I need help with this question: Let X and Y be jointly normal random variables with $X, Y \sim N(0, 1)$ and correlation 1/2. Evaluate $\mathbb{E}\left[ e^{X} \ | \ Y=1 \right]$.
Here's where I'm stuck: we know that:
$$ \mathbb{E}\left[ e^{X} \ | \ Y=1 \right] = \int_{ -\infty }^{\infty} e^{x} f_{X|Y}(x|1) dx $$ And that: $$ f_{X|Y}(x|y) = \frac{f_{X,Y}(x, y)}{f_{Y}(y)} $$ $$ f_{X,Y}(x, y) = \frac{1}{\pi \sqrt{3}} e^{-\frac{2}{3}(x^2 + y^2 - xy)} $$ $$ f_{Y}(y) = \frac{1}{\sqrt{2\pi}} e^{-\frac{y^2}{2}} $$ Plugging all the numbers, I get to the following integral: $$ \mathbb{E}\left[ e^{X} \ | \ Y=1 \right] = \sqrt{ \frac{2}{3\pi} } \int_{ -\infty }^{\infty} e^{-\frac{1}{6} (4x^2 - 10x + 1) } dx $$ and I don't know what's the best approach to solve it (actually any approach). Wolfram tells me that this magically solves to $e^{7/8}$. Can anyone help?
We are told $X,Y\sim N(0,1)$ are jointly normal with correlation $\rho$. Observe that $X-\rho Y\sim N(0,1-\rho^2)$ and $Y$ are jointly normal and uncorrelated and hence independent. Thus,
$$\begin{align}E[e^X|Y=1]&=E[e^{X-\rho Y}e^{\rho Y}|Y=1]\\ &=E[e^{X-\rho Y}e^{\rho}|Y=1]\\ &=e^{\rho}E[e^{X-\rho Y}]\quad (\text{by independence})\\ &=e^{\rho}e^{\frac{1}{2}(1-\rho^2)}\quad (\text{using MGF})\\ \end{align}.$$
For $\rho=1/2$, this simplifies to the desired solution, $e^{7/8}.$