How to show $P(b | a,e) = P(b)$ given some conditions?

Question

$A,B,C$ are binary r.v.'s such that $B$ and $C$ are independent and $P(a | b,e) = P(a | \overline{b}, e) = 1$. I'm trying to show that $P(b | a,e) = P(b)$.

We have $P(b | a,e) = P(a,b,e) / P(a,e)$.

The denominator can be rewritten:

$P(a,e) = P(a,e,b) + P(a,e,\overline{b}) = P(a|e,b)P(e,b) + P(a|e,\overline{b})P(e,\overline{b}) = P(e,b) + P(e,\overline{b}) = P(e)$.

So we now have $P(b | a,e) = P(a,b,e) / P(a,e) = P(a,b,e)/P(e) = P(a,b|e).$

Is there some way to show that $P(a,b|e) = P(b|e)$ ?

Answer 1

Is there some way to show that $P(a,b|e)=P(b|e)$ ?

You are given that : $P(a\mid b, e) = 1$

Then as $P(a, b\mid e) = P(a\mid b, e) P(b\mid e)$ so...