Forcing the best fit line to pass through a certain point (Octave)

[Wrichik Basu]

I have the following code in Octave:

h = [29.3 25.3 19.7 16.0 10.9];
v = [0.53 0.47 0.37 0.29 0.21];
plot(h,v,'obk')
hold on
p = polyfit(h,v,1);
y = polyval(p,h);
plot(h,y,'-bk')

And I get a good graph:

I can extrapolate the best fit line using the following code:

x = -1:0.01:11;
>> y = polyval(p,x);
>> plot(x,y,'--r')

and if I zoom on the graph, I get this:

Evidently, the line doesn't pass through (0,0).

But I have to make it pass through the Origin. In that case, it will no longer be the best fit line, but nevertheless it will serve my purpose.

Any idea on how to do this?

 

[jedishrfu]

You could add 0,0 to your collection of points.

You might find something in this discussion on Octave

https://octave.1599824.n4.nabble.com/fit-data-trendline-through-zero-td1610448.html
or this one for MATLAB but you'd need to find equivalent functions (google search maybe):

https://www.mathworks.com/matlabcen...ce-the-intercept-of-a-regression-line-to-zero

 

[marcusl]

Find the angle of the line through origin that minimizes mean squared error.

 

[RPinPA]

Evidently, the line doesn't pass through (0,0).

But I have to make it pass through the Origin. In that case, it will no longer be the best fit line, but nevertheless it will serve my purpose.

Any idea on how to do this?

Yes. You're fitting a different model. Instead of $y = mx + b$ you want to fit $y = mx$. I don't know if you can do it with polyfit(), but the math is pretty simple.

Minimize the square error E
$E = \sum_i (y - y_i)^2 = \sum_i (mx_i - y_i)^2 = \sum_i m^2 x_i^2 - 2 \sum_i mx_i y_i + \sum_i y_i^2$

$dE/dm = 0 \\ \Rightarrow 2m \sum_i x_i^2 - 2\sum_i x_i y_i = 0 \\ \Rightarrow m = (\sum_i x_i y_i )/ (\sum_i x_i^2)$

That is the best fit value of $m$ for the model $y = mx$, and you should interpolate / extrapolate using that model.

 

[FactChecker]

In the MATLAB function fitlm, you can specify the desired model that does not have a constant term using a modelspec like 'Y ~ A + B + C – 1' . See https://www.mathworks.com/help/stats/fitlm.html#bt0ck7o-modelspec
I believe that their other linear regression tools have similar capabilities. I don't know about Octave.

 

[Wrichik Basu]

I don't know about Octave.

Octave says that fitlm has not yet been implemented.

 

[Wrichik Basu]

@RPinPA Thanks, that works fine. I plotted the function using fplot, and I am getting the desired results.

You could add 0,0 to your collection of points.

Good idea.

 

[FactChecker]

You could add 0,0 to your collection of points.

That would draw the line toward (0,0). You may need to add it many times to get it as close as you want and then all the statistical calculations would be messed up.

 

[jedishrfu]

That would draw the line toward (0,0). You may need to add it many times to get it as close as you want and then all the statistical calculations would be messed up.

Sometimes cheap solutions work but not as well as one would like that’s why they’re cheap.

Octave apparently doesn’t have the fitlm() function but does have some linear regression methods.

https://octave.sourceforge.io/optim/function/LinearRegression.html

 

[Wrichik Basu]

Sometimes cheap solutions work but not as well as one would like that’s why they’re cheap.

Absolutely, I don't expect any better. How much can one provide for free? There are some major differences between Matlab and Octave. For example, for symbolic math, Octave depends on SymPy, while Matlab was created much before Python.

 

[jedishrfu]

I sometimes use freemat when I need to compute a quick plot. It has much of the core Matlab functionality and is easy to install.

More recently, Julia from MIT has come online to challenge Matlab in performance. Much of its syntax is similar to Matlab with notable differences in how arrays are referenced ie parens in Matlab vs square brackets in Julia. Many folks are extending the Julia ecosystem with new packages on github everyday. It’s main weakness is its IDE which is cobbled together using Juno or using Jupyter notebooks. The notebooks are preferred over the IDE but Matlab users have a great IDE that’s hard to give up.

 

[statdad]

In general forcing a regression line to go through the origin is not a good idea -- it renders the traditional correlation and r^2 values meaningless, for example. You should only do it if you have a valid reason (theoretical or other) do do so.

 

[Wrichik Basu]

In general forcing a regression line to go through the origin is not a good idea -- it renders the traditional correlation and r^2 values meaningless, for example. You should only do it if you have a valid reason (theoretical or other) do do so.

I know that. I do it only when I am forced to plot a graph of $y=mx$ rather than $y=mx+c$ because of the nature of the underlying equation.