Talk:Conditional expectation

message: This article has several significant mathematical errors.  In the 'Conditioning as factorization' section, X must take values in a field so that the integral makes sense at all.  And X and Y do not need to be both U-valued.  In fact, U doesn't have to be a field, and so the current definition is broken.  I suggest making X real valued, leaving Y U-valued, and patching things from there.  —Preceding unsigned comment added by 140.247.149.155 (talk) 14:19, 14 February 2009 (UTC)
replies: text: I have tried to fix this error. --Bdmy (talk) 16:27, 14 February 2009 (UTC)
replies: text: Looks good, thanks.  —Preceding unsigned comment added by 140.247.142.164 (talk) 06:05, 16 February 2009 (UTC)
replies: text: The section 'Conditioning as factorization' is still really problematic. Defining E(X|Y) as a mapping over U is wrong, it is a random variable on the probability space (Omega,F,P). This section should simply mention that there exists a function g over U such that E(X|Y) = g(Y), the latter being the composition of g with Y, thus defined over Omega. Note that this is currently redundant with the previous section. 132.169.4.223 (talk) 14:53, 28 June 2017 (UTC)
replies: text: Yes. See "#section 4.2 error?" below. Boris Tsirelson (talk) 18:54, 28 June 2017 (UTC)

text: Right! Corrected. Thank you. Boris Tsirelson (talk) 20:56, 2 November 2013 (UTC)
replies: text:  Thank you. 80.98.239.192 (talk) 22:47, 2 November 2013 (UTC)
thread_title: Errors

message: Does anybody agree to put the fact that conditional expectation is the best regression in the mean square error sense? --Memming (talk) 15:51, 20 September 2008 (UTC)
replies: text: This has been done since 2008. AVM2019 (talk) 13:58, 23 May 2022 (UTC)
thread_title: {{strike|Link to regression}}

message: Um, this has to be one of the worst Wikipedia pages I've ever read. It is written like a probability theory lecture or a textbook, not an encyclopedia article. For example, "In order to handle the general case, we need more powerful mathematical machinery." has no place here. I've studied significant amounts of math and statistics, and it's somewhat hard for even me to read, I'd imagine it is completely inaccessible to a layperson. Compare, for example, to the article on conditional probability. And at the same time, it attempts to explain things like discrete random variable right there. A complete (ok, you can keep the first paragraph) rewrite is needed.Zalle 18:41, 26 March 2007 (UTC)
replies: text: I agree; this page is extremely arcane, and could use a major rewrite.  Cazort (talk) 17:59, 30 March 2008 (UTC)

text: Definitely it needs improvement, but the early section titled "special cases" shouls be easy to read for anyone who's had "significant amounts of math and statistics".  But the topic of the article necessarily requires that at least some of what is said will not be accessible to the "layperson". Michael Hardy 19:21, 26 March 2007 (UTC)
replies: text:  And the first sentence really should be accessible to anyone (except those who don't know what conditional probability distributions are, and it's not appropriate to ask that this be made accessible to those people, since "conditional probability distribution" is quite naturally a prerequisite to this topic), and for some purposes, says most of what need to be said. Michael Hardy 19:24, 26 March 2007 (UTC)

text:  Well, looking at your "revert" I think you've failed to comprehend the difference between a mathematics textbook and an encyclopedia.  "It can be shown that" is nothing but silly mathematical jargon and does not "signify that there is more to this argument". As this is an encyclopedia, it is by definition true (at least in an ideal situation) that all the claims contained can be "shown" to be true, and it is obvious without stating it that such a proof exists. I'd argue and edit more, but apparently this is a pet project so I don't think it's really worth the bother. Have fun, good luck.88.112.25.211 21:49, 26 March 2007 (UTC)
thread_title: One of the worst

