Expectation of conditional exponential distribution

Question

I was doing some reading and read this phrase without anyting else written down.

Suppose $Y\mid Z $ is distributed according to an exponential distribution, $f_{Y\mid Z}(y)=ze^{-zy}$.

Now I wanted to do some calculations with conditional expectations. First $E((E(Y\mid Z))=E\left(\frac{1}{Z}\right)$ since $Y\mid Z$ is exponential distributed. But second we have by tower porperty $E(E(Y\mid Z))=E(Y)$. But in general it is not true that $E(Y)=E\left(\frac{1}{Z}\right)$, so what was I doing wrong here? I can not find the error or did I missunderstood the phrase, but how should one understand this then?

Answer 1

Your calculations are right. Using law of total expectation you have

$$\mathbb{E}[Y]=\mathbb{E}\{\mathbb{E}[Y|Z]\}=\mathbb{E}[1/Z]$$

This is true in this case because $(Y|Z=z)\sim \exp(z)$. In general you have to derive the conditional expectation $\mathbb{E}[Y|Z]$ which is always a function of $Z$

now to calculate $\mathbb{E}[Y]$ you have to know the distribution of $Z$

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just as an example, if $Z\sim U(1;2)$ you get

$$\mathbb{E}[Y]=\int_1^2 \frac{1}{z}dz=\log 2$$