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.


1 Answer 1


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.


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