How to statistically determine how much two curves are different?

Hi,

I have two curves with diferent X and Y but the shape are similar. I would like to determin in percentagem or some number how they look like.

I tried Kolmogorov-Smirnov test in matlab but the problem is that my curve has different dimensions as well. One is 1x1001 and the other is 1x1000000.

Attached are the two data if it is helpfull.

Thanks for any help.

Regards,

Taynara

[Ive J]

I don't understand the question, what do you mean exactly by "to calculate how diff they are in numbers" ?

Why do you wanna use KS-test? what are you assumptions and research questions here?

Regardless, for "One is 1x1001 and the other is 1x1000000" you still can use kstest2.

[Tay]

Hi Ive,

I want to use ks-test because is a method to see how much those two curves are alike.

There are no assumptions here. Let say, that want to check how much my curve desviates from the reference.

Thanks, Ive. I did not know that we had a comand kstedt2, i was programing by myselft the calculations.

[Tay]

When say in number, it means that I dont want to check just byr eyes that they are diff but i want to quantify this.

[Ive J]

Every test (roughly) has some assumptions and KS is no exception here. So, in your case you're interested in testing if those two samples come from a same distribution, which is of course depends on your research questions and you cannot choose a test simply because it gives you some p-value without knowing the underlying assumptions. Say, you've measured refractive index in two different experimental settings (CT and Fraunhofer) and you are asking if both come from the same distribution (H0 = P_CT == P_Fraunhofer). But what happens if you reject H0 here? You should be able to discuss and interpret this.

That being said, if you just simply wanna see how much they differ with respect to your dependent variable (RI), the answer below would work for you.

[Tay]

Hi Ive,

Thank you for your reply.

Those curve are simulations, Franhoufer is a reference (not my simulation) and CIT is from my algorithum.

I am tryng to get close to Franhoufer curve. So the real question is how much CIT differ with respect to Franhoufer curve.

Hope was clear.

[Ive J]

In this case I guess you would be able to use a Wilcoxon Signed Rank Test (your data doesn't look fairly normal) test, but you must keep paired samples:

% height_CIT and CIT match
% height_Franhoufer and Franhoufer match
[f_cit, f_franhoufer] = ismember(hieght_CIT, height_Franhoufer); % under the assumption that height_CIT and height_Franhoufer have overlaps, otherwise you may use a cutoff for height to find their paired samples
f_franhoufer(f_franhoufer < 1) = [];

CIT = CIT(f_cit); Franhoufer = Franhoufer(f_franhoufer); % only keep paired samples
[p, h, stat] = signrank(CIT, Franhoufer); % does their underlying median significantly differ?

Nevertheless, as I also pointed out before, your research questions should be clear and of course the reason(s) why you intend to perform a hypothesis test. I suspect your data comes from some sort of simulation/modeling. In this scenario, you don't have any limitations in sample size, and as the result, you can easily increase the power to reject the null hypothesis (for a more complete discussion look here) rendering any p-value practically useless. But if the data come from some experiments (say you measured RI using different devices or experimental settings), then hypothesis tests may make sense.

You may also use correlation coefficients (your data looks monotonic, so you can use Spearman for instance) to see how well those two graphs correlate, but that does not measure the magnitude of their difference though.

[Tay]

Hi Ive,

Thank you very much for your comments. Vere revelevent and important reflection.

If you don`t mind let me do some questions...

In this particular case, what would be exalcly research questions for you? If you can give me an example will be very nice.

When you say that my data does not look normal, what does that mean?

Now related to your code...

CIT variable are the Y values from my data and Franhoufer variable is the Y values from Franhoufer data?

What is exacly height_CIT and height_Franhoufer? Would be the maximum value minus the minum value?

Thank you again.

Regards,

Taynara

[Ive J]

As an example, assume we measured blood pressure of some patients before and after some treatment. We wonder if this treatment has significanly influeced the blood pressure, so we compare the blood pressure between same patients before and after this treatment. We do that by comparing the median (only for the sake of using Wilcoxon Signed Rank Test, if we have enough sample size or normally distributed variables, we do a paired t-test instead) of two samples:

load hospital
[p, h] = signrank(hospital.BloodPressure(:, 1), hospital.BloodPressure(:, 2))
p =
   3.7965e-18
h =
  logical
   1

Small p-value here suggests we are safe to reject null hypothesis (our alpha is usually 0.05, which of course should be set before running any hypothesis test), and therefore, the median of two samples are not equal, so the treatment indeed has some effect on blood pressure values.

For normality, you should plot the histogram of your variable (see histogram) to see if your variables are normal or not (if not, you cannot use more powerful tests like t-test).

height_CIT and height_Franhoufer would be height value (on X-axis) for each value of your response (on Y-axis). For paired tests you need to compare the same samples together (look at my example above, each patient was measured twice, before and after the treatment). So your two vectors should be something like:

   sample height CIT Franhoufer
      1    5     1.51    1.55
      2    10    1.45    1.43
      .     .     .       .
   

[Stephen23]

https://web.archive.org/web/20211124063637/https://www.mathworks.com/matlabcentral/answers/736062-how-to-statistically-determine-how-much-two-curves-are-different

[Amanda Botelho Amaral]

I have a similar problem. How did you solve it?

Answer 1

[darova]

Use interp1 or spline to interpolate data. Here is an example

clc,clear

x1 = 0:0.2:4;               % fine mesh
x2 = 0:0.5:4;               % coarse mesh
y1 = sin(x1);               % fine data
y2 = sin(0.9*x2);           % coarse data
y22 = interp1(x2,y2,x1);    % create better data (interpolation)

plot(x1,y1,'.-r')
hold on
plot(x2,y2,'.-b')
plot(x1,y22)
plot([x1; x1],[y1; y22],'-b')
hold off

[Tay]

Thank, darova

I will try this one

[darova]

Please try and let know how it went