message: This page needs substantial reworking. 
* Motivation involving finite probability spaces 
* Intuitive generalizations
* CLear statements of abstract framework general theorems and the general probability framework
* Cond Exp as factorization (important for defining sufficient statististics)  
* References.
If nobody objects, I'll do it in the next few hours CSTAR 15:45, 9 May 2004 (UTC)
replies: text: The conditional expected value E( X | Y ) is a random variable in its own right, whose value depends on the value of Y.  Notice that the conditional expected value of X given the event  Y = y is a function of y (this is where adherence to the conventional rigidly case-sensitive notation of probability theory becomes important!).  If we write E( X | Y = y) = g(y) then the random variable E( X | Y ) is just g(Y). Similar comments apply to the conditional variance.
Charles, is that the intuitive explanation that went by you? Michael Hardy 00:01, 9 Oct 2004 (UTC)

text: Is the previous paragraph an example of a clear explanation? Is it too impolitic to say it doesn't seem to me to be hardly an improvement?  I'm also curious as to what you would point to as being more of an Augean stable in wikipedia, though I do agree that there are many, many articles which I think fit this bill.CSTAR 00:28, 9 Oct 2004 (UTC)
thread_title: Reworking suggested

message: Does anybody else feel as if the table of contents is too down in the article, and some more splitting in sections could be done at the top? I don't know what is a good way of splitting it myself. Oleg Alexandrov 20:07, 14 Apr 2005 (UTC)
replies: text: Yeah I agree; but don't look at me for changes..CSTAR 20:44, 14 Apr 2005 (UTC)
thread_title:  Split into sections? 

message: ===Can somebody provide the profes for conditional expectation is contractions? or should the reference for the profes be provided? I can add a good profe for Jensen's Inequality later. —Preceding unsigned comment added by Pondyyuan (talk • contribs) 18:09, 31 December 2006
:  Please describe what statement you want to prove regarding contractions in more detail. Jmath666 18:32, 28 March 2007 (UTC)
replies: text: Is not contraction an obvious application of Jensen's inequality for f(x)=|x|^s? (Unconditional) Jensen's inequality has already some proof on Jensen%27s_inequality#Proofs. 22:59, 2 November 2013 (UTC)
thread_title: Proofs

message: This article is pretty good, surely better than what I could find in any book I looked and I found it of great help when I needed to clarify this stuff. Compared with the treatment in the classical books, such as Feller, Varadhan, Levy,.. I found it really lucid. I ended up writing notes for myself and few others, which are hopefully even more lucid and some may find more satisfactory. Any comments welcome.
I plan on merging the notes with the article in future. For now, my original lives in LaTeX so any edits will be overwritten next time when I run the translator. Jmath666 22:44, 27 March 2007 (UTC)
thread_title: Pretty good, proposal to make it even better

message: The diagram 
   -------- X --------> 
 Ω                      R
   --Y--> U --E(X|Y)-->
you draw is not correct, because also E(X|Y): Ω --> R. It should look like:
   -------- X --------> 
 Ω                      R
   ---Y---> U ----g--->
   ---E(X|Y) = g(Y) -->
Sorry, wasn't logged in.Nijdam 11:51, 25 April 2007 (UTC)
thread_title: Composition

message: properties of conditional expectation should be list in the article. We can find them in any textbook. such as E(X|X)=?, tower rule, independent rule, ... Jackzhp (talk) 20:46, 11 March 2008 (UTC)
thread_title: properties

message: I would like to request someone adding material on using conditional expectation to "improve" estimators (like in the Rao-Blackwell theorem) to this page; I would add the material, except that I feel I don't understand it 100%.  Cazort (talk) 17:58, 30 March 2008 (UTC)
thread_title: Relation to estimators

message: Since conditional probability is a special case of conditional expectation, it will be good to treat the conditional probability in this article, rather than an independent article. Jackzhp (talk) 19:56, 12 March 2011 (UTC)
thread_title: relation with Conditional probability

message: We have defined condition expectation on a sigma field and on a (measurable) set, the latter one is very messy. Conditioning on a random variable has not been defined yet.
As for conditioning on a set, let's show  P(A|B)=P(AB)/P(B)) satisfies the conditional expectation definition. Here, the sigma field is \left\{ \emptyset,\Omega,B,\bar{B}\right\}, so we have to show 4 equalities. Can someone please elaborate this? Jackzhp (talk) 16:10, 12 March 2011 (UTC)
thread_title: Different Conditioning

