Tell me more ×
Cross Validated is a question and answer site for statisticians, data analysts, data miners and data visualization experts. It's 100% free, no registration required.

In a multivariate context, that is with at least X or Y being a random vector, are there formulae or theorems that link (even remotely) the forward and inverse regression, $\text{E}(X|Y)$ and $\text{E}(Y|X)$ ? Or alternatively $\text{E}(X_i|Y)$ and $\text{E}(Y|X_i)$, where $X_i$ is a given component of $X$.

Similarly, can something be said about $\text{var}[\text{E}(X|Y)]$ and $\text{var}[\text{E}(Y|X)]$ ? Or alternatively $\text{var}[\text{E}(X_i|Y)]$ and $\text{var}[\text{E}(Y|X_i)]$, where $X_i$ is a given component of $X$.

Both $X$ and $Y$ are supposed to be random. The typical application case is that $X$ follows a chosen distribution (for instance a multigaussian) and $Y$ is a function of $X$. The link function is too complicated to be used or even unknown (it is typically a numerical model). The distribution of $X$, however, is controlled. Thus, one should formulate hypothesis on $X$ if needed.

*Note : this question is related to dimension reduction trhough the sliced inverse regression technique.

It is also related to sensitivity analysis, as $\text{var}[\text{E}(Y|X)]$ is the numerator of Sobol' index (surprisingly, the tag sensitivity-analysis does not exist -- if someone with sufficient reputation read this, he might want to create it).*

share|improve this question
Unless both $X$ and $Y$ are random, either $E(X|Y)$ or $E(Y|X)$ is essentially meaningless, so I take it that both are supposed to be random... Have you studied errors-in-variables models? They are used for studying $E(Y|X)$ regression and, by symmetry when both the response and the explanatory variables are random, thus also $E(X|Y)$. – MånsT May 9 '12 at 15:31
I added some precisions regarding your first remark. I will have a look at what you suggested. – Alfred M. May 9 '12 at 15:46
Yves is right and has answered your question. There is no relationship between E(Y|X) and E(X|Y) without specifying the joint distribution. – Michael Chernick May 22 '12 at 1:11
@Yves, could you fomulate your comment into an answer ? What about the variance of conditional expectations ? What minimal assumptions on the joint distributions would allow to state some resulsts ? – Alfred M. May 22 '12 at 5:50
@Alfred.M Thank you. Yet I have no more to write... except a few details about the example? – Yves May 22 '12 at 14:32
show 2 more comments

1 Answer

up vote 2 down vote accepted
+50

Note that $\mathbb{E}(X \vert Y)$ heavily depends on this joint distribution, which is not determinated by $\mathbb{E}(Y \vert X)$. Consider the simple case where $X$ and $Y$ are scalar r.vs and $Y=X+\varepsilon$ where $\varepsilon$ has mean zero and is independent of $X$. Depending on the distribution of $\varepsilon$, one can find examples where the function $\mathbb{E}(X \vert Y=y)$ is respectively increasing, constant or decreasing in $y$ for large $y$. Yet in all cases, $\mathbb{E}(Y \vert X=x) \equiv x$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.