I am attempting to determine closed form equations for several integrals. Suppose $X=N(\mu,\sigma)$ is normally distributed with PDF $f(x)$ and CDF $F(x)$.
- $$\int_{T}^{\infty} xf(x)dx $$
- $$\int_{T}^{\infty} x^2f(x)dx $$
I am unsure whether my application of the truncated normal distribution is correct in this situation. Define $\alpha = \frac{T-\mu}{\sigma}$, and $\lambda(\alpha) = \frac{\phi(\alpha)}{1-\Phi(\alpha)}$, and $\delta(\alpha) = \lambda(\alpha)(\lambda(\alpha) - \alpha)$.
For (1), since $$E(X|X > T) = \int_{T}^{\infty}\frac{xf(x)dx}{1-F(T)}$$
I arrive at $$\int_{T}^{\infty} xf(x)dx = [1 - \Phi(\alpha)] [\mu + \sigma \lambda(\alpha) ]$$
For (2), since $$ Var(X|X > T) = \sigma^{2}(1 - \delta(\alpha))$$
I arrive at $$ \int_{T}^{\infty} x^2f(x)dx = [1 - \Phi(\alpha)][\sigma^{2}(1-\delta(\alpha)) + (\mu + \sigma\lambda(\alpha))^{2} ]$$
I'm wondering whether this is correct. Is there a simpler formula? Thanks.