I am trying to solve the following exercise:
$$\mathbb E[e^Y\textbf 1_{X<T}]$$
where,
- X and Y $\sim \mathcal N(0,\sigma)$, $\sigma = 0.5$
- $\operatorname{cor}(X,Y) = \rho =-0.98$
- $T=-3$
I solved it analytically below but I feel like I overcomplicated things and have probably made mistakes. There has to be a smoother way to do it. Without the decomposition of $Y?$
Also I would like to know if there is a way to estimate it with numerical methods. I had access to a scripting language to solve it. I tried it but even with $10$ billions simulations $X$ is never below $-3$.
I add what I have done analytically:
Y can be written as:
$Y = \rho X + \sqrt{1-\rho^2}Z $ where $Z\sim \mathcal N(0,\sigma)$ independent from $X$
\begin{align} \mathbb E[e^Y\textbf 1_{X<T}] & = \mathbb E[e^{\rho X} e^{\sqrt{1-\rho^2}Z}\textbf 1_{X<T}] & \\ &= \mathbb E[e^{\rho X} \textbf 1_{X<T}] \mathbb E[ e^{\sqrt{1-\rho^2}Z}] & \\ &= \left( \int_{-\infty}^{-3} e^{\rho x} \exp\left(-\frac {x^2}{ 2\sigma^2} \right)\, dx \right) \exp\left(\sigma^2 \frac{1-\rho^2}{2}\right) & \text{MGF for Gaussian } \\ \end{align}
I compute the second integral with a classic change of variable
$$ \begin{align} \int_{-\infty}^{-3} e^{\rho x} \exp\left(-\frac {x^2}{ 2\sigma^2} \right)\, dx & = \exp\left(\frac{ \sigma^2 \rho^2}2\right) \int_{-\infty}^{-3} \exp\left(-\frac {(x -\sigma^2 \rho )^2}{ 2\sigma^2} \right)\, dx \\ & = \exp\left(\frac{\sigma^2 \rho^2}2\right) \sigma^2 \int_{-\infty}^{\frac{-3-\sigma^2 \rho}{\sigma^2}} \exp\left(-\frac {u^2}{2} \right)\, du \\ &= \exp\left(\frac{\sigma^2 \rho^2}2\right) \sigma^2 \phi\left(\frac{-3-\sigma^2 \rho}{\sigma^2}\right), \end{align} $$
where $\phi$ is the CDF of a standard Gaussian.