Frage

I recently went through a video which said that in the relation x->W<-Y, X does not influence y.X has causal relationship to W and W has evidential relationship to Y .So will X not affect Y ?

War es hilfreich?

Lösung

Let's let W represent "The Lawn is Wet," X represent "It rained recently," and Y represent "The automatic sprinklers were on recently."

Clearly, X influences W: If it rained recently, it is likely that the lawn is wet. Clearly, Y influences W: If the sprinklers were on recently, it is likely that the lawn is wet. Clearly, knowing W, we can make inferences about X and Y.

But, does X directly influence Y? Put differently, does the fact of recent rain (or not) influence whether the automatic sprinklers were on recently? No. If we know nothing about the state of the lawn, because we didn't look outside, the chance of recent rain is independent of the chance of recent sprinkler activity.

Once we look outside, though, and determine the state of the lawn, then we can draw inferences between rain and sprinkler activity.

Andere Tipps

If W is not observed, then x and y are independent.

One such v-structured CPD (that is entirely deterministic) looks like this.

Draw X and Y independently as binary variables, and then W is the sum. If you know only X=1, then Y could be 0 and W would be 1, or Y = 1 and W = 2. Knowing X doesn't let you distinguish between those two possibilities.

In general, I think reasoning about v-structures is much easier with deterministic functions than probabilistic ones. Think about what happens when the v-structure is ADD, XOR, and AND and you can usually get specific insights about general claims.

something that helped me understand this was a concrete example that I found in Sections 1.3 and 1.4 of this online book:

It goes through all the cases you have listed above, and for each case explains w/ the running example described in Section 1.4. Please have a look here:

http://people.cs.aau.dk/~uk/papers/pgm-book-I-05.pdf

Lizenziert unter: CC-BY-SA mit Zuschreibung
Nicht verbunden mit StackOverflow
scroll top