I can't come up with an efficient solution of the following problem.
Let $(X, Y)$ be a 2-dimensional random variable vector which follows the 2-dimensional Gaussian distribution, and its probabilistic density function is, \begin{align*} f(x, y|\ \rho) = \frac{1}{2\pi\sqrt{1-\rho^2}} \exp\left(-\frac{x^2-2\rho xy+y^2}{2(1-\rho^2)}\right) \end{align*} where $-\infty<x, y<\infty;\ |\rho|<1$.
Find the following conditional expectation, \begin{align*} E\left[Y^2\frac{\partial}{\partial \rho}\log{f(X, Y|\ \rho)\ \Big|\ X= x}\right] \end{align*}
My solution is straightforward. I calculated the inside of the big parentheses, used the additive formula of expectation, and then find 2, 3, and 4 dimensional probability moment of $Y$.
The problem is that my solution is time-consuming, and induces mistakes in calculation. The good properties of the score function may be useful, in my opinion, but I couldn't make use of them in the end.
Please tell me more efficient way of solving.
First note that $f(x,y;\rho) = f(y|x ; \rho) f(x;\rho)$, where \begin{equation} f(y|x;\rho) = \frac{1}{2 \pi \sqrt{1-\rho^2}} e^{-(y-\rho x )^2/2(1-\rho^2)}, \end{equation} and \begin{equation} f(x; \rho)=f(x) = \frac{1}{2 \pi} e^{-x^2/2} . \end{equation}
Note that $f(x)$ does not depend on $\rho$, therefore \begin{align} \frac{\partial \log f(x,y;\rho)}{\partial \rho} &=\frac{1}{f(x,y;\rho)}\frac{\partial f(x,y;\rho)}{\partial \rho}\\ &= \frac{f(x)}{f(x,y;\rho)} \frac{\partial f(y|x;\rho)}{\partial \rho}\\ &= \frac{1}{f(y|x;\rho)} \frac{\partial f(y|x;\rho)}{\partial \rho} \end{align}
Now, \begin{align} \mathbb{E}\left(Y^2 \frac{\partial \log f(x,y;\rho)}{\partial \rho} |X=x\right)&=\int_{-\infty}^\infty f(y|x;\rho) y^2 \frac{\partial \log f(x,y;\rho)}{\partial \rho} dy \\ &=\int_{-\infty}^\infty f(y|x;\rho) y^2 \frac{1}{f(y|x;\rho)} \frac{\partial f(y|x;\rho)}{\partial \rho} dy \\ &=\int_{-\infty}^\infty y^2 \frac{\partial f(y|x;\rho)}{\partial \rho} dy \\ &=\frac{\partial }{\partial \rho}\int_{-\infty}^\infty y^2 f(y|x;\rho) dy \\ &=\frac{\partial }{\partial \rho} \mathbb{E}\left(Y^2|X=x\right), \end{align}