Not Getting Anywhere

[JeffM]

As everyone providing assistance here knows, I am neither a student in a class nor doing homework. This came up in a project I am working on, and I cannot figure out whether there is no answer or whether I am missing something obvious.

$$P(X)= \text {a known positive value } a < 1;\\ P(Y) = \text {a known positive value } b < 1;\\ P(Z) = \text {a known positive value } c < 1;\\ P(Y \ | \ X) = 1;\\ P(Y \ | \ Z) = 1, \text { and}\\ P(Z \ | \ Y) = \text {a known positive value } d < 1.$$
The question is whether a, b, c, and d are sufficient to compute $P(Z \ | \ X)$, and if so how?

I cannot prove a, b, c, and d are sufficient nor that they are not. I am feeling like an idiot and keep going in circles.

 

[The Highlander]

Please show us what you have already tried. ???

 

[Dr.Peterson]

As everyone providing assistance here knows, I am neither a student in a class nor doing homework. This came up in a project I am working on, and I cannot figure out whether there is no answer or whether I am missing something obvious.

$$P(X)= \text {a known positive value } a < 1;\\ P(Y) = \text {a known positive value } b < 1;\\ P(Z) = \text {a known positive value } c < 1;\\ P(Y \ | \ X) = 1;\\ P(Y \ | \ Z) = 1, \text { and}\\ P(Z \ | \ Y) = \text {a known positive value } d < 1.$$
The question is whether a, b, c, and d are sufficient to compute $P(Z \ | \ X)$, and if so how?

I cannot prove a, b, c, and d are sufficient nor that they are not. I am feeling like an idiot and keep going in circles.

I believe the answer is no.

The circles I went in are Venn diagrams, starting with the subset relationship implied by the 1's (ignoring possible measure-zero effects). I also observed that d is redundant.

But I'm not willing to declare my answer as proved, because I've only thought about it informally with little pictures.

 

[blamocur]

I agree with @Dr.Peterson. Treating $X,Y,Z$ as sets, we can show that $X\subset Y$ and $Z\subset Y$ (again, ignoring zero measure subsets). The question can now be reduced to whether the knowledge of $P(Z|Y)$ is enough to compute $P(Y|Z)$, which it is not in the general case.
It should be possible to construct a counterexample with a Venn diagram, but I feel too lazy for that.