# Ensemble mean of a fraction

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.

• 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 Commented Sep 8, 2023 at 11:18
• @Ggjj11 Thanks. I don't have any simulation results or histogram, however I do have the equation relating these variables: $Y=ae^{\alpha t}/(1+ae^{\alpha t}\int\limits_0^tXe^{a\tau}d\tau)$.
– AtoZ
Commented Sep 8, 2023 at 12:26
• Because there is no general formula, please state the problem you actually have rather than some abstract generalization of it.
– whuber
Commented Sep 8, 2023 at 13:01
• And that's exactly the problem: to get beyond the definition you need a particular distribution for $(X,Y).$
– whuber
Commented Sep 8, 2023 at 17:22
• 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.
– whuber
Commented Sep 9, 2023 at 17:14