Assume that I'm running a path analysis and I have discovered that certain relationships, empirically, are quadratic rather than linear. In order to model the data as such, I want to transform the relationship so that it becomes linear. Then, do I transform both variables involved through taking their square roots and then running the path analysis again (in an SEM software such as AMOS), or do I just transform one of the variables?

For example, Imagine I have three variables arranged like this:

A -> B -> C

The relationship between A and B is linear, and the relationship between B and C is quadratic. if I want to model the data as such, do I then only transform variable C through taking it's square root? What is the general principle for which variables to transform, when what I actually want to transform is the relationship between variables?


1 Answer 1


There are a number of issues.

Issue 1: the relationship can be quadratic, yet be such that a square-root won't linearize it.

Example: $E(C|b) = \beta_0 +\beta_1 b + \beta_2 b^2$

Here's a plot of $E(C|b)$ and its square root for a specific $(\beta_0, \beta_1, \beta_2)$:

enter image description here

As you see, it doesn't actually make the relationship linear. If a particular relationship between the coefficients holds, then a square root transformation will linearize it, but otherwise it may not help much - it might even make things worse. In many cases it can improve things, but you can't expect it to always work, even approximately.

If the relationship's actually quadratic, you're better off including a quadratic function to fit the mean.

Issue 2: transformation affects the error-distribution; if the variance is roughly constant before transformation, it may be very far from constant after transformation, if it was close to normal before, it may be quite skew after (similarly, additive errors will no longer be additive).

Issue 3: expectations don't transform. If the aim is to 'model the mean' of C, a model for the expectation of $\sqrt C$, when squared, is biased (downward) for $C$. This is a consequence of Jensen's inequality.

Transformation should be used carefully, and with a clear understanding of the issues it presents. It can be a valuable tool, but it's often misused.

  • $\begingroup$ This is a helpful list of concerns, but it doesn't address the basic parts of the question. From your answer I deduce that it is just the predictor that you take the square root out of. Is this correct? Further, you don't address how you include this in a path analysis - or is that not possible? $\endgroup$
    – histelheim
    Commented Jan 20, 2014 at 2:36
  • 1
    $\begingroup$ histelheim - (i) the point of the answer is to illustrate that the question seemingly relies on faulty assumptions (that transformation in any form is a good approach for this problem); (ii) Correct, I took the square root of C (as you should be able see from the axis labels), but there's no simple transformation of both variables either, unless you know the very thing you're trying to estimate; (iii) I believe it should be possible to include $b$ and $b^2$ as predictors of $C$ in your path analysis, but it depends on what exactly you're doing and what you're using to do it. $\endgroup$
    – Glen_b
    Commented Jan 20, 2014 at 5:16

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