I found this formula written on a blackboard entering a math class
$$E[X|Y]= \frac{E[XY]Y}{E[Y^2]} $$
At first glance, this seemed too good to be true to me, but it could be due to my ignorance.
I have not seen this before and was wondering if this was true or under which conditions is it true?
Here's one case: $X,Y$ are bivariate Normal, each with mean $0$.
\begin{align*} E[X \mid Y] &= EX + \rho \frac{\sigma_X}{\sigma_Y}[Y - EY] \\ &= \frac{E[XY]}{EY^2 - [EY]^2}Y\\ &= \frac{E[XY]Y}{E[Y^2]}. \end{align*}
$E[X \mid Y]$ is a random variable that is a function of the random variable $Y$ (not $X$), say $g(Y)$ and it is the minimum mean-square error (MMSE) estimator of $X$ in terms of $Y$. That is, $g(\cdot)$ is the (measurable) function that minimizes the mean-square error $E[(X-h(Y))^2]$ over all choices of (measurable) functions $h(\cdot)$ that we might use to estimate $X$ in terms of $Y$. Thus, the property in question $$ E[X \mid Y] = \frac{E[XY]Y}{E[Y^2]} = \left(\frac{E[XY]}{E[Y^2]}\right)Y\tag{1}$$ holds whenever the function $g(\cdot)$ is just a scalar multiple of $Y$. Now, if we restrict the functions $h(y)$ to be linear functions (more correctly, affine functions) of the form $ay+b$, then the function $g_L(Y)$ that minimizes $E[(X-h(Y))^2]$ over all choices of linear functions $h(\cdot)$ is the linear MMSE estimator of $X$ in terms of $Y$ and is known to be $$E[X] + \left.\left.\rho\frac{\sigma_X}{\sigma_Y}\right(Y - E[Y]\right) = E[X] + \left.\left.\frac{\operatorname{cov}(X,Y)}{\operatorname{var}(Y)}\right(Y - E[Y]\right).\tag{2}$$ Thus, $(1)$ holds when the MMSE estimate of $X$ given $Y$ is proportional to $Y$ (is a linear estimate) and so is the same as the linear MMSE estimate. But, $(1)$ has no constant term as in $(2)$ and so it must be that $$E[X] = \frac{\operatorname{cov}(X,Y)}{\operatorname{var}(Y)}E[Y]\tag{3}$$ to make the constant term in $(2)$ vanish. But more is needed since whenever $(3)$ holds, $(2)$ becomes $$\left(\frac{\operatorname{cov}(X,Y)}{\operatorname{var}(Y)}\right)Y = \left(\frac{E[XY]-E[X]E[Y]}{E[Y^2]-(E[Y])^2}\right)Y \neq \left(\frac{E[XY]}{E[Y^2]}\right)Y $$ unless $E[X]=E[Y]=0$ (and so $(3)$ continues to be satisfied). In short, $(1)$ holds whenever
and
A standard example when these conditions hold is when $X$ and $Y$ are zero-mean random variables enjoying a bivariate normal distribution as in @Taylor's answer, but this is by no means the only possible example. Consider $(X,Y)$ uniformly distributed on the interior or the parallelogram with vertices $(-1,-1)$, $(0,-1)$, $(1,1)$ and $(0,1)$. It is easily verified that $E[X] = E[Y] = 0$. Note also that for any $y \in (-1,1)$, the conditional distribution of $X$ given that $Y = y$ is $\mathcal U\left(-\frac 12 + \frac y2, \frac 12 + \frac y2\right)$ and so $E[X\mid Y] = Y$. Verification that $E[XY] = E[Y^2]$ is left as an exercise for the skeptical reader.
