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I have some data which I think can be fitted using the following function:

$k(x,y)=\beta \cdot e^{-\gamma y} \cdot \mathrm{ln} \left( \frac{x - a}{b} \right) \wedge \beta, \gamma \in \mathbb{R}_0^+$

Where $a$ and $b$ are known values, and $\beta$ and $\gamma$ are the parameters I am trying to find using least squares and I'm pretty sure they're not related to one another. Not looking ahead, I first started calculating the separate partial derivatives and soon saw that the expression for $\gamma$, because it is inside an exponent, was not very easy to find by hand. So I started over by looking at the logarithmic error which made $\gamma$ a linear term.

When I worked everything out, I first found an expression for $\beta$, from which I had hoped to figure out $\gamma$:

Derivation of beta

EDIT: The last bit has to be $\frac{1}{n}$ times the big summation with the two logs.

But two problems arise already: (1) all of my data for $f_i$ is negative and (2) all of my data for $\mathrm{ln} \left( \frac{x_i - a}{b} \right)$ will also turn out to be negative due to the nature of the original experiment. So even if I could use this expression for $\beta$ to find a potential expression for $\gamma$, I wouldn't be able to calculate either of them using my data.

I read up on log fitting a bit and I noticed that a lot of people just add constants to the negative values to make their sample data positive like so:

$\mathrm{ln}\left( \beta + \lambda_0 \right)$

$\mathrm{ln} \left( \mathrm{ln} \left( \frac{x_i-a}{b} \right) + \lambda_1 \right)$

But I get different results for $\beta$ and $\gamma$ when I use different values of $\lambda_i$, which adds another level of uncertainty.

I do not have a formal education in statistics or mathematics, so I can't really think of any other methods to proceed from here.

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  • $\begingroup$ It will help immensely if you post your progress on calculating the derivatives. I think you’re not calculating the correct derivative. $\endgroup$
    – Dave
    Apr 24, 2020 at 15:37
  • $\begingroup$ I see, I have worked them out on paper so I will convert them to a more digital format first. $\endgroup$ Apr 24, 2020 at 15:39
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    $\begingroup$ I don’t want to see your calculation of the derivative. What I really want to know is exactly what you’re taking the derivatives of. (It should be the loss function, not the original model.) $\endgroup$
    – Dave
    Apr 24, 2020 at 16:03
  • $\begingroup$ Oh sorry, I just saw your comment, well I added the full derivation anyways :) $\endgroup$ Apr 24, 2020 at 16:18
  • $\begingroup$ Sorry @Dave, but I think I'm going to find the nonlinear least squares solution programatically as per dlnB's answer. Thanks for looking into it though! $\endgroup$ Apr 24, 2020 at 16:27

2 Answers 2

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Why not try nonlinear least squares? Denote the left-hand side, $f(x,y)$ as $z$.

$$(\hat{\gamma}, \hat{\beta}) = \text{arg} \min_{(\gamma,\beta)} \sum_{i=1}^N(z_i - \beta e^{-\gamma x_i} \ln (\frac{y_i-a}{b})^\beta)^2.$$

If you cannot work out the solution analytically, which would involve solving $$ \frac{\partial \sum_{i=1}^N(z_i - \beta e^{-\gamma x_i} \ln (\frac{y_i-a}{b})^\beta)^2}{\partial \gamma} = 0$$ $$ \frac{\partial \sum_{i=1}^N(z_i - \beta e^{-\gamma x_i} \ln (\frac{y_i-a}{b})^\beta)^2}{\partial \beta} = 0$$

for $\gamma$ and $\beta$, you can estimate these values using a grid search.

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  • $\begingroup$ I've added my derivation of $\beta$ in my question, indeed I was trying to do this, but I was not aware that a thing like grid search existed. I'll give that a shot! $\endgroup$ Apr 24, 2020 at 16:19
  • $\begingroup$ It's straightforward to implement. What software are you using? $\endgroup$
    – dlnB
    Apr 24, 2020 at 16:20
  • $\begingroup$ Uh, I was doing everything by hand and writing some Python snippets for the calculations. $\endgroup$ Apr 24, 2020 at 16:21
  • $\begingroup$ Are you comfortable using R or MATLAB? If so I can help you with the code. $\endgroup$
    – dlnB
    Apr 24, 2020 at 16:21
  • $\begingroup$ I've never used R and I haven't used MATLAB in ages, but I'm sure there are plenty of tutorials which I can use to write out a grid search in Python. Thanks for your help! $\endgroup$ Apr 24, 2020 at 16:22
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I anticipate questions, so please do ask away in the comments.

Let's start out by discussing what ordinary least squares does.

We have some data and want to fit a line. Let's say that we have $x$-values $1$, $2$, and $3$, and $y$-values $4$, $5$, and $7$.

These are points in the plane: $(1,4)$, $(2,5)$, and $(3,7)$.

What least squares says is that the line of best fit is the line that minimizes square loss. "Least" comes from the minimization; "squares" comes from squaring something.

Square loss is just the squares of the true values minus the predictions.

