So I have fundamental confusion about graphical models. Suppose the following graphical model is given:enter image description here

Now the question is do we have the following equality?: $$p(m_2|\alpha,\beta,y_3)=p(m_2|\alpha,\beta)$$

If so, why exactly and could you please introduce me a source explaining it?


4 Answers 4


Yes, this is because the $\alpha, \beta$ node $d$-separates $m_2$ and $y_3$. See Probabilistic Reasoning in Intelligent Systems for an explanation of $d$-separation.


Another intuitive example of why two sibling nodes are independent given their parent:

  • Imagine $A$ and $B$ are two guys living in the same city.

  • Whether A gets wet or not depends on whether it rains or not. Same for B.

Now, if don't know whether it rained or not but we observed that $A$ is wet, we are going to think that it rained, and therefore it will be likely that $B$ will be wet as well. That is, when we do not know the value of the parent node (rain) the information about $A$ gives also some information about $B$.

However, if we know it rained (the parent node is observed) then to guess whether $B$ is wet or not we do not need to now nothing about $A$; we don't care, since we already know it rained.

That is, when you now the state of the parent you don't care about the rest of nodes since you only depend on your parent. If you don't know the state of your parent then yes, the rest of the nodes can give you some hint on his state.


I like to think about it like this:
What additional information does knowing $y_3=X$ actually give you?

Now the additional is key here. Imagine you knew nothing but $y_3=X$. Then in order:

  1. We can learn about $m_3$ by asking what it needs to make $y_3=X$ more likely
  2. We can learn about $\alpha,\beta$ by asking ourselves what they need to be so to make $m_3$ more likely to be what we think it should be from step 1
  3. The knowledge about $\alpha,\beta$ from step 2 can help us guess what the distribution of $m_2$ is.

So that's $p(m_2|y_3)$.

But now do the same process, already knowing for certain what $\alpha,\beta$ are, that is $p(m_2|\alpha,\beta,y_3)$.
Well it turns out 2 is useless. Step 2 is useless because we already know $\alpha,\beta$, since they are given and no new information changes that. So step 3 happens regardless of whether $y_3$ is known or not.

Hopefully this wasn't more confusing.


\begin{align} p(m_2|\alpha,\beta,y_3)&=\frac{p(m_2,y_3|\alpha, \beta)}{p(y_3|\alpha,\beta)}\\ &= \frac{p(m_2|\alpha,\beta)p(y_3|\alpha, \beta)}{p(y_3|\alpha, \beta)}\\ &=p(m_2|\alpha,\beta) \end{align}

$p(m_2,y_3|\alpha, \beta)=p(m_2|\alpha,\beta)p(y_3|\alpha, \beta)$ holds because $m_2$, $y_3$ and $\alpha, \beta$ form a common cause trail. That's given $\alpha$ and $\beta$ the correlation between $m_2$ and $y_3$ disappears, and $m_2$ can influence $y_3$ only when $\alpha, \beta$ and $m_3$ are all not observed.


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