I have a very general question, and maybe stupid one. I was wondering if $\mathbf{E}[x|x>a]$ can be expressed as $1-\mathbf{E}[x|x<a]$.
The reason why I am asking this question in the following: $\mathbf{E}[x|x>a]$ can be written as
$$\frac{\int_{a}^{\infty}xf(x)dx}{1-F(x)}.$$
Now if I assume a logistic distribution and I integrate by parts, it diverges and goes to infinity. Being able to rewrite it in the form $1-\mathbf{E}[x|x<a]$ would solve my problem.
Thank you and best
No, that wouldn't make sense since $\mathbb{E}[x\mid x>a]$ certainly has to be at least $a$, and in many cases $1-\mathbb{E}[x\mid x<a]$ clearly won't be.
Since these are expectations, not probabilities, there is no significance to $1$. You might hope that the $1$ should be $\mathbb{E}[x]$, but even that would not be correct - the correct relationship along these lines is $$\mathbb{E}[x\mid x>a]\mathbb{P}[x>a]=\mathbb{E}[x]-\mathbb{E}[x\mid x\leq a]\mathbb{P}[x\leq a].$$
