Suppose that $X, Y$ are random variables,both from a probability space to $(0,\infty]$, such that X and $1+Y$ have the same distribution,when $Y=Z_1$ or $Y= \frac{Z_1Z_2} {Z_1+Z_2}$ with equal probability and $Z_1,Z_2$are iid and both are copies of $X$.
what can be said about $X$?is $X$ finite almost sure?what about $E(X),Var(X)$?
what i've tried: $E(X)=1+E(Y)=1+\frac{1}{2}E(Z_1)+\frac{1}{2}E(\frac{Z_1Z_2} {Z_1+Z_2})$, because $E(X)=E(Z_1)$,we get $E(X)=2+ E(\frac{Z_1Z_2} {Z_1+Z_2})$.
So $E(X-\frac{Z_1Z_2} {Z_1+Z_2})=2$,it is finite.Therefore $X-\frac{Z_1Z_2} {Z_1+Z_2}$ is almost sure finite.but what about $X$?
Is there any hint how to understand $X$?
Thanks!
Not really an answer, but maybe it is useful some way anyhow.
Assume that $\Pr\left(X=\infty\right)=p$.
Then also $\Pr\left(Y=\infty\right)=p$ since $X$ and $1+Y$ have the same distribution.
Formally we need a proper definition of $W\left(\omega\right):=\frac{Z_{1}\left(\omega\right)Z_{2}\left(\omega\right)}{Z_{1}\left(\omega\right)+Z_{2}\left(\omega\right)}$ on set $\left\{ Z_{1}=\infty\vee Z_{2}=\infty\right\} $.
Actually we are only interested in $\Pr\left(W=\infty\right)$ and on base of limits (as you suggested in a comment) we find:
$W\left(\omega\right)=\infty$ iff $ Z_{1}=Z_{2} =\infty$.
Then $\Pr\left(W=\infty\right)=\Pr\left(Z_{1}=\infty\wedge Z_{2}=\infty\right)=p^{2}$.
Now we can find $\Pr\left(Y=\infty\right)$ on another way.
Let $Y=Z_{1}U+W\left(1-U\right)$ where $\Pr\left(U=1\right)=\frac{1}{2}=\Pr\left(U=0\right)$
Here $U,Z_{1}$ are independent and $U,W$ are independent.
Then $\Pr\left(Y=\infty\right)=\Pr\left(Y=\infty\mid U=1\right)\Pr\left(U=1\right)+\Pr\left(Y=\infty\mid U=0\right)\Pr\left(U=0\right)$ leading to
$\Pr\left(Y=\infty\right)=\frac{1}{2}\Pr\left(Z_{1}=\infty\right)+\frac{1}{2}\Pr\left(W=\infty\right)=\frac{1}{2}p+\frac{1}{2}p^2$.
So we have the equation:
$p=\frac{1}{2}p+\frac{1}{2}p^2$ leading to $p=0\vee p=1$.