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For $X\sim N(\mu,\sigma^2)$ ,

$$E[X|X>\alpha] = \mu +\sigma \frac{\phi\left(\frac{\alpha-\mu}{\sigma}\right)}{1-\Phi\left(\frac{\alpha-\mu}{\sigma}\right)} $$

Is there an analogous expression for when $X$ is distributed logistically?

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You are asking for the conditional expectation $$ \DeclareMathOperator{\E}{\mathbb{E}} \E \left[ X \mid X>\alpha \right] $$ when $X$ has a logistic distribution. I will only do the case when $X$ has the standard logistic distribution, leaving for you to add location and scale parameters. The density function is then $f(x)=\frac{e^{-x}}{(1+e^{-x})^2}$ and the cdf $F(x)=\frac1{1+e^{-x}}$.

We first find the conditional probability $$ \DeclareMathOperator{\P}{\mathbb{P}} \P\left( X>t \mid X>\alpha \right) =\frac{\P(X>t)}{\P(X>\alpha)} $$ (for $t>\alpha$). This expression simplifies to $$ \P\left( X>t \mid X>\alpha \right) =\frac{e^{-t}(1+e^{-\alpha})}{e^{-\alpha}(1+e^{-t})} $$ and by differentiation we find the conditional density function as $$ f_\alpha(t)= \frac{{\mathrm e}^{-t} \left(1+{\mathrm e}^{-\alpha}\right)}{{\mathrm e}^{-\alpha} \left(1+{\mathrm e}^{-t}\right)}-\frac{\left({\mathrm e}^{-t}\right)^{2} \left(1+{\mathrm e}^{-\alpha}\right)}{{\mathrm e}^{-\alpha} \left(1+{\mathrm e}^{-t}\right)^{2}} $$ Then, as usual by integration, we find the conditional expectation as $$ \E \left[ X \mid X>\alpha \right] = \ln \! \left(1+{\mathrm e}^{-\alpha}\right) {\mathrm e}^{\alpha}+\ln \! \left(1+{\mathrm e}^{-\alpha}\right)+\alpha $$

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Mills Ratio of a continuous random variable is m(x) = (1-F(x))/f(x). This is the inverse of the hazard, used primarily in survival analysis, where x is survival time.

For x a real number the logistic distribution is

F(x) = 1/(1+e(-(x-a)/s))

where a is the mean and s is the standard deviation.

The logistic density is

f (x) = (-F^2)e(-(x-a)/s)(-1/s) = F(x)(1-F(x))/s

so Mills Ratio is a relatively simple expression

m(x) = (1-F(x))/f(x) = s/F(x).

The hazard is

h(x)= F(x)/s.

The logistic distribution or function is one of a family of distributions where f=c*F(1-F), for a constant c, which gives a relatively simple form to Mills Ratio or the hazard. The normal distribution is not part of that family.

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  • $\begingroup$ Thank you for this answer. I think my question title may be imprecise. I'm asking if there's an expression for $E[X|X>a]$ when $X$ is logistically distributed. Does this follow immediately from your answer? I apologize if I'm just not seeing it. $\endgroup$ Commented Aug 30, 2021 at 7:30

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