0
$\begingroup$

In general, we can calculate the joint distribution of the endogenous variables of a structural casual model (SCM) as follows $$ P(X, X_1, \dots, X_n) = \prod_i = P(X_i \mid \text{parent}(X_i)) $$ where $\text{parent}(X_i)$ is the parent node of $X_i$ in the graphical representation of the SCM. Essentially, this is the chain rule, and it is also called "product decomposition". In this scenario, nodes or variables without endogenous parents are expressed as simply $P(W)$, where $W$ does not have any parent in the causal graph.

For example, we may have the following SCM

enter image description here

Then the joint probability of the endogenous variables $X$, $Y$ and $Z$ is $$P(X, Y, Z) = P(Z \mid X, Y) P(Y) P(X)$$

In this case, we can see that both $Y$ and $X$ do not have endogenous parents. Note that in this case $U_X, U_Z$ and $U_Y$ are exogenous variables (variables which are external to the model and that we do not describe the causes of).

But why do we care about the joint distribution or probability of the endogenous variables? Why do we care about all these happening? Is it because we can calculate the probability of some combination of the realisation of these variables?

$\endgroup$
  • $\begingroup$ Think about it: if you did not know the structure of the SCM (which gives you the product decomposition), how could you tell how to estimate $P(Y|do(X))$? $\endgroup$ – Carlos Cinelli Jan 4 at 4:31
  • $\begingroup$ @CarlosCinelli Can you please elaborate a little more? It is not clear exactly to me what the relation between your interventional distribution and the joint distribution I am mentioning is. $\endgroup$ – nbro Jan 19 at 20:32
0
$\begingroup$

In general, we care about joint distributions because we may be looking for answers to questions like "What is the probability that today will rain and my friend will visit me".

In the context of causality, we are also interested in this type of questions. For example, suppose we have the causal model $X \rightarrow Y \rightarrow Z \rightarrow W$ (a chain), where $X$ stands for clouds/no clouds, $Y$ stands for rain/no rain, $Z$ stands for wet pavement/dry pavement, and $W$ stands for slippery pavement/unslippery pavement.

Based just on our experience of the world, how plausible is it that $\mathbb{P}(\text{clouds}, \text{no-rain}, \text{dry pavement}, \text{slippery pavement}) = 0.23$? Of course, this is not a very easy question to answer. Imagine if the the number of variables were 100? How could we estimate such joint distributions?

In the context of causality, we can use the product rule to more easily estimate such probabilities. Using the product rule, we can express $\mathbb{P}(\text{clouds}, \text{no-rain}, \text{dry pavement}, \text{slippery pavement})$ as follows

$$\mathbb{P}(\text{clouds}) \mathbb{P}(\text{no rain}\mid \text{clouds}) \mathbb{P}(\text{dry pavement}\mid \text{no rain}) \mathbb{P}(\text{slippery pavement}\mid \text{dry pavement})$$

Of course, it is easier for humans to estimate probabilities like $\mathbb{P}(\text{clouds})$ or $\mathbb{P}(\text{slippery pavement}\mid \text{dry pavement})$ than to estimate the joint probability above. For example, we may estimate $\mathbb{P}(\text{clouds})$ to be $0.5$ if we live in a region where it often rains. Likewise, we can estimate $\mathbb{P}(\text{slippery pavement}\mid \text{dry pavement})$ to be low, given that if the pavement is dry then it is unlikely is also slippery.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.