If $p(a \mid b, c) = p(a \mid b) = p(a \mid b, d)$, then is $p(a \mid b, c, d) = p(a \mid b)$?

Question

Both my intuition, and a graphical model of the situation suggest that the following statement might be true:

$$ p(a \mid b, c) = p(a \mid b) = p(a \mid b, d)\quad\Rightarrow\quad p(a \mid b, c, d) = p(a \mid b) $$

Yet, I have struggled to show, algebraically, that this is the case. Is it true? If so, how can one show it algebraically?

Answer 1

Not true in general. Here is my example.

Then \begin{align} P(A\mid B) &= \frac{P(AB)}{P(B)} = \frac{4/10}{6/10} = \frac{2}{3} \\ P(A\mid BC) &= \frac{P(ABC)}{P(BC)} = \frac{2/10}{3/10} = \frac{2}{3} \\ P(A\mid BD) &= \frac{P(ABD)}{P(BD)} = \frac{2/10}{3/10} = \frac{2}{3} \\ P(A\mid BCD) &= \frac{P(ABCD)}{P(BCD)} = \frac{1/10}{1/10} = 1 \end{align}