Solving $E(X)$ $$=E(X-a+a)$$ (By linearity) $$=E(X-a)+E(a)$$ $$=E(X-a)+E(a)$$ $$=E(X-a)+a$$
Does this hold for all probability distributions?
The place where this seems counter-intuitive to me is that, expected value for any distribution is $E(X|X>a)$ is $E(X-a|X>a)+a$. So there is an addition of $a$ to the expected irrespective of the distribution.
Where I am confused is that the difference in answer to $E(X)$ and $E(X-a)$ is dependent on the just $a$ not distribution.