I have a fairly basic question that relates to understanding a particular derivation.
I have the following function $Q(x) = E\left[I(F(x+\varepsilon)>c)\right]$, where $x \in R$, $\varepsilon \sim N(0,\sigma^2)$ for a known $\sigma>0$, $c$ is a scalar constant, $F(\cdot)$ is a known, bounded function, and $I(\cdot)$ is an indicator function.
This can be written as:
$E[I(F(x+\varepsilon)>c)] = \frac{1}{{\sqrt{2\pi}}\sigma} \int{I(F(x+\varepsilon)>c) \exp\left(-\frac{\varepsilon^2}{2\sigma^2} \right)}d\varepsilon $
Let $y = x + \varepsilon$ and rewrite the above as:
$E[I(F(x+\varepsilon)>c)] = \frac{1}{{\sqrt{2\pi}}\sigma} \int{I(F(y)>c) \exp\left(-\frac{(y-x)^2}{2\sigma^2} \right)}dy$
I am interested in calculating $\frac{\partial Q(x)}{\partial x}$. The derivative is:
$\frac{\partial E[I(F(x+\varepsilon)>c)]}{\partial x}= \frac{1}{{\sqrt{2\pi}}\sigma} \int{I(F(y)>c) \exp\left(-\frac{(y-x)^2}{2\sigma^2} \right)} \frac{(y-x)}{\sigma^2}dy \quad \quad $ (1)
Substituting back in again and rearranging slightly:
$\frac{\partial E[I(F(x+\varepsilon)>c)]}{\partial x} = \frac{1}{\sigma^2} E[I(F(x+\varepsilon)>c)\varepsilon]$
Finally, here are my two questions: (1) is the above derivation correct? (2) If so, I am trying to understand why in eq(1) I do not need to take the derivative of $I(F(y)>c)$ with respect to $x$. I know there has been a change of variables, but $y$ is obviously a function of $x$. Yet I was told that "once you do the substitution, $y=x+\varepsilon$, then y is just a dummy variable (i.e. an index) which runs from -infinity to infinity, and one cannot take the derivative in a dummy variable." I know you can't take the derivative of an indicator function, but somehow I can't get this clear in my head.
Thanks.
More broadly, in my actual problem, $x$ and $\varepsilon$ are high dimensional and I am trying to compute the derivative via Monte Carlo integration....