers 2


Looking at the plot, it is quite clear (as you wrote it) that you face a problem of multilinear regression.

Let me suppose that the data points are $(x_i,y_i,z_i)$. The simple multilinear model is $$z=a+b x+c y+d x y$$

Try that (it is quite simple) and detect outliers (there will be a few of them looking at the plot). Remove them and rerun.

  • $\begingroup$ Hm, I'm not sure, whether you have understood my question. So, the plot is only a partial regression plot and show the relationship between one independent and the dependent variable. (The other relationships for each independent variable look quite similar.) $\endgroup$
    – KiRa
    Commented Jul 26, 2019 at 7:09
  • $\begingroup$ See the addition to my own answer. $\endgroup$
    – JJacquelin
    Commented Aug 16, 2019 at 11:22

I am not sure to well understand the problem. From the plot joint to the question the function $y(x)$ represented looks like a multivalued function on the kind : $$y(x)=a\,x+b \, \nu$$ where $a$ and $b$ are some parameters to be evaluated thanks to least mean square regression. The integer $\nu$ associated to a point $(x,y)$ characterizes the branch on which this point lies.

In a preliminary inspection (on a simplistic graphical approach) one can find this approximate result with : $$a\simeq \frac38 \quad;\quad b\simeq\frac18$$

enter image description here

Of course the above values for $a,b$ are not accurate but they can be used as initial "guess" for a global least mean squares regression.

Given the data $$(x_1,y_1),(x_2,y_2), … , (x_k,y_k), … , (x_n,y_n)$$ one computes the respective $\nu_k$ : $$\nu_k=\text{Round}\left(\frac{y_k-ax_k}{b}\right)$$ Round(X) is the integer the closest to the real $X$. In this formula the values of $a,b$ are the above "initial guess".

Then a linear regression is carried out for $a$ and $b$ , according to $$y_k=a\,x_k+b\,\nu_k+\epsilon_k$$ so that $\sum_{k=1}^n(\epsilon_k)^2$ be minimum.

Since the data is not published in the question, an approximate data has been obtained thanks to the scanning of the published plot. The result is : $$a=0.377007\quad;\quad b=0.125498$$ This is very close to the "guessed" values : $a=0.375\quad;\quad b=0.125$

enter image description here The root mean square error is :$\quad\text{RMSE}\simeq 0.014$


The pertinent comment of Claude Leibovici draw to not forget a constant parameter $c$ in the multivalued function : $$y(x)=a\,x+b \, \nu+c$$ With the same initial guess $a=\frac38$ , $b=\frac18$ and $c=0$ the three parameters regression gives : $$a=0.374254\quad;\quad b=0.125079\quad;\quad c=-0.011177$$

enter image description here

RMSE is significantly improved : $\quad\text{RMSE}\simeq 0.009$

  • $\begingroup$ Hi Jean ! What do you think about what I wrote in my deleted answer ? $\endgroup$ Commented Aug 15, 2019 at 6:44
  • $\begingroup$ Hi Claude ! This is another interpretation of the question (which is a bit ambiguous on my opinion). Except I would have consider an even simpler multilinear model as $z=a+bx+cy$ I would not have deleted your answer. The comment from KiRa let me hesitant. $\endgroup$
    – JJacquelin
    Commented Aug 15, 2019 at 8:02
  • $\begingroup$ The full multilenar problem is (to me) what I wrote. Since you have the data, do you mind to run it ? Or, can you send me the data (as a txt file) ? I am curious to see what is $d$. I shall undelete my answer. Thanks & cheers. $\endgroup$ Commented Aug 15, 2019 at 8:08
  • $\begingroup$ OK. I will send you the data (which is not from KiRa, but only approximates from scanning his plot). $\endgroup$
    – JJacquelin
    Commented Aug 15, 2019 at 8:22
  • 1
    $\begingroup$ Hi Jean ! My previous (deleted) comment was wrong (typo's in the $\nu_k$ values. Jus to add how significant are your parameters $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ a & -0.011177 & 0.00141 & \{-0.014036,-0.008318\} \\ b & +0.374254 & 0.00173 & \{+0.370753,+0.377755\} \\ c & +0.125079 & 0.00034 & \{+0.124388,+0.125770\} \end{array}$$ $\endgroup$ Commented Aug 19, 2019 at 4:14

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