I have proven that: $X⊥Y|Z\ {\rm iff}\ p(x,y|z)=p(x|z)p(y|z)$ for all $x,y,z$ such that $p(z)>0$.

The next question is to prove an alternative definition: $X⊥Y|Z$ iff there exist functions $g$ and $h$ such that $p(x,y|z)=g(x,z)h(y,z)$ for all $x,y,z$ such that $p(z)>0$.

I'm thinking that I need to integrate the function somehow...?

  • 1
    $\begingroup$ Is this a problem for a class? If so, please add the [self-study] tag & read its wiki. $\endgroup$ – gung - Reinstate Monica Nov 25 '14 at 19:39
  • $\begingroup$ You have already proven one part of the iff. To prove the converse, you have indeed to integrate the second equality. $\endgroup$ – Xi'an Nov 25 '14 at 20:14
  • $\begingroup$ I've tried to integrate the second equality, but got stuck along the way. I could use some help with how to set up the integration procedure. $\endgroup$ – MatildaB Nov 25 '14 at 20:19
  • $\begingroup$ If you are given $p(x,y|z)=g(x,z)h(y,z)$, by integrating out the equality in $x$ or in $y$, you obtain the marginal conditional of $y$ or $x$ given $z$, hence deduce proportionality between $g(x,z)$ and $p(x|z)$ and similarly between $h(y,z)$ and $p(y|z)$. $\endgroup$ – Xi'an Nov 25 '14 at 21:35
  • $\begingroup$ Is $g(x, z)$ a pmt function for the joint distribution of $z$ and $x$? So the integral is over both $z$ and $x$? What do you mean by integrating out the equality? $\endgroup$ – MatildaB Nov 25 '14 at 21:54

Since $X$ and $Y$ are independent given $Z$ iff $p(x,y|z)=p(x|z)p(y|z)$, all you need to prove is that, if $p(x,y|z)=g(x,z)h(y,z)$, then $p(x,y|z)=p(x|z)p(y|z)$.

Starting from the equality $p(x,y|z)=g(x,z)h(y,z)$. one can integrate both sides in x: $$ \int_\text{X}p(x,y|z)\text{d}x=\int_\text{X}g(x,z)h(y,z)\text{d}x $$ This implies $$ p(y|z)=\int_\text{X}g(x,z)h(y,z)\text{d}x=h(y,z)\int_\text{X}g(x,z)\text{d}x $$ and tells you that $$ h(y,z)\propto p(y|z) $$ [where the proportionality sign is for a function of $y$, meaning that the proportionality constant can depend on $z$, i.e., $h(y,z)=\nu(z) p(y|z)$]. A symmetric argument leads to $$ g(x,z)\propto p(x|z) \quad\text{i.e., } g(x,z)=\eta(z) p(x|z) $$ Therefore, $$ g(x,z)h(y,z)=\eta(z) p(x|z)\nu(z) p(y|z)\,, $$ and since both sides integrate to $1$ (when integrating both in $x$ and $y$), we conclude with $$ \eta(z) \nu(z)=1 $$ Hence,$$p(x,y|z)=p(x|z)p(y|z)$$

  • $\begingroup$ Could you please clarify the following: > and since both sides integrate to 1 (when integrating both in x and y), we conclude with $$ \eta(z)v(z)=1 $$ $\endgroup$ – Richard Aug 26 '15 at 5:00
  • $\begingroup$ If I integrate both sides of$$g(x,z)h(y,z)=\eta(z) p(x|z)\nu(z) p(y|z)$$I get$$1=\int p(x,y|z)\text{d}x\text{d}y=\int g(x,z)h(y,z)\text{d}x\text{d}y=\int \eta(z) p(x|z)\nu(z) p(y|z)\text{d}x\text{d}y=\eta(z)\nu(z)\times 1$$ $\endgroup$ – Xi'an Aug 27 '15 at 5:37

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