How to find these ensemble means

Question

I am trying to calculate the ensemble mean of two correlated RVs. I am not sure if I am doing it correctly or not. I am using a general formula, if X is a RV then $X=E[X]+X'$, where $E[X]$ is the mean of $X$ and $X'$ is the fluctuation, moreover $E[X']=0$. I do not have any given densities associated with these. I have an expression which contains three ensemble means. The calculation I have done so far is as follows:

1. $E[\frac{dX'}{dt}]$ I have set this to zero, based on the reason that since $E[X']=0$, means average fluctuation of $X$ is centered around zero and if it is symmetric around zero then the on average, rate of change will balance out to zero $\implies E[\frac{dX'}{dt}]=0$. I am not sure if the reasoning is correct.

2. $E[X'Y']$. Since $X$ and $Y$ are correlated then I utilized the covariance formula. $cov(X',Y')=E[X'Y']-E[X']E[Y']=E[X'Y']$. Now I don't know what to do next? $X,Y$ are correlated but their densities are not known.

3. $E[X'^2Y']$. I used the same as above, to reach $E[X'^2Y']=cov(X'^2,Y')$.

Lastly, I would like to know if $X$ & $Y$ are not independent (i.e., correlated) then does the "mean operator" $E$ act linearly? i.e., $E[X+Y]=E[X]+E[Y]$?

Answer 1

1. No, in general you cannot interchange the differentiation with expectation. In order for $$E[\frac{dX'(t)}{dt}]=\frac{dE[X'(t)]}{dt}$$ to hold, you have to impose extra conditions like dominated convergence .

2 is fine, but I have no idea what you want to do next. 3 is fine if $E[Y']=0$.

Lastly, it is true that $E[X+Y]=E[X]+E[Y]$, which is a very basic property of the expectation. You ought to check the facts more carefully.