Basic math explanation (related to estimating linear regression with no intercept)

Question

I have a question on the Cross Validated Stack Exchange site where I ask how to update the exponential regression coefficient of a vertically translated depreciation curve.

A Cross Validated community member has been kind enough to provide an answer to my question.

The solution has been explained as follows:

...the equation you need to estimate is $$y=21-e^{ax},$$

which is equivalent to

$$21-y=e^{ax}.$$

If you take logarithms both sides (you can do it because $y<21$), then $$log(21-y)=ax.$$

Renaming $log(21-y)=z$, this is of the form $$z=ax,$$

which is a linear regression with no intercept that can be estimated with many standard software packages.

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I think I understand everything up to, and including this part:

If you take logarithms both sides (you can do it because $y<21$), then $$log(21-y)=ax.$$

However, as someone who has limited math skills, I'm having difficulty understanding a couple of things:

1. Why would I want to rename $log(21-y)=z$ --as-- $z=ax$ ?

2. How do I estimate the "linear regression with no intercept"?

Could #1 and #2 be explained, in layman's terms?

Answer 1

In general, people like to work with linear equations, as it is easier and more intuitive. The equation $y = e^x$ for example, one could be tempted to transform the data $ x \to \ln{x}$, to get to $ y = e^ {\ln(x)} = x $ and have a nice linear equation.

This is I believe what happens in your case. You have to deal with $$ \log (21-y) = a x$$ and she suggested a change of variable (i.e. an operation to be performed on each $y$-datapoint) that will give you a linear equation.

Analogously, in many cases people like to present log-log plots (so, plots considering the logarithms of each variable), which under certain circumstances appear linear.

As a trivial example, you might have the dataset $x = (1,2,3)$ and $y = (2.718, 7.389, 20.085)$. Just take the logarithm of the $y$-datapoints and it will be more transparent: the transformed $y$-datapoints are now $(1,2,3)$, very handy indeed.

Your second point is a litte bit trickier. Given datapoints $(x,y)$, you have to find the best line $y = ax$ such that some error is minimised. Lines described as $y= ax$ go through the origin (for $x = 0$, you get $y=0$), so the intercept is zero.

Would you like the algorithm to be described? In practice, a tool as simple as Excel will do it for you. Place a trendline and in the dialogue box, tick the "select intercept" option, and in the related cell, input "0".

Answer 2

Here's a supplement to An aedonist's great answer. I've stated things slightly differently, which you may find helpful.

Why would I want to rename $\log(21−y)=z$ as $z = ax$?

The way you have phrased this is just a little confusing. What you are really doing is renaming $\log(21 - y)$ as $z$ so that the equation $\log(21 - y) = ax$ becomes $z = ax$. The reason for doing this is because in this form it is clear to see that $z$ depends linearly on $x$ and you have now restated the problem of finding a fit of the form $y = 21 - e^{ax}$ as a problem of finding a fit of the form $z = ax$.

How do I estimate the "linear regression with no intercept"?

What you would like to do is perform a linear regression on the data points $(x_i, z_i)$. You will need to take each of your current data points, $(x_i, y_i)$ and transform them to $(x_i, z_i) = (x_i, \log(21 - y_i))$. Then on these data points you perform a linear regression, which can be done in Excel, for example. At this point you will have a line of best fit that looks like $z = ax$. The $a$ value is the parameter you are looking for. You can plug it in to the original $y = 21 - e^{ax}$ to get the original fit you were looking for.