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What is the intuition of $\frac{1}{1+f(x)}$ ? I often see formulas of such style in statistics. For example sigmoid function is of form $\frac{1}{1+e^{-x}}$.

Why is $\frac{1}{1+f(x)}$ used frequently? What is its intuition?

mallea
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    For positive $f(x)$, this always yields a function that takes values in $(0,1)$. – Alex Provost Aug 19 '17 at 20:02
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    Every function $g$ without zeros can be written in that way. Specifically, $$g(x)=\frac{1}{1+\left(\frac{1-g(x)}{g(x)}\right)}$$ –  Aug 19 '17 at 20:10

2 Answers2

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If $f(x)$ is a positive valued function, then $g(x)=\frac{1}{1+f(x)}$ will always take on values between $0$ and $1$. It kind of flips things around, because when $f$ gets big, $g$ gets small, and vice versa. The "$1+$" part of the denominator keeps the fraction from blowing up when $f$ gets very small.

In other words, if you want a function that is close to $0$ when $f$ is very large, and close to $1$ when $f$ is very small, then this one fits the bill!

You might want a function that takes values between $0$ and $1$ for various reasons, e.g., if you want to interpret an answer as a proportion, or as a probability. You can also place the range between $0$ and any other positive value simply by replacing the $1$ on top with whatever you want your maximum value to be.

G Tony Jacobs
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This is a way of taking a function whose values can become infinitely large and transforming it into a function whose values lie in a finite range like $[0,1]$.

For example, if $f$ is a function whose value ranges from $0$ to $\infty$, then

$$g(x) = \frac{1}{1+f(x)}$$

is a function whose value ranges from $0$ to $1$.

  • The reciprocal part $1/\ldots$ has the effect of making large values very small. This transforms infinitely large values into manageably small values.

  • Furthermore, adding some small constant in the denominator prevents a divide by zero error in case $f(x) = 0$.

  • When $f(x) = \exp(-x)$ in particular, adding one in the denominator makes the symmetric midpoint of the function occur at $g(0) = \frac{1}{2}$.

Other functions with similar properties include $\exp{(-x)}$, which sends values in $[0, \infty)$ to $[0,1]$, and $\arctan{x}$, which sends values in $(-\infty, \infty)$ to $(-\frac{\pi}{2},\frac{\pi}{2})$.

The choice of function may depend on the range of values you want to transform, as well as what kind of symmetry you want the transformed function to have. For example, $1/(1+\exp{(-x)})$ is symmetric as a function of $x$.

user326210
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