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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!

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  • $\begingroup$ "is $X$ finite almost sure?" What exactly do you mean? Put differently: if $X:\Omega\to E$ then what does $E$ look like? Does it contain elements that are not finite? Usually $E=\mathbb R$. $\endgroup$ – drhab Aug 14 '16 at 13:29
  • $\begingroup$ @drhab No.$X$ goes to $(0,\infty]$ $\endgroup$ – user115608 Aug 14 '16 at 13:40
  • $\begingroup$ So $E=(0,\infty]$ and not $E=(0,\infty)$ as you said first. Do I understand you correct? $\endgroup$ – drhab Aug 14 '16 at 13:42
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    $\begingroup$ There is no limit. $\omega\in\Omega$ might exist such that $Z_1(\omega)=\infty$ and e.g. $Z_2(\omega)=5$. In that case the expression $\frac{Z_1Z_2}{Z_1+Z_2}$ needs a definition. If this lacks then the expression cannot be recognized as a random variable. $\endgroup$ – drhab Aug 14 '16 at 14:16
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    $\begingroup$ More things are wrong. Also your conclusion $\mathbb E(X-\frac{Z_1Z_2}{Z_1+Z_2})=2$ based on the former $\mathbb EX=2+\mathbb E(\frac{Z_1Z_2}{Z_1+Z_2})$ is only legal under the extra condition that $\mathbb EX<\infty$ which at that stage is not proved yet. $\mathbb EX=2+\mathbb E(\frac{Z_1Z_2}{Z_1+Z_2})$ is also true if both sides take value $\infty$. $\endgroup$ – drhab Aug 14 '16 at 14:22
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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 $p=0\vee p=1$.

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  • $\begingroup$ thank u a lot!but why do u assume $z>1$ in line 7 ? $\endgroup$ – user115608 Aug 14 '16 at 16:55
  • $\begingroup$ If $z_1<1$ then $\lim_{z_2\to\infty}\frac{z_1z_2}{z_1+z_2}=0$ and if $z_1=1$ then the limit takes value $1$. So in both cases not $\infty$. $\endgroup$ – drhab Aug 15 '16 at 8:00
  • $\begingroup$ i think if we want the fraction to be infinity , both $Z_1,Z_2$ have to be infinity.note the comments we discussed. $\endgroup$ – user115608 Aug 15 '16 at 9:45
  • $\begingroup$ Why both? If $1<z_1<\infty$ then also $\lim_{z_2\to\infty}\frac{z_1z_2}{z_1+z_2}=\infty$. $\endgroup$ – drhab Aug 15 '16 at 9:58
  • $\begingroup$ no in this case limit equals $z_1$.note the comments ewe discussed about it. $\endgroup$ – user115608 Aug 15 '16 at 10:03

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