I do not have a background of Maths, but I met an equation when I read a paper,
$$E(X\mid Y)-E(X)=(Y-E(Y))\frac{\operatorname{cov}(X,Y)}{E(Y^2)}$$
Could anyone tell me how to prove this? I have tried a lot but failed...
Sorry that, I did not put all the original info from the paper.
This is the original context in the paper "Endogenous versus exogenous shocks in systems with memory":
"To quantify the response in such case, we recall a standard result of stochastic processes with finite variance and covariance that the expectation of some process $X(t)$ conditioned on some variable $Y$ taking a specific value $A_0$ is given by [22]
$$E[X(t)\mid Y=A_0]-E[X(t)]=(A_0-E[Y])\frac{\operatorname{Cov}(X(t),Y)}{E[Y^2]}$$"
Citation[22] is the book "Limit Theorems for Stochastic Processes" which is too huge for me to find this out.
It's certainly not true in general. Some context might help. Where did you get this?
EDIT: For example, consider a case where $X = Y^2$, $E[Y] = 0$, $E[Y^2] = 1$. Then the equation says $$ A_0^2 - 1 = A_0 E[Y^3]$$ which is true for at most two possible values of $A_0$.
Perhaps this is supposed to be for Gaussian processes?