# Covariance, and the Taylor expansion for the expected value of a linear function of random variables

Suppose I have two correlated random variables $$X$$ and $$Y$$ and am interested in the quantity $$\theta = U(X) - U(Y)\ ,$$ where U is a smooth function. I am trying to determine the correct Taylor expansion of $$\mathbb{E}[\theta]$$ to order $$\sigma_i^2$$. Using the linearity of the expected value, I can write \begin{align} \mathbb{E}[\theta] &= \mathbb{E}[U(X)] - \mathbb{E}[U(Y)]\\ &\approx \left( U(\mu_X) + \frac{U''(\mu_X)}{2}\sigma_X^2 \right) - \left( U(\mu_Y) + \frac{U''(\mu_Y)}{2}\sigma_Y^2 \right)\ . \end{align}

Is this correct? If so, how and why does $$Cov[X,Y]$$ not enter the picture?

To be concrete, I am comparing this result to the formula for the Taylor expansion of a general non-linear function of two variables: $$\mathbb{E}[\theta(X, Y)] = \theta(\mu_X, \mu_Y) + \frac{1}{2}\left[ \partial_{X}^2 \theta \sigma_{X}^2 + 2\partial_X\partial_Y\theta Cov[X, Y] + \partial_{Y}^2 \theta \sigma^2_{Y} \right]\ ;$$ substituting $$\theta = U(X) - U(Y)$$ into this formula gives the result above if and only if $$\partial_X\partial_Y \theta = 0$$. But can this be true if $$X$$ and $$Y$$ are correlated? The answer seems to be in the affirmative, but I am having trouble justifying this to myself. Any intuition would be appreciated.

Expectation is linear, and therefore $$\mathbb{E}[\theta] = \mathbb{E}[U(X)]-\mathbb{E}[U(Y)]$$ regardless of whether $$U,Y$$ are independent or not. They could be arbitrarily correlated, this would not change the expectation.
Note that, to answer your last specific point, with your $$\theta$$ you do have $$\partial_X\partial_Y \theta = \partial_X\partial_Y U(X) - \partial_X\partial_Y U(Y) = 0 - \partial_X\partial_Y U(Y) = 0$$ since $$\partial_Y U(Y)$$ does not depend on $$X$$.