# Thread: Expected value of the logistic distribution?

1. ## Expected value of the logistic distribution?

How would you show that the expected value of the logistic distribution is equal to it's mean? ie E(X)=mu

$E(X) = \int_{-\infty}^{\infty} {\frac{xe^{-(x-\mu)/s}} {s\left(1+e^{-(x-\mu)/s}\right)^2}} \!$

slightly simplified:

$E(X) = e^{\mu/s}/s \int_{-\infty}^{\infty} {\frac{xe^{-x/s}} {\left(1+e^{-(x-\mu)/s}\right)^2}} \!$

Various substitutions I've tried don't seem to work, $u={1+e^{-(x-\mu)/s}}$, etc haven't worked to well for me.

2. Originally Posted by evidor
How would you show that the expected value of the logistic distribution is equal to it's mean? ie E(X)=mu

$E(X) = \int_{-\infty}^{\infty} {\frac{xe^{-(x-\mu)/s}} {s\left(1+e^{-(x-\mu)/s}\right)^2}} \!$

slightly simplified:

$E(X) = e^{\mu/s}/s \int_{-\infty}^{\infty} {\frac{xe^{-x/s}} {\left(1+e^{-(x-\mu)/s}\right)^2}} \!$

Various substitutions I've tried don't seem to work, $u={1+e^{-(x-\mu)/s}}$, etc haven't worked to well for me.
Since $\frac{e^{-A}}{\left( 1 + e^{-A}\right)^2} = \frac{1}{4} \, \text{sech}^2 A$ it follows that $\frac{e^{-(x-\mu)/s}} {s \left( 1 + e^{-(x-\mu)/s} \right)^2} = \frac{1}{4s} \, \text{sech}^2 \left( \frac{x - \mu}{2s}\right)$.

Therefore: $E(X) = \int_{-\infty}^{+\infty} \frac{x}{4s} \, \text{sech}^2 \left( \frac{x - \mu}{2s}\right) \, dx$.

Substitute $w = \frac{x - \mu}{2s}$ and split the integral up: $E(X) = s \int_{-\infty}^{+\infty} w \, \text{sech}^2 w \, dw + \frac{\mu}{2} \int_{-\infty}^{+\infty} \text{sech}^2 w \, dw$.

The first integral is equal to zero since the integrand is an odd function of w. The second integral is simple to perpetrate and is equal to 2.

Then $E(X) = \mu$.

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### finding the expectation of logistic distribution

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