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Short version:

I would like to calculate the expected value if you apply the sigmoid function $\frac{1}{1+e^{-x}}$ to a normal distribution with expected value $\mu$ and standard deviation $\sigma$.

If I'm correct this corresponds to the following integral:

$$\int_{-\infty}^\infty \frac{1}{1+e^{-x}} \frac{1}{\sigma\sqrt{2\pi}}\ e^{ -\frac{(x-\mu)^2}{2\sigma^2} } dx$$

However, I can't solve this integral. I've tried manually, with Maple and with Wolfram|Alpha, but didn't get anywhere.

Some background info (why I want to do this):

Sigmoid functions are used in artificial neural networks as an activation function, mapping a value of $(-\infty,\infty)$ to $(0,1)$. Often this value is used directly in further calculations but sometimes (e.g. in RBM's) it's first stochastically rounded to a 0 or a 1, with the probabililty of a 1 being that value. The stochasticity helps the learning, but is sometimes not desired when you finally use the network. Just using the normal non-stochastic methods on a network that you trained stochastically doesn't work though. It changes the expected result, because (in short):

$$\operatorname{E}[S(X)] \neq S(\operatorname{E}[X])$$

for most X. However, if you approximate X as a normal distribution and could somehow calculate this expected value, you could eliminate most of the bias. That's what I'm trying to do.

Community
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Michiel DM
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3 Answers3

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Apart from the the MacLaurin approximation, the usual way to compute that integral in Statistics is to approximate the sigmoid with a probit function. More specifically $\mathrm{sigm}(a) \approx \Phi(\lambda a)$ with $\lambda^2=\pi/8$. Then the result would be: $$\int \mathrm{sigm}(x) \, N(x \mid \mu,\sigma^2) \, dx \approx \int \Phi(\lambda x) \, N(x \mid \mu,\sigma^2) \, dx = \Phi\left(\frac{\mu}{\sqrt{\lambda^{-2} + \sigma^2}}\right).$$

Dalek
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korkinof
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    Unless I'm mistaken somewhere, $\lambda = \pi / 8$ isn't a particularly good value of $\lambda$ to use. The best value, in terms of max-norm convergence, seems to be about 0.588 (about $\pi / 5.35$). – Danica Jan 19 '14 at 19:43
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    I should also note for posterity that, since the equality wasn't obvious to me, here's a nice proof. – Danica Jan 19 '14 at 20:38
  • Could you post the result of the more general form $\int \text{sigm}(ax+b) N(x \vert \mu, \sigma^2) dx$ ? – bourneli Jul 08 '18 at 13:17
  • @korkinof I have not seen this used before. Can you provide some references? – user144410 Jul 18 '19 at 22:29
  • @Danica note that korkinof's answer defines $\lambda^2 = \pi/8$ which gives $\lambda = \sqrt{\pi/8} \approx 0.63$ which is much closer to your numeric solution. – STJ Apr 25 '22 at 12:05
  • https://arxiv.org/abs/1703.00091 has an in-depth look at expectations of the sigmoid. That article suggests $\lambda \approx 0.61$ based on Monte Carlo estimates. – STJ Apr 25 '22 at 12:11
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    Hah, yes, you're definitely right @STJ, a silly mistake. Also, my 8-year-old notebook above roughly minimizes the absolute distance between the CDFs, which might not be quite the best approximation for the purpose of computing this integral. – Danica Apr 25 '22 at 19:01
  • @bourneli For the general case: $\int_{-\infty}^{+\infty} \operatorname{sigmoid}(ax + b) \cdot \mathcal{N}(x | \mu, \sigma^{2}) dx \approx \Phi \left(\frac{\mu + \frac{b}{a}}{\sqrt{\frac{1}{a^{2} \lambda^{2}} + \sigma^{2}}}\right)$, where $\lambda = \sqrt{\frac{\pi}{8}}$. The original question corresponds to: $a = 1$ and $b = 0$, which coincides with the result here. – Cuong Jul 21 '25 at 13:50
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Since I do not have enough reputation to comment, I'll instead add a new answer. @korkinof's answer is almost correct. The final integral evaluates to the following: \begin{equation} \int \operatorname{sigmoid}(x) \mathcal{N}(x; \mu, \sigma^2) \mathrm{d}x \approx \int \Phi(\lambda x) \mathcal{N}(x; \mu, \sigma^2) \mathrm{d}x = \Phi\left(\frac{\lambda \mu}{\sqrt{1 + \lambda^2 \sigma^2}}\right). \end{equation} I verified my answer through simulation.

STJ
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Smoke
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I doubt that there's a closed-form solution. However, here's a series in powers of $\sigma$:

$$ \left( {{\rm e}^{-{\mu}}}+1 \right) ^{-1}+{\frac { \left( { {\rm e}^{-{\mu}}}-1 \right) {{\rm e}^{-{\mu}}}}{2\, \left( {{\rm e} ^{-{\mu}}}+1 \right) ^{3}}}{{\sigma}}^{2}+{\frac { \left( { {\rm e}^{-3\,{\mu}}}-11\,{{\rm e}^{-2\,{\mu}}}+11\,{{\rm e}^{-{ \mu}}}-1 \right) {{\rm e}^{-{\mu}}}}{8\, \left( {{\rm e}^{-{\mu} }}+1 \right) ^{5}}}{{\sigma}}^{4}+{\frac {{{\rm e}^{-{\mu} }} \left( {{\rm e}^{-5\,{\mu}}}-57\,{{\rm e}^{-4\,{\mu}}}+302\,{ {\rm e}^{-3\,{\mu}}}-302\,{{\rm e}^{-2\,{\mu}}}+57\,{{\rm e}^{-{ \mu}}}-1 \right) }{48\, \left( {{\rm e}^{-{\mu}}}+1 \right) ^{7}}}{{ \sigma}}^{6}+{\frac {{{\rm e}^{-{\mu}}} \left( {{\rm e}^{-7\,{\mu}}}-247\,{{\rm e}^{-6\,{\mu}}}+4293\,{ {\rm e}^{-5\,{\mu}}}-15619\,{{\rm e}^{-4\,{\mu}}}+15619\,{ {\rm e}^{-3\,{\mu}}}-4293\,{{\rm e}^{-2\,{\mu}}}+247\,{{\rm e}^{ -{\mu}}}-1 \right) }{384\, \left( {{\rm e}^{-{\mu}}}+1 \right) ^{9}}} {{\sigma}}^{8}+O \left( {{\sigma}}^{10} \right) $$

EDIT: To obtain this, first do the change of variables $x = \mu + \sigma t$. The integral becomes $$ \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty \dfrac{e^{-t^2/2}}{1 + e^{-\mu - \sigma t}}\ dt $$ Now take the Maclaurin series $$\frac{1}{1+e^{-\mu - \sigma t}} = \frac{1}{1+e^{-\mu}} + \frac{e^{-\mu} \sigma t}{(1+e^{-\mu})^2} + \frac{e^{-\mu} ( e^{-\mu} - 1) \sigma^2 t^2}{(1+e^{-\mu})^3} + \ldots$$ and integrate term by term.

Robert Israel
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    Hi Robert, could you comment on the derivation of that series? Where does it come from? – wvoq Dec 18 '12 at 21:52