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I want to convert an integral from $(0, 1)$ range to $(-\infty, \infty)$ range by change of variable. What is the best transform function to do this - one that is simple, monotonic with $f(-\infty)=0$ and $f(\infty)=1$?


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There is no best way, it depends on the integral. You could try $f(x) = \frac{1}{\pi}(\arctan x+\frac{\pi}{2})$. – copper.hat Jul 21 '12 at 19:03
Try a sigmoid function, the one that better fits your problem – FKaria Jul 21 '12 at 19:03
It will depend on the integral; if you have a particular example in mind, don't hesitate! – Davide Giraudo Jul 21 '12 at 19:52
I hope what Davide Giraudo means is don't hesitate to tell us what the integral is. – Michael Hardy Jul 21 '12 at 20:18
The examples are all good ones. The reason the choice depends on your integral, as copper.hat says, is that you will get $1+x^2$ in the denominator from the arctangent, and exponentials from the logit function. One or the other may fit better with your integrand. – Ross Millikan Jul 22 '12 at 17:14
up vote 2 down vote accepted


$$ z=\logit p = \log \frac{p}{1-p}. $$ As $p$ goes from $0$ to $1$, $z$ goes from $-\infty$ to $+\infty$ (provided the base of the logarithmic function is more than $1$; most often it is $e$.). The logit of $1/2$ is $0$. The graph is symmetric about $(1/2,0)$, so that, for example, $\operatorname{logit}\ 0.2= -\operatorname{logit}\ 0.8$.

This function is used in statistics.

The first syllable is pronounced with the "long o" sound as in "low"; the "g" like the "j" in "jet".

The inverse is the logistic function $$ p = \frac{e^z}{1+e^z}\ = \frac{1}{1+e^{-z}}. $$

If you're applying this to an integral, then which of many such functions should be chosen would depend on what integral it is.

Later edit: I occurs to me that if one wishes to be aware of the logit function, one show know this fact about probability: $$ \logit \Pr(A\mid D) = \logit \Pr(A) + \log\frac{\Pr(D\mid A)}{\Pr(D\mid \text{not }A)} $$ (and the letter $D$ may be take to stand for "data".)

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One rather regular function is $\arctan : \mathbb R \to (-\pi/2, \pi/2)$, so you can take $f(x)=\frac{1}{\pi} \arctan x + \frac{1}{2}$.

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