Let $X$ and $Y$ be continuous random variables with $Y \sim \mathcal{N}(\mu_Y,\sigma_Y^2)$.
I want to approximate $E[g(YX)]$ where $g(x) = e^{-x^2}$.
Specifically, my plan is to use a Taylor expansion of $Y$ at $\mu_Y$, which I expect should be accurate when the variance $\sigma_Y^2$ is small.
If we had $E[g(Y)]$ instead of $E[g(YX)]$ we could use the Taylor expansion directly to get \begin{align} E[g(Y)] &\approx g(\mu_Y) + \frac{\sigma_Y^2}{2}g''(\mu_Y). \end{align}
But the $X$ in $E[g(YX)]$ complicates matters. I made an attempt at it and obtained the following expression: \begin{align} E[g(YX)] & \approx E[g(\mu_Y X)] + \frac{\sigma_Y^2}{2}E[g''(\mu_Y X)]. \end{align} This seems to match up well intuitively with what we have for the simpler expecation $E[g(Y)]$ above. But I'm not certain whether its correct. Here's what I did (note that I'm still just getting to grips with conditional probability): \begin{align} E[g(YX)] & = E[E[g(YX)|X]] \\ & = \int E[g(YX)|X=x] p_X(x) dx \\ \end{align} Since the expectation is conditioned on $x$, I treat the $X$ in $g(XY)$ like a constant and Taylor expand with respect to $Y$. That is, I define $h(y;x) = e^{- y^2 x^2}$, so that $h(y;x) = g(yx)$, and then write: \begin{align} g(Yx) = h(Y;x) &\approx h(\mu_Y;x) + \frac{h''(\mu_Y;x)}{2} (Y-\mu_Y)^2. \end{align} Taking the conditional expectation gives \begin{align} E[g(YX)|X=x] = E[h(Y;X)|X=x] &\approx E\bigg[h(\mu_Y;X) + \frac{h''(\mu_Y;X)}{2} (Y-\mu_Y)^2\bigg|X=x\bigg] \\ & = h(\mu_Y;x) + \frac{h''(\mu_Y;x)}{2} \sigma_Y^2. \end{align}
Then I substitute this expansion into the integral from earlier to get \begin{align} E[g(YX)] & \approx \int \bigg(h(\mu_Y;x) + \frac{h''(\mu_Y;x)}{2} \sigma_Y^2\bigg) p_X(x) dx \\ & = E[h(\mu_Y;X)] + \frac{\sigma_Y^2}{2}E\bigg[h''(\mu_Y;X)\bigg] \\ & = E[g(\mu_Y X)] + \frac{\sigma_Y^2}{2}E[g''(\mu_Y X)]. \end{align}
I'm not sure whether what I've done is valid, and if so, if I have fully justified what I have done.
If this is not valid, is it possible to correct it such that we can obtain an approximation $E[g(YX)]$ by way of taking a Taylor expansion of $Y$..or is what I am trying to fundamentally impossible?
The second-order Taylor expansion of $g(xy)$ around $y=\mu_Y$ is $$ g(xy)=g(x\mu_Y)+g'(x\mu_Y)x(y-\mu_Y)+\frac{1}{2}g''(x\mu_Y)x^2(y-\mu_Y)^2+R. $$ Thus, since $X$ and $Y$ are independent, $$ \mathsf{E}g(XY)=\mathsf{E}g(X\mu_Y)+\frac{\sigma_Y^2}{2}\mathsf{E}[g''(X\mu_Y)X^2]+\mathsf{E}R. $$