Assume that I uniformly sample data points from an interval [-$a$; $a$], then pass these data points to a logistic function of the form

$f(x) = \frac{1}{1+e^{-\beta\cdot x}}$,

and fit a linear line to these data samples (x, f(x)). Can the expected slope of this linear fit be expressed as a function of $\beta$ and $a$?

(Bonus: tangens hyperbolicus function $f(x)=\tanh(\beta x)$ instead of the logistic function)

  • $\begingroup$ Fit a line to which points? If you have "sample from logistic distribution" then presumably the points are i.i.d., so there is no "line". Moreover, this would be univariate data. Do you mean that you sample $x$'s from some distribution and then transform them using logistic function? $\endgroup$
    – Tim
    Aug 17, 2020 at 12:21
  • $\begingroup$ ... and then fit your linear regression to the logistic function, or to the 0-1-valued data that comes from running the Bernoulli trials with the parameter coming from the logistic function? $\endgroup$ Aug 17, 2020 at 12:22
  • $\begingroup$ Sorry, the question did not make sense in the original form. I now clarified. $\endgroup$
    – monade
    Aug 17, 2020 at 12:29
  • $\begingroup$ The answer is trivially yes, because $a$ and $\beta$ are the only unknowns in the situation and the expectation of the slope exists. This leads me to suspect you intended to ask a different question--but what would it be? Do you want to compute that expectation? If so, could you explain what its statistical or theoretical applications might be? $\endgroup$
    – whuber
    Aug 17, 2020 at 13:45
  • $\begingroup$ I'd like to know the formula that describes the relationship between the expected slope and $a$ / $\beta$. At this point it's more mathematical curiosity how one would approach this. $\endgroup$
    – monade
    Aug 17, 2020 at 18:55

1 Answer 1


You have data $X_1,\dots,X_n\stackrel{\text{iid}}\sim\text{Unif}(-a,a)$ and obtain $Y_i = \sigma(X_i)$ where $\sigma(z) = \frac{1}{1+e^{-bx}}$.

The population slope of a simple linear regression is$\newcommand{\E}{\operatorname{E}}$ $$ \beta := \frac{\text{Cov}(X_i, Y_i)}{\text{Var}(X_i)} = \frac{\E(X\sigma(X)) - (\E X) (\E \sigma(X))}{\text{Var}(X)}. $$ $X$ is symmetric about $0$ and the variance is just that of a uniform RV which is known, so all we really need to compute is $\E(X\sigma(X))$. This is $$ \E(X\sigma(X)) = \frac 1{2a} \int_{-a}^a \frac{x}{1+e^{-bx}}\,\text dx. $$ We can note that $$ \int \frac{1}{e^{-bx} + 1}\,\text dx = \int \frac{e^{bx}}{1 + e^{bx}}\,\text dx \\ = \frac 1b \int \frac 1u \,\text du = \frac 1b \log (e^{bx} + 1) $$ so we can integrate by parts to get $$ \E(X\sigma(X)) = \frac{1}{2ab}x\log(e^{bx}+1)\bigg\vert_{-a}^a - \frac 1{2ab}\int_{-a}^a \log(e^{bx} + 1)\,\text dx. $$ With the first term (aside from some scaling constants) we end up with $$ \log(e^{ab}+1) + \log(e^{-ab}+1) = \log\left[e^{ab}(e^{-ab}+1)\right] + \log(e^{-ab}+1) \\ = ab + 2\log(e^{-ab}+1) $$ so $$ \frac{1}{2ab}x\log(e^{bx}+1)\bigg\vert_{-a}^a = \frac a2 + \frac 1b \log(e^{-ab}+1). $$ For the other term, we can let $u = -e^{bx}$ so $\frac 1{bu}\,\text du = \text dx$ which means$\newcommand{\Li}{\operatorname{Li}_2}$ $$ \int_{-a}^a \log(e^{bx} + 1)\,\text dx = -\frac 1b \int_{-e^{ab}}^{e^{ab}} \frac{\log(1-u)}u\,\text du \\ = \frac 1b\left[\Li(e^{ab}) - \Li(-e^{ab})\right] $$ where $\Li$ is the dilogarithm function. All together this means $$ \E(X \sigma(X)) = \frac a2 + \frac 1b \log(e^{-ab}+1) + \frac{\Li(-e^{ab}) - \Li(-e^{-ab})}{2ab^2}. $$ I have something of the form $\Li(z) - \Li(1/z)$. Using the fact that $$ \Li(z) + \Li(1/z) = -\frac{\pi^2}6 - \frac 12 \log^2(-z) $$ I can rewrite this with just a single $\Li$ in it: $$ \E(X \sigma(X)) = \frac a2 + \frac 1b \log(e^{-ab}+1) + \frac{-\frac{\pi^2}6 - \frac 12 a^2b^2 - 2 \Li(-e^{-ab})}{2ab^2} \\ = \frac a4 + \frac 1b \log(e^{-ab}+1) - \frac{\pi^2 + 12 \Li(-e^{-ab})}{12ab^2}. $$ This means $$ \beta(a,b) = \frac{\frac a4 + \frac 1b \log(e^{-ab}+1) - \frac{\pi^2 + 12 \Li(-e^{-ab})}{12ab^2}}{a^2/3} \\ = \frac{3}{4a} + \frac{3}{a^2b}\log(e^{-ab}+1) - \frac{\pi^2 + 12 \Li(-e^{-ab})}{4a^3b^2}. $$