message: Is the formal definition given actually correct?  As stated, it says that that E[X|\Beta] = X a.e.  I'd imagine you should explicitly state what the who probability measures are, and clarify that E is a Radon–Nikodym derivative times the random variable.  —Preceding unsigned comment added by 128.176.122.34 (talk) 11:52, 8 May 2011 (UTC)
thread_title:  Formal definition

message: Regretfully, an incorrect paragraph was introduced by User:3mta3 on 14:55, 14 March 2009, and is unnoticed during almost 6 years. In the end of Section "Calculation" we see:
:: \operatorname{E} (X | Y=y) \operatorname{P}(Y=y) = \sum_{x \in \mathcal{X}} x \ \operatorname{P}(X=x,Y=y), 
:and although this is trivial for individual values of y (since both sides are zero), it should hold for any measurable subset B of the domain of Y that:
:: \int_B \operatorname{E} (X | Y=y) \operatorname{P}(Y=y) \ \operatorname{d}y = \int_B \sum_{x \in \mathcal{X}} x \ \operatorname{P}(X=x,Y=y) \ \operatorname{d}y. 
This is a nonsense. Just "Since both sides are zero" in the former, the integrands are zero in the latter. Thus, we still have 0=0. The author of this paragraph believes naively, that integration is able to gather a continuum of zeros into a positive number. I understand his intuition, but no, it does not work this way. This formulation could be acceptable in Wikipedia of 18-th century, but not now. :-) Boris Tsirelson (talk) 12:13, 15 January 2015 (UTC)
thread_title:  Error 

message: Should it not be [0,1] instead of (0,1)? After all, A's probability could be zero or one.  — Preceding unsigned comment added by Doubaer (talk • contribs) 08:11, 3 February 2015 (UTC)
replies: text: Really, both are a nonsense. The indicator is a random variable. That is, measurable w.r.t. the sigma-algebra given on the relevant probability space. Now fixed. Boris Tsirelson (talk) 12:02, 3 February 2015 (UTC)
thread_title:  Definition of conditional probability 

message: Several times, conditional expectation is discussed without assuming integrability of the given random variable (while in fact it is essential). Boris Tsirelson (talk) 16:12, 30 December 2015 (UTC)
thread_title:  Integrability forgotten 

message: "The definition of \operatorname{E}(X \mid \mathcal{H}) may resemble that of \operatorname{E}(X \mid H) for an event H but these are very different objects.  The former is a \mathcal{H}-measurable function \Omega \to \mathbb{R}^n, while the latter is an element of \mathbb{R}^n.  Evaluating the former at H yields the latter."
However, \operatorname{E}(X \mid \mathcal{H}) is a function \Omega \to \mathbb{R}^n (which is explicitely stated in this paragraph); how can one evaluate this function at an arbitrary event H\in\mathcal{H}, when \mathcal{H}\neq \Omega?  — Preceding unsigned comment added by 84.114.30.251 (talk) 21:10, 16 December 2018 (UTC)
replies: text: You are right. If the σ-algebra \mathcal{H} is finite (which is elementary case), then it corresponds to a finite partition of \Omega into parts of positive measure, and on every part H the function \operatorname{E}(X \mid \mathcal{H}) is constant (almost sure), and this constant is indeed \operatorname{E}(X \mid H). But generally that function is not constant on H, and \operatorname{E}(X \mid H) is rather its mean (average) value over H according to the given probability measure, provided that H\in\mathcal{H} and P(H)\ne0. Boris Tsirelson (talk) 22:28, 16 December 2018 (UTC)
replies: text:  This inconsistency has been resolved. "Evaluating the former at H yields the latter." means taking an integral of the former over H. AVM2019 (talk) 13:54, 23 May 2022 (UTC)
thread_title:  {{strike|Domain conflict}}