If $P(A) \neq Q(A)$, then is $P(A|B) \neq Q(A|B)$?

Question

The title pretty much sums it up. Although I'm curious as to the general answer, I'm specifically referring to two different $P_\theta,\, \theta \in \Theta$ from a family of probability measures $(P = \{P_\theta\})$ on a statistical experiment with space $(\mathcal X,\mathcal B, P)$. So, for any $B \in \mathcal B$, if $P_{\theta_1}\neq P_{\theta_2}$, does that mean $P_{\theta_1}(A|B) \neq P_{\theta_2}(A|B)$?

Answer 1

For an intuitive example. Consider the probability to get 6 when throwing either a standard 6-sided die and a 12-sided die. They are obviously not the same. Then let the B event be that the throw is strictly below 7 -- then the 12-sided die becomes a 6-sided die in terms of probabilities.

Answer 2

Let $\Bbb P_{\theta} $, $\theta > 0$ have the following density $$f_{\theta} (x) = \frac 1 2 1_{(-1,0)}(x) + \frac 1 2 \theta e^{-\theta x } 1_{(0,\infty)}(x)$$ Clearly, for $\theta_1 \neq \theta_2$ we have $\Bbb P _{\theta_1} \neq \Bbb P _{\theta_2}$, but for $B = (-1,0)$ we have $$\Bbb P _{\theta_1} (\cdot \vert B) = \text{Uni}(-1,0) = \Bbb P _{\theta_2}(\cdot \vert B)$$ where $\text{Uni}(-1,0)$ denotes the uniform distribution on $(-1,0)$.

Answer 3

Or take $B=A$ to get $P(A\mid B) = Q(A\mid B) = 1$ even if $P(A) \ne Q(A)$.