# $P(A|B,C) = P(A|B) * P(A|C) *$ …?

In his book "Probabilistic Robotics", Thrun has the following equation: (Context here) - $\eta$ is supposed to be a "normalizer"

As I see it, this boils down to:

$P(A|B,C) = \dfrac{P(A|B) \cdot P(A|C)}{P(A)}\cdot constant$

I have tried to convert the left side to the right side, but failed. I need to know whether he assumes that P(A|B) and P(A|C) are stochastically independent, since in my case, they aren't. Can anybody point me in the right direction on how to solve this? I tried using the chain rule and (obviously) Bayes' theorem.

• where does $\eta$ come from? – Alex Jul 3 '17 at 11:18
• The following answer seems to indicate that the equation in the book is nonsense - Kind of weird, since it is its third edition and widely used in the field of robotics: math.stackexchange.com/questions/638635/… – Thomas Jul 3 '17 at 11:34
• Well, it seems the book assumes $m$ is conditionally (on $x_t$) independent of $u_t, x_{t-1}$. I take it $m$ is some external variable to the robot, so it seems like a reasonable assumption. – Slug Pue Jul 3 '17 at 11:51
• @Thomas independence is not enough, but conditional independence is. – Slug Pue Jul 3 '17 at 12:03
• I have closed a duplicate of this question (Probabilistic Robotics: Map-based motion model) over on Robotics. Please feel free to migrate it to Robotics if you don't think this question is appropriate here. – Mark Booth Jul 3 '17 at 15:26

Basically if $P(A\mid B,C)= \eta\cdot P(B\mid A)P(C\mid A)\div P(A)$ and if $B,C$ are conditionally independent given $A$, then we can find $\eta$. \begin{align}P(A\mid B,C) ~& = \dfrac{P(B,C\mid A)P(A)}{P(B,C)} \\[1ex] &=\dfrac{P(B\mid A)P(C\mid A)~P(A)}{P(B,C)} \\[1ex] &=\dfrac{P(B\mid A)P(C\mid A)~P(A)^2}{P(B,C)~P(A)} \\[1ex] &= \dfrac{P(A\mid B)P(B)~P(C\mid A)P(C)}{P(B,C)~P(A)}\\[1ex] &= \dfrac{P(B)P(C)}{P(B,C)}\cdot\dfrac{P(A\mid B)P(A\mid C)}{P(A)}\\[3ex]\therefore\qquad\eta ~&= \dfrac{P(B)P(C)}{P(B,C)}\\[1ex]&=\dfrac{P(B)}{P(B\mid C)}\end{align}
So in this case if we have $~p(x_t\mid \mu_t, x_{t-1}, m) = \eta\cdot\dfrac{p(x_t\mid \mu_t,x_{t-1})p(x_t\mid m)}{p(x_t)}~$ then it is likely because $\{\mu_t,x_{t-1}\}$ and $\{m\}$ are conditionally independent given $\{x_t\}$ and that $$\eta =\dfrac{p(\mu_t,x_{t-1})~~~~~~}{p(\mu_t,x_{t-1}\mid m)}$$