$\Li$ is a special function and can't be expressed in terms of elementary functions in general, and since this could be solved for $\Li$ that means there's no general elementary expression for $\beta$. But we can evaluate it in terms of elementary functions for particular values of $a$ and $b$. One such value is $$ \Li\left(-\varphi\right) = -\frac{\pi^2}{10} - \log^2\varphi $$ so if $ab = -\log(\varphi)$ then we'll be able to evaluate $\beta$ in terms of elementary functions. As an example of this, we could take $$ a = \sqrt 2 \\ b = -\frac 1{\sqrt 2}\log(\varphi). $$ Although this is still a really messy expression even if it's only in terms of elementary functions.

This was a lot of work just to express the integral $\int_{-a}^a \frac{x}{1+e^{-bx}}\,\text dx$ in terms of a different integral, but I think the value is that it shows that there isn't a closed form for $\beta(a,b)$ that we're missing, and it relates it to a well-studied special function that has high quality implementations available.

Here's a simulation to check.

nsim <- 1e6
a <- 2.34
b <- 1.2
x <- runif(nsim, -a, a)
y <- 1 / (1 + exp(-b * x))

curve(plogis(b*x), -a, a, 500, col=4)
points(y[1:100]~x[1:100], cex=.5)

# I'm just integrating to avoid needing to load other libraries
Li2 <- function(z) -integrate(function(u) log(1-u)/u, 0, z)$value
slopefunc <- function(a,b) {
  3/(4*a) + 3 / (a^2 * b) * log(exp(-a*b) + 1) - (pi^2 + 12*Li2(-exp(-a*b))) / (4*a^3*b^2)

cov(x, y) / var(x)
  • $\begingroup$ Thanks a lot for this wonderful derivation! I learned a lot on the way. Just one follow-up: what does the $\varphi$ refer to? $\endgroup$
    – monade
    Aug 17, 2020 at 20:15
  • $\begingroup$ @monade glad this helped! And I’m using $\varphi=\frac{\sqrt 5 + 1}2$, the golden ratio (which is why that’s an interesting particular value) $\endgroup$
    – jld
    Aug 17, 2020 at 22:45
  • $\begingroup$ Ah I see, thanks! $\endgroup$
    – monade
    Aug 18, 2020 at 9:31

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