Fabian Werner
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 Dec 18 comment Formal definition of conditional probability Yes, sorry, whenever I wrote r.c.p. I actually meant the more general r.c.d. (without fixing G). Now it becomes at least clear to me that $\mu$ (if defined on $\sigma(X)$) and $P^X$ if only evaluated on sets $A \in \sigma(X)$ coincide in the sense that $\mu = P^X(X^{-1}, X)$ and $P^X$ on $(A, x)$ is $A = X^{-1}(B) = X^{-1}(B')$ then $X \in B == X \in B'$ and thus $E[1_{X \in B}|\sigma(x)] = E[1_{X \in B'}|\sigma(x)]$ and factorizing this into $f_A \circ X$ yields $P^X = f_A(x)$. I.e. they coincide 'on' $\sigma(X)$. Dec 18 comment Formal definition of conditional probability The thing is that the case $\mathcal{G} = \sigma(X)$ is boring as one uses this often with $\mathcal{G} = \sigma(Y)$ from another random variable $Y$. Anyhow, thanks for the answer. Dec 18 comment Formal definition of conditional probability @StefanHansen Careful: There is an additional twist: $\mu$ does not accept sets from $\mathcal{F}$ but rather from the target sigma algebra $\mathcal{B}(\mathbb{R})$. Thats why I inserted that $X^{-1}(...)$ in the twist. Hmm... But if we can easily construct their $\mu$ from your $P^X$ then I have the feeling that your defn. should be more general (although as said, I can only show that $\mu = P^X(X^{-1}, X)$ if $\mathcal{G} = \sigma(X)$... weird... Dec 18 comment Formal definition of conditional probability @StefanHansen: sorry, $\mathcal{F}$ was an unclever special case. Generally, for example, in math.duke.edu/~rtd/PTE/PTE4_1.pdf 5.1.3 on p. 197 they define thr r.c.p. w.r.t. any sub-sigma-algebra $\mathcal{G}$ as a map $\mu : \Omega \times \mathcal{B}(\mathbb{R}) \to [0,1]$, i.e. going from the target sigma algebra times the original space to the reals while in your case, $P^X$ goes from the original sigma algebra times the target space... see the twist [original <-> target]? Does your definition coincide somehow with 5.1.3 in the reference? Dec 17 comment Conditional Probability… (?) I think (3) is answered as follows: in wwwmath.uni-muenster.de/statistik/lehre/SS13/WT/Skript/… Satz 8.16 they show that E(h(X,Y)|Y=y) = E(h(X,y)). Fix $\lambda$ and put $X_\lambda = \tilde{X}(\cdot, \lambda)$. In my case I want to compute $E[1_{X_\lambda \in \{x\}}|\Lambda=\lambda]$ and $1_{X_\lambda \in \{x\}} = 1_{\{x\} \times \mathbb{R}} \circ (X_\lambda, \Lambda)$, i.e. $h(x,y) = 1_{\{x\} \times \mathbb{R}}(x,y)$ and $E(h(X_\lambda, \lambda)) = E(1_{\{x\}}(X(\omega))) = P[X=x]$ Dec 17 comment Conditional Probability… (?) erm... but both things you mentioned are vital for my question... my questions are: (1) what does P mean? [i.e. does P(...) mean this and that?] (2) [at least partially]: Why are we allowed to insert $\lambda$ into the density? So its important to keep them separated... dont you think? Dec 17 asked Conditional Probability… (?) Dec 17 comment Formal definition of conditional probability @StefanHansen: A question concerning the definition of r.c.p.: All the other people do it like this: $\nu: \Omega \times \mathcal{B}(\mathbb{R}) \to \mathbb{R}$ and demand that $\nu(\cdot, B) = E[1_{X \in B}|\mathcal{F}]$ i.e. we should suspect that $\nu(\omega, B) = P^X(X^{-1}(B), X(\omega))$ but this does not work: when integrating over an arbitrary set $A$ then $1_A$ is not factorizable... Could you elaborate? Are these two different definitions of r.c.p.? Dec 10 answered Conditional Expectations: Minimal Square error? Dec 2 revised Conditional Expectations: Minimal Square error? added 1 character in body Dec 2 asked Conditional Expectations: Minimal Square error? Jun 2 accepted Translation:Bayes Classificator -> precise math? Jun 2 comment Translation:Bayes Classificator -> precise math? Ok, Im not happy with it but this seems close to the truth. Im gonna accept. May 30 comment Translation:Bayes Classificator -> precise math? In fact, you do, Writing Pr(...) means that a value <=1 must occur... please decide: do you want to use Pr(X=x|T) [then the value must be somewhere between 0 and 1] or do you want to use this obscure p (little p) instead of capital P? May 29 comment Translation:Bayes Classificator -> precise math? To be honest: this answer does not make me happy. I can understand the assumption of independence as an 'accepted' and natural one... To assume, however, that this strange function $x \mapsto P(X=x|T)$ is even measurable seems very unnatural for me. May 29 comment Translation:Bayes Classificator -> precise math? What do you mean by 'Pr(X|T) follows a Gaussian distribution'? I guess you mean that some function $f:someSet \to \mathbb{R}$ follows a Gaussian distribution. What function f do you mean? May 29 comment Translation:Bayes Classificator -> precise math? That is precisely the reason why I stick to the math stuff... I think they actually mean P(Z=C_k|X=x) by writing p(C_k|x)... May 29 comment Translation:Bayes Classificator -> precise math? @simonzack but in the beginning they write things like p(x|C) and so forth... this means p must be the usual P... May 29 comment Translation:Bayes Classificator -> precise math? 1) Ill check... I just dont feel well without it: if you say 'random variable' you say 'measurable' and if you say 'measurable' you must say which sigma algebra to take... 2) was it about the word 'normal'? I removed it... the question is about why P(X=single point) is not zero... May 29 revised Translation:Bayes Classificator -> precise math? added 56 characters in body