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

Equation...

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.

Thomas
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    where does $\eta$ come from? – Alex Jul 03 '17 at 11:18
  • It is supposed to be a "normalizer", something that is constant. See the full text here: https://books.google.de/books?id=2Zn6AQAAQBAJ&lpg=PA140&ots=vHnEODAD_A&dq=probabilistic%20robotics%205.48&pg=PA140#v=onepage&q=probabilistic%20robotics%205.48&f=false – Thomas Jul 03 '17 at 11:33
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    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: https://math.stackexchange.com/questions/638635/if-a-and-b-are-independent-what-can-we-say-about-px-mid-a-b?rq=1 – Thomas Jul 03 '17 at 11:34
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    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. –  Jul 03 '17 at 11:51
  • Even if this were true, it would not be enough? This link has a good counterexample were A&B are independent: https://math.stackexchange.com/questions/638635/if-a-and-b-are-independent-what-can-we-say-about-px-mid-a-b?rq=1#comment1346757_638643 – Thomas Jul 03 '17 at 12:00
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    @Thomas independence is not enough, but conditional independence is. –  Jul 03 '17 at 12:03
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    I have closed a duplicate of this question (Probabilistic Robotics: Map-based motion model) over on [robotics.se]. Please feel free to migrate it to Robotics if you don't think this question is appropriate here. – Mark Booth Jul 03 '17 at 15:26

1 Answers1

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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 $$\eta =\dfrac{p(\mu_t,x_{t-1})~~~~~~}{p(\mu_t,x_{t-1}\mid m)}$$

Graham Kemp
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