Expectation normal distribution

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.

• Can you write down the joint PDF of $\left(X,Y\right)$, say, $f_{X,Y}(x,y)$, and$$\mathbb{E}\left(e^Y\mathbb{1}_{\left\{X<T\right\}}\right)=\int_{\mathbb{R}^2}e^y\mathbb{1}_{\left\{x<T\right\}}f_{X,Y}(x,y){\rm d}x{\rm d}y?$$Anyway, this seems not to contribute to any simplification... Apr 30, 2018 at 16:05
• @hypernova with your method I get something like this $$\int_{\mathbb R} \int_{-\infty}^{-3} - \frac 12 e^y e^{\sigma^2 (x,y) \begin{bmatrix}1 & \rho\\\rho & 1\end{bmatrix}(x,y)^T } dx dy$$ But I don't think I know how to integrate this if you could help me Apr 30, 2018 at 16:32
• One note: you said, "I tried it but even with 10 billions simulations $X$ is never below $−3$". The probability of $X$ being below $-3$ should be about 1 out of 1 billion, The chances that you don't see this happen at all in 10 billion simulations are quite low ($< 0.0001$). So, I suspect there may be some issue with whatever process you're using to generate this random variable that may have trouble with extraordinarily rare occurrences, or maybe there's a rounding issue somewhere, etc. Apr 30, 2018 at 16:42
• @AaronMontgomery Ah yes you are right my bad. I meant millions not billions : $P(Z<6.109) = 0.999999999$ $Z$ standard Gaussian Apr 30, 2018 at 17:13