C.28.MR:5.1 - Observing an alarm and forcing its output

Let H mean binary high load and S mean a binary alarm. Take independent uniform inputs U_H,U_S on [0,1] and the supplied mechanisms

H = 1[U_H < 0.5]

S = 1[U_S < 0.1 + 0.8 H].

Here 1[condition] is one when the condition holds and zero otherwise. In high-load cases, the alarm sounds with probability 0.9. In low-load cases, it sounds with probability 0.1.

The observational probability of no alarm is 0.5*0.1 + 0.5*0.9 = 0.5. High load with no alarm has probability 0.5*0.1 = 0.05. Thus P(H=1 | S=0)=0.05/0.5=0.1: the observation changes what we infer about the load.

For do(S=0), replace the second mechanism by S=0. The equation for H and the input law are retained. Therefore P(H=1 | do(S=0))=0.5. Forcing this indicator removes its information about H without changing H in the supplied model.

The result answers a current-load question under the stated absence of an S-to-H influence. If alarm output controls subsequent cooling, a later-load query needs that later mechanism and the duration of the override. The current calculation remains usable for the current question; the new temporal question has an additional required input.