I got a distribution of variable "y" (unit=mm). Similarly, I have got two more distributions (x1 and x2 unit =mm/s). These two distributions are same but one (i.e x1) is distributed closer to y while other is distributed far from y. However, both x1 and x2 follows same distribution. I want to fit them corresponding to "y" by introducing a constant "C (unit (s))". However, I want to prefer the variable (i.e. x1) which is closer to y and I want to skip x2 (which is far from "y"). Simply, I want to get the result that "x1" has more capability to fit y variable in comparison to "x2". For this I want to conduct a statistical test. Can anyone help me in this regard? Thanks and Regards
Distributions are shown pictorially as attached below.
Hello Ajay,
What you need is a test for goodness of fit.
Generally, it is performed on one distribution vs any standard reference distribution, but in your case you take Y as your reference, and run goodness of fit for x1.
Or, you can analyze qq plots between Y and X1/X2, and check which one has a higher R2.
Another method is, calculate pearson correlation between Y and X1/X2. Whichever has higher R, select that.
I hope this helps.
Anubhav Das
Pearson correlation and R² is meant for linear correlation which is not the case here. (this also applies to Mercedes Orús-Lacort as linear regression assumes linear dependency of the dependent variable on the independent one).
Ajay Goyal
I think you shouldn't need a statistical test to argue that x1 fits Y better than x2 when you can present the graphic above. It's obvious. If you still need some stats to undermine your decision, compute differences of data-points of Yi-x1i and Yi-x2i, with i between -1 and 1 with 0.2 increments. In total there should be high differences.
It seems that the lines shown in the figure are functions of the form
f(x; a) := a * 2*x0.5*(1-x)0.5
where a is a parameter specifying the height at the maximum (which is at x=0.5, y=a). The parameter values for your data is a = 1.0 for "y", a = 0.9 for "x1" and a = 0.2 for "x2".
It's only the parameter a in what these curves differ, and this is the maximum. So you can easily find the most similar curves at those which have the most similar maximum values (that is, the value at x=0.5).
If you want to find the "best (closest) match", there is no statistics involved. It's only about finding this curve with the smallest difference of its parameter value a to 1. In your example thei difference is 1-0.9 = 0.1 for "x1" and 1-0.2 = 0.8 for "x2". Since the difference for "x1" is the smallest, "x1" is the best match to "y".
NB: the curves are not distributions. The curves are functions (or graphs of functions).
Dear Anubhav Das and Mercedes Orús-Lacort
Thank you very much for your valuable time and reply. However, with due respect, R2 will give equal correlation coefficient (i.e.=1.) as I mentioned that both distributions are same like dependent variable. However, one is close to dependent variable and the other (x2) is far from it. One other example that may be considered is "three concentric circle with different radius where one circle (say with radius=1) represents dependent variable and I have to select a circle (out of two one with radius =0.9 and other r=0.1) closer to it for fitting the reference circle of radius 1". Similar observations is explained by Jochen Wilhelm and Alexander Steiger . So, R2 may not be solve my purpose. However, I will try a goodness of fit test (Say KS or AD test).
Dear Alexander Steiger and Jochen Wilhelm
Thank you very much for giving the plot idea. I can also compute SSE. I think both will work. Goodness of fit test can also be a solution to this problem. Thanks again.
Best Regards to all.
Ajay
The objective of the study is not clear.
Test the goodness of fit is to be performed for each of the two distributions in order to select which one is better.
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