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If I want to compute the probability $p(y,z|\theta,\lambda)$,then how to? I know the answer is $p(y|z,\theta)p(z|\lambda)$, but I do not know how to? Please help me, thanks a lot

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    $\begingroup$ As you said, you know the answer, so what is not clear? $\endgroup$
    – Tim
    Dec 8, 2015 at 9:49
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    $\begingroup$ This sounds like a self-study question so please add the tag and tell us what you tried to solve the question, the background, and why it is important. $\endgroup$
    – Xi'an
    Dec 8, 2015 at 9:57
  • $\begingroup$ I think this is an easy problem about joint probability computation. I just can not be sure if there are any other tricks here $\endgroup$
    – DuFei
    Dec 10, 2015 at 12:13

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What you want to do compute is the joint probability of $p(y,z|\theta, \lambda)$. Therfore you need to multiply $p(y|z,\theta)$ and $p(z|\lambda)$.

If you look at the graph you can see that $y$ has $z$ and $\theta$ as parent nodes. This means that $y$ depends on the values of $z$ and $\theta$. In terms of probabilities this is expressed as $p(y|z, \theta)$. The vertical bar expresses conditional probabilities.

The same principle applies to $z$ which depends only on $\lambda$ and can therefore be expressed as $p(z|\lambda)$.

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  • $\begingroup$ I see. $p(y,z|\theta,\lambda) = p(y|z,\theta,\lambda)p(z|\theta,\lambda)$ Since $y$ only depends on $z$ and $\theta$ and $z$ only depends on $\lambda$. Thus we obtain $p(y,z|\theta,\lambda) = p(y|z,\theta)p(z|\lambda)$. Am I right? I thought $\lambda$ influences $y$ through $z$. So I am not sure if $ p(y|z,\theta,\lambda)=p(y|z,\theta) $ $\endgroup$
    – DuFei
    Dec 10, 2015 at 12:10

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