Suppose random variable $ Y_{il} $ has the expectation:
$E(Y_{il}) = \int E(Y_{il} | Z_i= z)P^{Z_i}(dz)$
I am given that
$ E(Y_{il} | Z_i= z) = 1-exp(-\lambda_izB) $
Thus we get end up with the following equation for the Expectation.
$E(Y_{il}) = \int (1-exp(-\lambda_izB)P^{Z_i}(dz))$
Skipping over the constants $\lambda_i, B$ because they are irrelevant to the question.
Suppose that $Zi$ ~ $\gamma(\theta,\frac{1}{\theta}) $, and $P^{Z_i} $ is the probability distribution for Zi.
My Question: Because in the conditional expectation we condition on $Z_i= z$, Does this mean that our probability distribution also in terms of $Z_i= z$ or just a generic random variable $Z_i$?
Apologies if this is a stupid question.
z should be lowercase, because you want to integrate over all its possible values.
You have $$E(Y)=\int E(Y|Z=z)p(Z=z)dz \text { integral rule}\\ =\int(1-e^{-\lambda Bz})p(Z=z)dz\\ =\int_{-\infty}^\infty(1-e^{-\lambda Bz})\frac{(1/\theta)^\theta}{\Gamma(\theta)}z^{\theta-1}e^{-\frac z \theta}dz$$
Indeed in your very first line you should write $P^{Z=z}$ not $P^Z$ otherwise that equality is not technically true.