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Is there a satisfactory notation for marginal distributions? This would be one that clearly distinguishes the marginal distribution of $X$ relative to the joint distribution $(X,Y)$ from $X$ as it would be envisaged had $Y$ not been introduced. One option might be $\mathbb{E}_Y(X|Y)$, but this seems to go beyond the basic definition of a marginal distribution.

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    $\begingroup$ I usually just use $p[x]$, i.e. there is no need to reference $y$ (or any other "nuisance variable" that may have been omitted from the analysis!). Is there a particular context you are interested in? (Also, did you mean $\mathbb{E}_y\big[p[x|y]\big]$?) $\endgroup$ – GeoMatt22 Jan 4 '17 at 4:28
  • $\begingroup$ If (X,Y) is a continuous bivariate distribution then the marginal for X denoted f is given for each x f(x) = integral of g(x,y) dy over the range of values of y. So is integrated out and similarly h(y) is defined by taking g(x,y) and integrating out x. $\endgroup$ – Michael Chernick Jan 4 '17 at 5:05
  • $\begingroup$ This is probably known by the OP and it illustrates why GeoMatt22 is exactly right. Y can be viewed as a nuisance variable that is removed by integration to give the marginal for X. Likewise for Y, Xis a nuisance variable that ids integrated out to get the marginal for Y. $\endgroup$ – Michael Chernick Jan 4 '17 at 5:15
  • $\begingroup$ This may not be standard but I suggest that the marginal for X could be given the notation g(x,.) and the Y marginal by g(.,y) This is at least analogous to the notation for marginal means in ANOVA. $\endgroup$ – Michael Chernick Jan 4 '17 at 5:19
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As pointed out by GeoMatt2, there is no need or even no meaning for a special notation.

"...clearly distinguishes the marginal distribution of $X$ relative to the joint distribution [of] $(X,Y)$ from $X$ as it would be envisaged had $Y$ not been introduced."

There is indeed some confusion in this question in that the marginal of $X$ is the marginal of $X$, no matter how it is derived: $$p(x)=\int_\mathfrak{Y} p(x,y)\,\text{d}y=\int_\mathfrak{Y\times Z} p(x,y,z)\,\text{d}y\text{d}z=...$$for any completion mechanism one can dream of.

Introducing an extra-notation thus does not make sense when the function is the same.

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    $\begingroup$ Thank you. I seem to have become confused as a result of being confused about adjustment for confounding. $\endgroup$ – Helmut Jan 4 '17 at 7:41
  • $\begingroup$ Didn't my comments from last night say the same thing? I was expecting that maybe you would want me to write up a shorthand formula for marginalization. Clearly there is nothing standard and the idea that it just meant integrating out X to get the marginal fo Y and vice versa was explained well before Xi'an gave this answer. $\endgroup$ – Michael Chernick Jan 4 '17 at 12:57
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In my comments I amplified on the comment of GeoMatt22. The point was made by both of us that the marginal distribution for X bore no relationship to Y because Y was integrated out. Prior to Xi'an giving his answer I suggested that I would like to summarize my comments in an answer because I did have a rational choice for a notation although it may not be standard. Xi'an answered before the OP answered my comment. I am giving this answer because I am the only one so far to directly address what would be a rational notation for a marginal distribution for X. Here is a summary of the comments and the previous answer:

  1. GeoMatt22 rejected any notation that included Y conditionally. He made the important point that Y is a nuisance variable and said that he just uses p(x). He probably didn't make it a question because his choice is not standard.

  2. I then made three comments elaborating on GeoMatt22. Since my third comment included a notation that may not be standard but gets around the problem with Y not being involved in the marginal for X. Please look at the comments above before judging my answer.

  3. Xi'an enters an answer that acknowledges the key point made by GeoMatt22 and then expresses the point differently that a marginal integrates out the nuisance variable(s). He does not suggest a notation because as he says it doesn't make sense.

  4. The OP accepts Xi'an's answer and gives a comment as to how he became confused but does not seem to have read my comments and possibly not GeoMatt22's either.

Now let me explain my answer illustrating the ANOVA analogy. When you do an ANOVA marginal sample averages are given a notation. To illustrate In the One-way ANOVA the observations are denoted as X_ij where i represents the ith observation from the jth group where i runs from 1 to n_j for a given j and j from 1 to p (n_j observations in group j with p groups).

So the notation X_i. represents the average of the observations for fixed i where the group effects are averaged out.

Likewise X_.j denotes for the jth group the average of the X_ij where each of the n observations for the jth group are averaged. So X_i. = set of (X_ij) for fixed i averaged over all j from 1 to p and X_.j is the set of (X_ij) for fixed j averaged over all i in the jth group.

So by analogy for f(x,y) denoting the joint density at X=x and Y=y let f(x, .) represent the marginal density at X=x with Y integrated out and f(.,y) the marginal density for Y at Y=y with X integrated out.

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    $\begingroup$ While the notation links to the ANOVA model, using $f(x,.)$ for the marginal density is confusing since, in mathematics, $a(.)$ denotes the function $a:\,x\to a(x)$ and thus $f(x,.)$ would denote the function $y\to f(x,y)$, that is, the joint density as a function of $y$ for a fixed value of $x$. This is the opposite of the intended meaning. $\endgroup$ – Xi'an Jan 4 '17 at 17:13
  • $\begingroup$ @Xi'an The notation is a stretch. I just thought it was an interesting idea that statisticians could relate to. I don't really think it will be adopted but it was important to make it clear to the OP that conditioning on Y didn't make sense. I think what is currently done is exactly what GeoMatt22 does completely eliminate the second argument. I guess you can see that I was a little miffed at the OP because he ignored or overlooked my comments and my question to him. No knock against your answer.It was as good as any and it did gett the idea across to the OP. $\endgroup$ – Michael Chernick Jan 4 '17 at 21:28
  • $\begingroup$ I suspect the central part of the question was to distinguish between what the OP believed were two separate definitions of the marginal, although they could not differ. Defining the marginal by integrating out an auxiliary variable did not seem to cause a problem to the OP. $\endgroup$ – Xi'an Jan 4 '17 at 21:38
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    $\begingroup$ Sorry if I seemed unresponsive, but to me @Xi'an's answer addressed the "root cause" of my question in that my question was based on a misconception/confusion. Thanks. $\endgroup$ – Helmut Jan 4 '17 at 23:21
  • $\begingroup$ @Helmut Okay. I got a little carried away because I thought my choice of notation was clever. But it is clearly not important as you can see from my above discussion with Xi'an. If your confusion was your main concern GeoMatt22 and I essentially addressed that in comments before Xi'an even wrote up his answer. You could check the history. What was it in his answer that rang true for you that wasn't already in our comments? $\endgroup$ – Michael Chernick Jan 5 '17 at 7:46

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