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I am working on a mathematical problem involving the evaluation of a double integral, and I am seeking an analytical solution or techniques to solve it. The integral I'm dealing with is as follows:

​$$\int_{-\infty}^\infty \int_{-\infty}^\infty [L(C1+0.5w+0.5z) - L(C2+0.5w+0.5z) ] \cdot p(W=w \mid X=x) p(Z=z \mid X=x) \, dw \, dz$$

Here, $C1$ and $C2$ are constants, $L(\cdot)$ represents the logistic function $L(u)=\frac{1}{1+\exp(-u)}$, and $W$ and $Z$ follow Gaussian distributions with mean $0.5x$ and variances $\sigma_W^2 \ll 1$ and $\sigma_Z^2 \ll 1$ respectively.

I am looking for an approach to analytically solve this integral or simplify it further, if possible. Any insights, techniques, or relevant references would be greatly appreciated.

Thank you for your time and assistance!

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  • $\begingroup$ there is no integration over $x$? $\endgroup$
    – Carlo Beenakker
    Commented Jan 10 at 11:28
  • $\begingroup$ Thank you for your comment. Yes. There is no integration over $x$. $\endgroup$
    – Charles
    Commented Jan 10 at 11:32

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Since the Gaussian distributions of $w$ and $z$ are narrowly peaked around $x/2$, you can expand the logistic function around that value; to second order in the variance $\sigma_W^2+\sigma_Z^2$ this gives for the integral $$I=(1+e^{-C_1-x/2})^{-1}-(1+e^{-C_2-x/2})^{-1}$$ $$+\left(\frac{e^{C_2+\frac{x}{2}} \left(e^{C_2+\frac{x}{2}}-1\right) }{8 \left(e^{C_2+\frac{x}{2}}+1\right)^3}-\frac{e^{C_1+\frac{x}{2}} \left(e^{C_1+\frac{x}{2}}-1\right) }{8 \left(e^{C_1+\frac{x}{2}}+1\right)^3}\right)(\sigma_W^2+\sigma_Z^2)+{\cal O}(\sigma_W^2+\sigma_Z^2)^2.$$ Other approximation strategies may be more appropriate depending on the parameters $x,C_1,C_2$.

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  • $\begingroup$ I am afraid that I don't quite follow your solution since I'm not familiar enough with calculus. Could you please detail the steps a little bit? Many thanks! $\endgroup$
    – Charles
    Commented Jan 10 at 12:05
  • $\begingroup$ expand the integrand in a power series in $w$ and $z$ around the point $x/2$, to second order; the zeroth order term is just $L(C_1+x/2)-L(C_2+x/2)$; the first order term vanishes upon averaging over $w$ and $z$; the second order term gives the quantity between large brackets times the average $\mathbb{E}[(W+Z-x)^2]=\sigma_W^2+\sigma_Z^2$. $\endgroup$
    – Carlo Beenakker
    Commented Jan 10 at 12:08
  • $\begingroup$ @CarloBeenakker : You wrote the exact equality, $I=\cdots$. As such, this equality is of course false. If the equality is approximate, then in what sense, and why does it hold? $\endgroup$
    – Iosif Pinelis
    Commented Jan 10 at 14:55
  • $\begingroup$ @IosifPinelis Thank you for your comment. Do you imply that the equality only holds when $\sigma_W \rightarrow 0$ and $\sigma_Z \rightarrow 0$? $\endgroup$
    – Charles
    Commented Jan 10 at 22:35
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    $\begingroup$ @CarloBeenakker Thank you very much! $\endgroup$
    – Charles
    Commented Jan 10 at 22:35

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