I'm struggling a little with this expansion:
Where $E$ is the expectation operator, $U$ is a function of $Y$ and $Z^~$ is a random variable. In the second passage why the expansion looks like what's written above? Shouldn't the function be expanded around the mean of $Z$? Maybe I'm missing something, any explanation will be greatly appreciated!
On
It seems that the taylor expansion has been made around $\tilde{Z}_0=0$. This might be the mean of $\tilde{Z}$.
Therfore $U(\tilde{Z}+Y)= U(\tilde{Z}_0+Y)+U_y(\tilde{Z}_0+Y)\cdot (\tilde{Z}-\tilde{Z}_0)+\frac{1}{2}\cdot U_{yy}(\tilde{Z}_0+Y)\cdot (\tilde{Z}-\tilde{Z}_0)^2+H(\tilde Z^3)$
$U(\tilde{Z}+0)= U(0+Y)+U_y(0+Y)\cdot (\tilde{Z}-0 )+\frac{1}{2}\cdot U_{yy}(0+Y)\cdot (\tilde{Z}-0)^2+H(\tilde Z^3)$
$U(\tilde{Z})= U(Y)+U_y(Y)\cdot (\tilde{Z})+\frac{1}{2}\cdot U_{yy}(Y)\cdot (\tilde{Z})^2+H(\tilde Z^3)$
With the expectation operator it is
$\mathbb E U(\tilde{Z})= \mathbb EU(Y)+U_y(Y)\cdot \mathbb E\tilde{Z}+\frac{1}{2}\cdot U_{yy}(Y)\cdot \mathbb E[(\tilde{Z})^2]+\mathbb EH(\tilde Z^3)$
And $Var(\tilde Z)=E[(\tilde{Z})^2]-E[(\tilde{Z})]^2$. Because of $E(\tilde{Z})=0$ we get
$\mathbb E U(\tilde{Z})=\mathbb EU(Y)+\frac{1}{2}\cdot U_{yy}(Y)\cdot Var(\tilde Z)+\mathbb EH(\tilde Z^3)$
Since Y is not a random variable it simplifies to
$\mathbb E U(\tilde{Z})= U(Y)+\frac{1}{2}\cdot U_{yy}(Y)\cdot Var(\tilde Z)+\mathbb EH(\tilde Z^3)$
Start with the Taylor expansion:
$$U(Y+\tilde Z)=U(Y)+\tilde ZU'(Y)+\frac 1 2 {\tilde Z}^2U''(Y)+H({\tilde Z}^3)$$
I take it from this approximation that $\tilde Z$ is small compared to $U$.
I also assume that $H({\tilde Z}^3)$ is saying "and expressions involving third powers and higher of $\tilde Z$" - also can be written as $O({\tilde Z}^3)$.
We then go on to find the expectation of this expression. I will take the $RHS$ in parts:
$E\left[U(Y)\right]=U(Y)$
$E\left[\tilde ZU'(Y)\right]=E\left[\tilde Z\right]\times U'(Y)=0\times U'(Y)=0$ - this relies on the mean of $\tilde Z$ being $0$, so I wonder if this is given somewhere in the question?
$E\left[\frac 1 2 {\tilde Z}^2U''(Y)\right]=E\left[\frac 1 2 {\tilde Z}^2\right]\times U''(Y)=\frac 1 2 \sigma_{\tilde Z}^2 \times U''(Y)$ - it appears that $\tilde Z$ is not a standardised normal distribution as then we would have $\sigma_{\tilde Z}^2=1$
$E\left[H({\tilde Z}^3)\right]$ just stays as it is - I expect that later on you can just say that this will be small enough to ignore...