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Take the cumulative distribution function (cdf)

$\Phi(z) = \frac{1}{1+\exp(-z)}$

and probability density function (pdf)

$\phi(z)=\Phi(z)(1-\Phi(z))$

of the logistic distribution.

How can one calculate the integral below?

$$\int^{\infty}_{-\infty}\Phi\left(\frac{w-a}{b}\right)\phi(w)\,\mathrm dw$$

Jenny Reininger
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  • Please specify what prompted this question. Is this homework? – Carl Jun 27 '19 at 19:11
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    It's actual research. I know it works for the Normal distribution but my problem is logistically distributed. I know that$$\int \mathrm{sigm}(x) \, N(x \mid \mu,\sigma^2) \, dx \approx \int \Phi(\lambda x) \, N(x \mid \mu,\sigma^2) \, dx = \Phi\left(\frac{\mu}{\sqrt{\lambda^{-2} + \sigma^2}}\right).$$ but would like to replace N by L. – Jenny Reininger Jun 27 '19 at 20:59
  • You have flipped their names; what you have called the density is the distribution and vice versa. – Glen_b Jun 28 '19 at 03:00

1 Answers1

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Too long for a comment....

This question mixes up the words density and distribution: $\Phi(z) = \frac{1}{1+\exp(-z)}$ is the distribution function of a logistic random variable, not the density as the OP calls it, while $\phi(z) = \Phi(z)(1-\Phi(z))$ is the corresponding density function even though the OP calls it the distribution function. (OK, I see that the OP has corrected the nomenclature in the question, making the above not inapplicable any more).

Be that as it may, the result $$\int_{-\infty}^\infty \Phi\left(\frac{w-a}{b}\right)\phi(w)\,\mathrm dw = \Phi\left(\frac{-a}{\sqrt{b^2+1}}\right)$$ for normal random variables has been proved in two different ways in the answers to this question. My answer to this question explicitly uses the fact that the difference of two independent normal random variables is itself normally distributed. I suspect that a corresponding result does not hold for logistic random variables and so it might not be possible to adapt my method to solve the OP's problem. On the other hand, the answer by whuber is based on solving partial differential equations, and might be more adaptable to what the OP is trying to do with logistic random variables.

Dilip Sarwate
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    Thanks for the shout-out: but unfortunately this integral doesn't appear to have any conventional closed form formula. It *does* have such formulas for *rational* values of $a$ and $b$ (it's a matter of making an appropriate substitution in the integral, which reduces it to the integral of a rational function, which one can find through contour integration or otherwise), but the lengths of those formulas increase with the sizes of their numerators and denominators. – whuber Jun 28 '19 at 14:20