Proability: 3 variables, one is partially observed, can we infer it

Question

We have 3 binary variables: x, s, and v. x causes s and v; and s causes v.

x -> s, v

s -> v

There is some data restriction when s = 0, and so we want to use x and v to learn more about s.

We know

P(x = 1) = .5

P(v = 1|x = 1) = .29

P(s = 1|x = 0) = .37

P(s = 1|x = 1) = .78

P(v = 1|s = 1) = .28

P(s = 1|v = 1) = .75

P(v = 1|s = 1, x = 1) = .32

P(v = 1|s = 1, x = 0) = .18

We want to know P(v = 1|s = 0)?

We can derive P(s = 0) = P(s = 0, x = 0) + P(s = 0, x = 1) and P(s, x) = P(x) * P(s| x).

Is this solvable and what is the solution?

Answer 1

As @lulu suggested framing this in terms of the 8 possible combinations of s, v, and x will allow you to determine either what you want to know or find out that the combinations you listed are inconsistent.

Using Mathematica's Reduce command (because in my old age I like to let computers do the thinking) one finds that the "facts" given cannot all be true. I've ordered s, v, and x in alphabetical order so that p101 = P(s=1, v=0, x=1).

Reduce[{p001 + p011 + p101 + p111 == 1/2, (* P[x=1] *)
  p000 + p010 + p100 + p110 == 1/2,   (* P[x=0] *)
  (p011 + p111)/(1/2) == 29/100, (* P[v=1 | x=1] *)
  (p100 + p110)/(1/2) == 37/100,  (* P[s=1 | x=0] *)
  (p101 + p111)/(1/2) == 78/100,  (* P[s=1 | x=1] *)
  (p110 + p111)/(p100 + p101 + p110 + p111) == 28/100, (* P[v=1 | s=1] *)
  (p110 + p111)/(p010 + p011 + p110 + p111) == 75/100, (* P[s=1 | v=1] *)
  p111/(p101 + p111) == 32/100, (* P(v=1 | s=1, x=1) *)
  p110/(p100 + p110) == 18/100  (* P(v=1 | s=1, x=0) *)
  },
 {p000, p001, p010, p011, p100, p101, p110, p111}]
(* False *)

If the last equation is removed (and probably removing any one equation), then a plausible solution is reached:

Reduce[{p001 + p011 + p101 + p111 == 1/2, (* P[x=1] *)
  p000 + p010 + p100 + p110 == 1/2,   (* P[x=0] *)
  (p011 + p111)/(1/2) == 29/100, (* P[v=1 | x=1] *)
  (p100 + p110)/(1/2) == 37/100,  (* P[s=1 | x=0] *)
  (p101 + p111)/(1/2) == 78/100,  (* P[s=1 | x=1] *)
  (p110 + p111)/(p100 + p101 + p110 + p111) == 28/100, (* P[v=1 | s=1] *)
  (p110 + p111)/(p010 + p011 + p110 + p111) == 75/100, (* P[s=1 | v=1] *)
  p111/(p101 + p111) == 32/100 (* P(v=1 | s=1, x=1) *)
  },
 {p000, p001, p010, p011, p100, p101, p110, p111}]
 (* p000 == 4223/15000 && p001 == 449/5000 && p010 == 251/7500 && 
    p011 == 101/5000 && p100 == 93/625 && p101 == 663/2500 && 
    p110 == 181/5000 && p111 == 78/625 *)