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In the Conditional expectation Wikipedia page, in the Classical definition: Conditional expectation with respect to an event section, there is a $P(dx | H)$ notation, whose meaning I don't understand:


enter image description here


According to the explanation below it should be $P(dx\cap H)/P(H)$, but the $dx$ part is still confusing.

What I want is a very clear, purely real analysis notation, i.e., one that specifies the domain over which to integrate, the corresponding $\sigma$-algebra, the measure and the integrand function, instead of the sorta elusive probabilistic formulation. It seems the integral domain is $\mathcal X$, but neither the $\sigma$-algebra nor the measure is readily apparent to me. Any help?

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  • $\begingroup$ Sorry for the blurred image. I always have this problem uploading images on my iPhone. This can't be helped. $\endgroup$ – Vim Sep 21 '16 at 15:34
  • $\begingroup$ Its a Radon Nykodym derivative. $\endgroup$ – Pedro Tamaroff Sep 21 '16 at 15:38
  • $\begingroup$ $E(X|H) = \int_X x f_{X|H}(x) dx$ where $f_{X|H}(x)$ is the pdf of $X | H$ and since $f_{X|H}(x) = \frac{d}{dx}[ P(X \le x|H)]$ you have $f_{X|H}(x) dx = d [P(X \le x|H)] $, and $P(dx | H)$ would be a short hand for that $\endgroup$ – reuns Sep 21 '16 at 15:39
  • $\begingroup$ @user1952009 well, thanks. But I'm still struggling with the basic concept of conditional expectation. Could you explain without it? $\endgroup$ – Vim Sep 21 '16 at 15:39
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    $\begingroup$ The notation is an abomination indeed, confusing measures on the source space $\Omega$ and on the image space $\mathbb R$. Correct formulas would be $$E(X\mid H)=\int_\Omega X(\omega)P(d\omega\mid H)$$ or $$E(X\mid H)=\int_\mathbb R x\mu_H(dx)$$ where $\mu_H$ would be the distribution of $X$ conditionallly on $H$, that is, a measure on $\mathbb R$. Since this conditional distribution would have to be defined carefully, I am not sure the WP formula explains anything at all. $\endgroup$ – Did Sep 21 '16 at 19:15
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Implicit in that explanation is 1) some state space $(\cal{X},\mathscr{X})$ in which $X$ takes values, and 2) that $P(\cdot\mid H)=P(\cdot,H)$ is a probability measure on $(\cal{X},\mathscr{X})$. The existence of a $P(\cdot,H)$ satisfying 2) and also satisfying the relation mentioned toward the end of the screenshot $P(A,H)=P(A|H)=P(A\cap H)/P(H)$ (also implicit is $A\in\mathscr{X}$) amounts to the existence of a regular conditional probability distribution for $X$ on $(\Omega,\mathcal{F},P)$, which will hold, e.g., when $X$ is real-valued. As pointed out in the comment below, when we pass to the more everyday bayesian formula definition of event $A$ conditional on event $B$ we actually want to write $P(X^{-1}(A)\cap H)/P(H)$

"What I want is a very clear, purely real analysis notation, i.e., one that specifies the domain over which to integrate," --> $\cal X$

the corresponding σσ-algebra, --> $\mathscr X$

the measure --> $P(\cdot,H)$

and the integrand function" --> the identity

Maybe some of the confusion is also due to the notation of the integrator $P(dx,H)$. This just means integrate with dummy variable $x$ using the measure $P(\cdot,H)$, e.g., for lebesgue measure I might write $\int_0^\pi cos(x)\lambda(dx)$ for $\int_0^\pi cos(x)dx$.

Another possible point of confusion: using the same letter $P$ to refer to the underlying measure on the probability space $(\Omega,\cal{F},P)$ and the rcpd $P(\cdot | \cdot)$.

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  • $\begingroup$ If $P(\ , H)$ is a measure on $\mathcal X$, then $P(A,H)=P(A\cap H)/P(H)$ is absurd since $A$ should be a subset of the source space $\Omega$ for $P(A\cap H)$ to make sense. (Not your fault, the absurdity is in the WP text.) $\endgroup$ – Did Sep 21 '16 at 19:19
  • $\begingroup$ @Did yeah, like the wikipedia article I muddle in my head the pushforward with the main probability. I will clarify shortly. $\endgroup$ – snarfblaat Sep 21 '16 at 19:48
  • $\begingroup$ Your edit misses the point, please refer to my first comment (or to my comment on main). As a result, this post is (still) misleading. $\endgroup$ – Did Sep 22 '16 at 5:37
  • $\begingroup$ @Did I interpret your comment to be that it is absurd to intersect subsets of different spaces, so I write $P(A,H)=P(X^{-1}(A)\cap H)/P(H)$ to avoid that absurdity. What remains misleading? $\endgroup$ – snarfblaat Sep 22 '16 at 18:33
  • $\begingroup$ The part "when we pass to the more everyday bayesian formula definition of event A conditional on event B" refers to nothing I can understand. The mention of "the existence of a regular conditional probability distribution for X" is offtopic as well. Recall that H is just an event (with positive probability), not a sigma-algebra. $\endgroup$ – Did Sep 22 '16 at 19:06

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