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Is interpreting log-log regression with log(1+x) as independent variable the same as having log(x) as independent variable? 1% increase in x results in beta% change in y?

What is x has both negative and positive values with the absolute value less than 1 so that 1+x are all positive?

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3 Answers 3

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Definitely not, except when $x$ is much larger than $1.$ This is one reason why the automatic reflex to "just add $1$ to values that might be zero before taking the log" is difficult to justify.

Let's see what is really going on. Suppose you are modeling a response $y$ in the form

$$\log y = \cdots + \beta \log(1 + x) + \cdots$$

where the missing stuff doesn't vary with $x.$ Then increasing $x$ by $100\delta\%$ changes $y$ to

$$\log y^\prime = \cdots + \beta \log(1 + x(1 + \delta)) + \cdots = \log y + \beta(\log(1 + x + x\delta) - \log(1 + x)),$$

showing that

$\log y$ changes by $\beta(\log(1 + x + x\delta) - \log(1 + x)).$

That's as nasty to understand as it looks, even when (as is usual) $\delta$ is taken to be very small. For small values of $x\delta$ (that is, $x$ isn't huge) we can approximate it by

$$\beta(\log(1 + x + x\delta) - \log(1 + x)) \approx \beta x\delta;\quad |x\delta| \ll 1+|x|.$$

This is approximately a $100 \beta x \delta \%$ change in $y$ itself. This covers the case of small negative values of $x,$ too -- but of course they cannot be $-1$ or less, for then $\log(1+x)$ would be undefined.

For large $x \gg 1$ we can approximate this change by neglecting $1$ in comparison to $x,$ giving

$$\beta(\log(1 + x + x\delta) - \log(1 + x)) \approx \beta \delta;\quad x \gg 1.$$

This is close to the usual log-log relation, reflecting approximately a $100 \beta \delta\%$ change in $y.$

Here, to illustrate and help the intuition, are log-log plots of two such relationships (with $\beta=1/2$):

Figure

The straight-line (dotted red) plot is $\log y = \beta \log(x).$ Because it is a line, we can interpret a change in $\log x$ as being related to a fixed multiple of that change in $\log y.$ But the black graph of $\log y = \beta\log(1+x)$ departs pretty strongly from this linear shape when $x$ is small to medium in size. Its changing curvature means that the relationship between any change in $\log x$ and the value of $\log y$ changes with $x:$ it's small for smaller $x$ but grows as $x$ gets larger.

Thus, the answer is it's complicated: when $x$ ranges from smallish to largish values, the change in $\log y$ ranges from a multiple $x\delta$ to a multiple $\delta$ of $\beta.$ The value depends on $x$ itself, at least until $x$ is sufficiently large. There is no fixed, simple relationship.

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$$\begin{array}{} \frac{y^\prime/y}{x^\prime/x} &=& a& \quad \text{if $y=x^a$}\\ \frac{y^\prime/y}{x^\prime/x} &=& a \frac{x}{1+x}& \quad \text{if $y=(1+x)^a$}\\ \end{array}$$


You can interpret it as following: a $q\%$ change in $x$ leads to a $q \frac{x}{1+x}\%$ change in $(1+x)$.

If $x$ is large they have a similar interpretation but not when $x$ is small.

For instance, when $x=1$, then a doubling of $x$ is the same as a $50\%$ increase of $1+x$. The $x$ changes from $1$ to $2$ and the $x+1$ from $2$ to $3$.

When $x=100$, then a doubling of $x$ is nearly the same as a doubling of $1+x$. The $x$ changes from $100$ to $200$ and the $x+1$ from $101$ to $201$.


Or the other way: a $q\%$ change in $1+x$ leads to a $q (1+1/x)\%$ change in $x$.

For instance, when $x=1$, then a doubling of $x+1$ is the same as a $200\%$ increase of $x$. The $x+1$ changes from $2$ to $4$ and the $x$ from $1$ to $3$.

When $x=100$, then a doubling of $x+1$ is nearly the same as a doubling of $x$. The $x+1$ changes from $101$ to $202$ and the $x$ from $100$ to $201$.


What is x has both negative and positive values with the absolute value less than 1 so that 1+x are all positive?

Transformations like $\log(x+c)$ are seen when $x$ can have zero or negative values. But it changes the interpretation as you see above.

It might be better to solve the problem directly without the transformation. E.g. if the goal is to do some sort of regression and you linearise the function for this with the transformation. Instead, one can also use non-linear least squares regression or a generalized linear model. You don't need to transform data in order to fit a function like $y=a x^b$.

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  • $\begingroup$ I suggested to use non-linear regression but I realize now that it are the x-values that are negative and not the y-values. In any case, one should have some good model about the data. What can cause negative x values, what sort of noise does this, and how should we model this? Simply linearizing the equation is more like a simplistic fitting solution that may not give a lot of insight into the underlying mechanism that governs the distribution of the data, the choice to shift by $1$ is a random choice. Why not use $x+2$ or $x+3$? Or possibly the data with large $x$ dominate the fit anyway? $\endgroup$ May 25 at 7:59
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I have always used the following intuitive explanation where log(1+x) (for positive x) is the transformation:

For small x (<< 1), $\log(1+x) \approx x$ . So the effects are still linear.

For large x (>> 1), $\log(1+x) \approx \log(x)$ . So the effects are multipliative.

Therefore, the transformation log(1+x) moves from being linear at one extreme to multiplicative at the other, and is somewhere between the two for "middle-sized" values of x.

By way of a concrete example, if the beta-value was +0.3, then this would be saying

"for small response variables, an increase of one unit of the predictor variable would lead to an additive increase of +0.3. For large response variables, an increase of one unit of the predictor would lead to a multiplication by 1.3 (or +30%). For response variables which are neither small nor large, the effect will be somewhere between the two."

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