Consider two random variables that are jointly normally distributed: $$ (T,C)\sim \text{BiNormal}[(\mu_t,\mu_c),(\sigma_t,\sigma_c),\rho] $$
I want to express $E[T|C<c^*]$ in terms of the gaussian parameters and the P.D.F and C.D.F of the standard normal.
Below is what I have – Is this correct?
- We know that for the bivariate normal, $E[T|C]=\mu_T + \frac {\sigma_T} {\sigma_C} \rho (C - \mu_C)$
- Using the law of iterated expectations, we can write $$E[T|C<c^*]=E[E[T|C]|C<c^*]=E[\mu_T + \frac {\sigma_T} {\sigma_C} \rho (C - \mu_C)|C<c^*]$$
- Taking out of the expectation the constants gives $$\mu_T -\frac {\sigma_T} {\sigma_C} \rho\ \left(\mu_c - E[C|C<c^*]\right)$$
- And given that $E[C|C<c^*]=\mu_c-\sigma_c\frac{\phi(\frac{c^*-\mu_c}{\sigma_c})}{\Phi(\frac{c^*-\mu_c}{\sigma_c})}$, we finally have $$ \mu_T - \rho\,\sigma_T \, \frac{\phi(\frac{c^*-\mu_c}{\sigma_c})}{\Phi(\frac{c^*-\mu_c}{\sigma_c})} $$
I do not know where your problem was - the approach looks correct to me and I like your method. Here is just another method based on the Cholesky decomposition which arrives with the same answer:
Let $Z_1, Z_2$ be independent standard normal random variables. Then $(T, C)$ has the same distribution as $(\mu_T + \sigma_T(\rho Z_1+\sqrt{1-\rho^2}Z_2), \mu_C+\sigma_CZ_1)$. Therefore,
$$ \begin{align} E[T|C < c^*] &= E\left[\mu_T + \sigma_T(\rho Z_1 + \sqrt{1-\rho^2}Z_2) \bigg|\mu_C + \sigma_C Z_1 < c^*\right] \\ &= \mu_T + \sigma_T \rho E\left[ Z_1 \bigg|Z_1 < \frac {c^* - \mu_C} {\sigma_C}\right] + \sigma_T\sqrt{1-\rho^2} E\left[Z_2 \bigg|Z_1 < \frac {c^* - \mu_C} {\sigma_C} \right]\\ &= \mu_T - \rho\sigma_T\frac {\displaystyle \phi\left(\frac {c^* - \mu_C} {\sigma_C} \right) } {\displaystyle \Phi\left(\frac {c^* - \mu_C} {\sigma_C}\right)} \end{align}$$
So you may better check again your numerical answer with your simulation code.