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For a project I'm looking for continuous distributions which have a somewhat simple closed form for upper-truncation expectation ($E[x|x>c]$). Here are two examples I've found so far:

Exponential distribution ($F(x)=1-e^{-\lambda x}$): $c+1/\lambda$

Uniform distribution on $(a,b)$: $\frac{c+b}{2}$

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Assuming positive $c$:

The logistic distribution with mean 0, scale parameter $s$ has truncated expectation $$-ck + s(1+k)\log(1+k),\text{ where }k=e^{c/s}$$

The Laplace distribution with mean 0, scale parameter $b$ has truncated expectation $$(b+c)(1+e^{-c/b})/2$$

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The class of distributions with this property is large (and not even completely defined - does an answer in terms of the gamma function, error function etc. count as closed form?). But note that

$$E(X|X > c) = \frac{\int_c^\infty x f(x) dx}{\int_c^\infty f(x) dx} $$

Therefore a closed form for $E(X|X>c)$ will exist so long as both integrals can be found in closed form.

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  • $\begingroup$ That makes sense. I guess I was looking for simplest examples since the expectation fits into a differential equation and that is ultimately what I hope to have a closed form solution of. $\endgroup$ – user180743 Oct 10 at 11:58

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