So this is an official Exam P practice problem:
The density of a manufacturer's loss, X, is given by
$$ f_{X}(x) = \frac{2.5(0.6)^{2.5}}{x^{3.5}} $$ for x > 0.6 and 0 otherwise.
The manufacturer has an insurance policy with a deductible of 2, after which point all loss is covered. Calculate the expected value of all loss not covered by the insurance company, Y.
I understand how to solve this problem; you split up the integral as such:
$$ E(Y) = \int_{0.6}^{2} xf_X(x)dx + \int_{2}^{\infty}2f_X(x)dx $$
and you end up getting 0.9343, the correct answer to the problem.
What I'm having trouble doing is connecting this problem back to the theory of conditional expectation that I know. Y is definitely conditional on X. Y|(0.6<X<2) = x and Y|(X>2) = 2 and 0 otherwise. How would you define the conditional density of Y given X, and how would you make sense of the conditional expectation formula in the context of this problem?
The conditional distribution of $\ Y\ $ given $\ X\ $ is always concentrated at a single atom, namely $\ X\ $ if $\ X<2\ $, or $2$ if $\ X\ge2\ $. It therefore doesn't have a density (unless you allow generalized functions as "densities"), and $$ E(Y|X)=\cases{X&if $\ X<2$\\ 2&if $\ X\ge2\ $.} $$ You then have \begin{align} E(Y)&=E(E(Y|X))\\ &=\int_{0.6}^\infty E(Y|X=x)f_X(x)dx\\ &=\int_{0.6}^2 xf_X(x)dx+\int_2^\infty2f_X(x)dx\ , \end{align} as you've already found.