Law of Total Probability in Bayes Theorem

While reading along the probabilistic robotics book, I came across these set of equations which bothers me. Let's take equation 2.13 for example.

P(Y) was expressed into a summation referred to as law of total probability. But should it be conditioned to x' only? Does not the law of total probability state?

 P(Y) = P(Y,X) + P(Y,X')


Then why is it only conditioned to x'?

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what do you mean by P(Y,X) and P(Y,X')? (and was the switch to upper case letters deliberate?) –  TooTone Feb 26 '14 at 16:10
@TooTone i think he means probability of y is equal to probability of y given x plus probability of y given x' –  user130512 Feb 26 '14 at 16:16
@user130512 thanks, I thought that might be the case but wanted to be sure... –  TooTone Feb 26 '14 at 16:17

The law of total probability basically says, if you can partition a sample space $Y$ into sets $X_1, ..., X_n$ (which can actually be countably infinite if necessary), then

$P(Y) = \sum_{n}P(Y \cap X_n) = \sum_{n}P{(Y\mid X_{n})}P(X_n)$

where the $2$nd equality holds by definition. Don't forget $P(Y|X)P(X) = P(Y \cap X)$

So,

$P(Y) = P(Y \cap X) + P(Y \cap X^c)$

is basically the law of total probability in the case where $n = 2$, since $X$ and $X^c$ form a partition of $Y$. Of course I'm assuming some things are nicely chosen (Like your $P(X) \neq 0$), but that's the gist of it.

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do you mean equation not inequality? –  TooTone Feb 26 '14 at 21:12
@TooTone I meant equality. thanks for pointing that out, i edited it. –  Tyler Feb 26 '14 at 21:23
Yes. That's the law of total probability. However, the denominator of the equation 2.13 (far right expression), only have conditioned to x'. Why is it expressed like that? The other partition (x) was not used in the formula. –  Xegara Feb 27 '14 at 15:21
That's a good point. Apparently the author isn't using ' to signify a set complement, they are just using it to differentiate between x' and x. I think it's really just a case of [extremely] poorly chosen notation on the author's part. Notice in my first equation I changed my summation to be indexed by n instead of to be indexed by x' –  Tyler Feb 27 '14 at 16:04
If that's the case, then my confusion is cleared. Thank you. –  Xegara Feb 27 '14 at 16:26