# Link between forward and inverse regression ($\text{E}(X|Y)$ and $\text{E}(Y|X)$ ;$\text{var}[\text{E}(X|Y)]$ and $\text{var}[\text{E}(Y|X)]$)

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).*

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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
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, $\mathbb{E}(Y \vert X=x) \equiv x$.