Does the integral below have a specific interpretation in statistics? It looks like the marginal expectation of y, without integrating over x. $$ \int_{y=-\infty}^{y=\infty} y f(x,y) dy$$
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Suppose $X,Y$ are cts rvs with joint density $f_{X,Y}(x,y)$. Then the conditional expectation of $X$ given $Y$, denoted as $E[X|Y=y]$, is defined as
\begin{align} E[X|Y=y] & = \int_{-\infty}^{\infty} x f_{X|Y}(x,y) dx \\ & = \int_{-\infty}^{\infty} x \frac{f_{X,Y}(x,y)}{f_Y(y)} dx \\ & = \frac{\int_{-\infty}^{\infty} x f_{X,Y}(x,y)dx}{f_Y(y)}, \end{align}
and this is defined for $f_Y(y) \neq 0$. Re-arranging when $f_Y(y) \neq 0$, we have
$$\int_{-\infty}^{\infty} x f_{X,Y}(x,y)dx = E[X|Y=y]f_Y(y).$$
So I guess you can think of that integral as a 'pre-normalized' conditional expectation. However, perhaps a more intuitive interpretation can be given. In particular, if you let $Y$ be discrete, then we have
$$\int_{-\infty}^{\infty} x f_{X,Y}(x,y)dx = E[X|Y=y]P(Y=y),$$
which you can think of the average effect of $X$ for $Y=y$ weighted by those with $Y=y$.
Note: I flipped $X$ and $Y$ relative to your question in my response, but this should not cause any confusion hopefully.
