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I want to compute the ensemble mean of the term: $\frac{Y^2}{X}$ Both $X$ and $Y$ are random variables that are not independent. I want to compute $E[\frac{Y^2}{X}]$. I proceed as follows, (Using the formula for splitting a RV into mean and fluctuation $A=E[A]+A'$)

$\frac{Y^2}{X}=\frac{(E[Y]+Y')^2}{E[X]+X'}=\frac{\left((E[Y])^2+Y'^2+2E[Y]Y'\right)}{E[X]\left(1+\frac{X'}{E[X]}\right)}$

$~~~~~=\frac{1}{E[X]}\left((E[Y])^2+Y'^2+2E[Y]Y'\right)*\left(1-\frac{X'}{E[X]}\right)$

Using Taylor series expansion upto linear order. This means the above relation is true for $|\frac{X'}{E[X]}|<<1$

This implies

$\frac{Y^2}{X}=\frac{1}{E[X]}\left[(E[Y])^2-\frac{(E[Y])^2}{E[X]}X'+Y'^2-\frac{1}{E[X]}X'Y'^2+2E[Y]Y'-\frac{2E[Y]}{E[X]}Y'X' \right]$

$E[\frac{Y^2}{X}]=\frac{1}{E[X]}\left[(E[Y])^2-\frac{(E[Y])^2}{E[X]}E[X']+E[Y'^2]-\frac{1}{E[X]}E[X'Y'^2]+2E[Y]E[Y']-\frac{2E[Y]}{E[X]}E[Y'X'] \right]$

also noting that $E[A']=0$, the above leads to

$E[\frac{Y^2}{X}]=\frac{1}{E[X]}\left[(E[Y])^2+E[Y'^2]-\frac{1}{E[X]}E[X'Y'^2]-\frac{2E[Y]}{E[X]}E[Y'X'] \right]$

Is there a way to get rid of $E[..]$ terms on the R.H.S? I don't have the distributions corresponding to $Y$ and $X$, but there is a complicated equation relating these two variables. Also, $X$ and $Y$ have a negative covariance relationship.

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    $\begingroup$ Could you add some simulation results for a histogram Y^2/X where you indicate the sample mean of this and the analytical expression? Just out of curiosity to see the agreement in an explicit situation $\endgroup$
    – Ggjj11
    Commented Sep 8, 2023 at 11:18
  • $\begingroup$ @Ggjj11 Thanks. I don't have any simulation results or histogram, however I do have the equation relating these variables: $Y=ae^{\alpha t}/(1+ae^{\alpha t}\int\limits_0^tXe^{a\tau}d\tau)$. $\endgroup$
    – AtoZ
    Commented Sep 8, 2023 at 12:26
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    $\begingroup$ Because there is no general formula, please state the problem you actually have rather than some abstract generalization of it. $\endgroup$
    – whuber
    Commented Sep 8, 2023 at 13:01
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    $\begingroup$ And that's exactly the problem: to get beyond the definition you need a particular distribution for $(X,Y).$ $\endgroup$
    – whuber
    Commented Sep 8, 2023 at 17:22
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    $\begingroup$ It's hard to tell, because your expression for $Y$ suggests $X$ is not a random variable, but rather a stochastic process. Otherwise, you would just factor it out of the integral. $\endgroup$
    – whuber
    Commented Sep 9, 2023 at 17:14

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