So what we want to minimize in a least squares problem isn't the model $\hat{y}=ax+b$ but this nasty equation, where $y$ is the true $y$-value, $\hat{y}$ is the predicted $y$-value, and $n$ is the number of observations of $y$:

$$L = \overset{n}{\underset{i=1}{\sum}}\big( y_i - \hat{y}_i\big)^2$$

$y_i - \hat{y}_i$ is by how much the prediction misses the true value; this is the residual. Then we square the residual. Then we add up all of our squared residuals to get a measure of by how much our line misses the data. The lower this number, the tighter the fit.

Since $\hat{y}= ax+b$, we can write the loss function like this:

$$L = \overset{n}{\underset{i=1}{\sum}}\big( y_i - (ax_i+b)\big)^2$$

For that example with the three points, that is this:

$$L = \big(4-(1a - b)\big)^2 + \big(5-(2a - b)\big)^2 + \big(7-(3a - b)\big)^2 $$

We then find the values of $a$ and $b$ that minimize $L$ and give us the least squares solution. For this type of model, calculus shows that $a$ and $b$ have a convenient formula that gets us out of having to do partial derivatives every time; we just plug the data into the formula and get our values of $a$ and $b$.

You, however, propose a more complex model. There's nothing wrong with that.$^{\dagger}$ You just lose the convenient formula to give $a$ and $b$, but if your data should not be modeled with a line, then that is just the price to pay for getting a model worth using.

You propose that your response variable, $z=f(x,y)$ is described by $f(x,y)=\beta \cdot e^{-\gamma y} \cdot \mathrm{ln} \left( \frac{x - a}{b} \right)$. Stick that into the loss function as your prediction!

$$L = \overset{n}{\underset{i=1}{\sum}}\big(z_i - \hat{z}_i)^2 =\overset{n}{\underset{i=1}{\sum}}\bigg[z_i - \beta e^{-\gamma x_i} \ln \bigg(\dfrac{y_i-a}{b}\bigg)^\beta\bigg]^2$$

I don't want to calculate the partial derivatives by hand, but it can be done. You could use some software like WolframAlpha to help you. The typical way to do this would be to use a computer to find the values of $\beta$ and $\gamma$ that minimize $L$. A grid search is one option. Instead of calculating the derivatives, the grid search picks out points in the $\beta\gamma$-plane and checks which one results in the smallest value of $L$. That point gives you your $\alpha$ and $\beta$ values.

So let's go through the questions in my comment to your original post.

What are you minimizing?

You're minimizing $L$, which is the square loss. Importantly, you do not minimize the model.

Why is this minimization not fundamentally different from ordinary least squares?

Instead of subtracting a prediction found from a linear model like in ordinary least squares, we subtract a prediction found by a different kind of model. However, all of the ideas about using calculus to find the point giving minimal loss are the same.

How is the grid search helping to do the optimization calculus?

The grid search proposes a grid of values as the minimizers of the loss function, then checks what loss results from each. Whichever point in the grid gives the minimal loss is declared the winner.

Note There is a technical point that I have skipped: the $\hat{\text{hat}}$ on variables in dlnB's post. That means an estimate that you have calculated from your data, instead of the true value. In real data analysis, you don't get to know the true value, but if you have good estimates, you can be confident that you're close.

$^{\dagger}$There are cautions throughout statistics about overly complex models. The gist is not to make a model complex just for the sake of being complex. I suspect that you have a scientific reason for wanting the model to take the form that you propose, so please do feel free to use your model.

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  • $\begingroup$ Thanks for the explanations, Dave! It seems the loss function is what I called $S(\beta, \gamma)$, except that you apply it to my original formula for $k$ rather than the one I've transformed by taking the logarithm. I was aware of least squares but I didn't know that it could be applied just like that for nonlinear regression. I think I used $k_i$ and $\hat{k}_i$ correctly here, but it was good to get a reminder of what I'm actually trying to do. Thanks! $\endgroup$ Apr 24, 2020 at 18:04
  • $\begingroup$ The point is that if you’re using the transformer model to get out of having to do nasty calculus, you don’t have to do that. Numerical optimization techniques like grid search do that for you, and then you get to use the original model that you proposed. (The accepted answer uses that original model, too.) $\endgroup$
    – Dave
    Apr 24, 2020 at 18:13
  • $\begingroup$ I like calculus though, figuring out derivatives can be very therapeutic :) But I agree, my goal isn't to find an analytic solution for my parameters, I'll be happy when the results are consistent. $\endgroup$ Apr 24, 2020 at 18:17
  • $\begingroup$ “Consistent” has a technical meaning in statistics. What do you mean when you use that word? And feel free to do the entire optimization by hand. $\endgroup$
    – Dave
    Apr 24, 2020 at 18:18
  • $\begingroup$ I meant consistent in the layperson's way, as in, I know what to expect from the data so if I change my inputs $x$ and $y$ only slightly, I shouldn't expect too much of a change in $k$ either. $\endgroup$ Apr 24, 2020 at 18:28

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