From Casella and Berger: I've run into an interesting property while doing exercises. The property being: $$f_Z(z) = 1-F_Z(z)$$ Would like to know if it is a result of a mistake or an actual property of a special set-up, and if so, if there are applications for this (visualizations would be amazing!).
The process of running into this property is as follows:
Suppose $U_1,...,U_X$ are a sample of uniform random variables. Let $Z\sim min(U_1,...,U_X)$. Finally X itself is a discrete random variable defined as: $$P(X=x)=\frac{1}{x!(e-1)}, \ \ x=1,2,3,...$$ Find the distribution of Z. Assuming they mean pdf, I decided to interpret $Z=min(U_1,...,U_X)=U_{(1)}$, as an order statistic problem.
Thus: $$P(U_{(1)}=z|X=x)=\frac{x!}{0!(x-1)!}f_U(z)[F_U(z)]^0[1-F_U(Z)]^{X-1}$$ By the pdf of any order statistic given continuous random samples. Simplifying: $$P(U_{(1)}=z|X=x)=x(1-z)^{x-1}$$ By the law of total probability: $$P(U_{(1)}=z)= \sum_{x=1}^{\infty}P(U_{(1)}=z|X=x)P(X=x)$$ $$= \sum_{x=1}^\infty \frac{x(1-z)^{x-1}}{x!(e-1)}$$ $$=\frac{1}{e-1}\sum_{x=1}^\infty \frac{(1-z)^{x-1}}{(x-1)!}$$ $$P(U_{(1)}=z)=P(Z=z)=\frac{e^{1-z}-1}{e-1}, 0<z<1$$
Hoping there aren't any stupid mistakes here. Now the conundrum begins when we derive $P(Z>z)$
$$P(Z>z)=\sum_{x=1}^\infty P(Z>z|X=x)P(X=x)$$ $$=\sum_{x=1}^\infty P(U_1>z,...,U_X>z|X=x)P(X=x)$$ By i.i.d uniform samples: $$=\frac{1}{e-1}\sum_{x=1}^\infty \frac{(1-z)^x}{x!}$$ $$P(Z>z)=1-F_Z(z)=\frac{e^{1-z}-1}{e-1}=f_Z(z), \ \ 0<z<1$$
They have the same distribution. Curious to whether there is a mistake anywhere or whether this is some special case, if so, are there any useful applications/visualizations for this?
Unfortunately there is a mistake in your calculation. The proof is correct up to the point of identifying $f_Z(z) = P(U_{(1)}=z)$; this should be
\begin{align*} f_Z(z) &= \sum_{x=1}^{\infty}P(U_{(1)}=z|X=x)P(X=x)\\ &= \sum_{x=1}^\infty \frac{x(1-z)^{x-1}}{x!(e-1)}\\ &=\frac{1}{e-1}\sum_{x=1}^\infty \frac{(1-z)^{x-1}}{(x-1)!}\\ & = \frac{e^{1-z}}{e-1} \end{align*}
I then agree with your derivation that leads to $$1 - F_Z(z) = \frac{e - e^{1-z}}{e -1},$$ which means the correct identity is in fact $$ 1 - F_Z(z) = \frac{e}{e-1} - f_Z(z),$$ or more simply $$f_Z(z) = \frac1{1-e} + F_Z(z).$$