Consider $$ \int_C^\infty (x - c) f(x) dx$$ where $f$ is the density of the normal distribution, mean 0 and variance $\sigma^2$.

I would like to simplify this formula and express it in terms of $\Phi$ and $\phi$, these being the cumulative and density functions respectively of the standard normal distribution.

Could I get a hint on how to proceed?

  1. Perform a change of variable. Let's pose $y = \frac{x}{\sigma}$, and hence $dx = \sigma dy$. Further, extrema are changed to $\frac{C}{\sigma}$ and $+\infty$:

$$\int_C^\infty (x - c) f(x) dx = \int_{\frac{C}{\sigma}}^{+\infty} (y\sigma - c)f\left(y\sigma\right)\sigma dy. $$

  1. Note that $\sigma f(y\sigma) = \phi(y)$, hence:

$$\int_C^\infty (x - c) f(x) dx = \int_{\frac{C}{\sigma}}^{+\infty} (y\sigma - c)\phi\left(y\right) dy. $$

  1. Working on the last integral, we get:

$$\int_{\frac{C}{\sigma}}^{+\infty} (y\sigma - c)\phi\left(y\right) dy = \sigma \int_{\frac{C}{\sigma}}^{+\infty} y \phi(y)dy - c \int_{\frac{C}{\sigma}}^{+\infty} \phi(y)dy = \\ = \sigma \int_{\frac{C}{\sigma}}^{+\infty} y \phi(y)dy - c\left(\Phi(+\infty)-\Phi\left(\frac{C}{\sigma}\right)\right) = \\ = \sigma \int_{\frac{C}{\sigma}}^{+\infty} y \phi(y)dy - c\left(1-\Phi\left(\frac{C}{\sigma}\right)\right). $$

  1. Finally, since $y\phi(y) = -\phi'(y)$, hence:

$$\sigma \int_{\frac{C}{\sigma}}^{+\infty} y \phi(y)dy - c\left(1-\Phi\left(\frac{C}{\sigma}\right)\right) = \\= -\sigma \left(\phi(+\infty)-\phi\left( \frac{C}{\sigma}\right)\right) - c\left(1-\Phi\left(\frac{C}{\sigma}\right)\right) = \\= \sigma \phi\left( \frac{C}{\sigma}\right) - c\left(1-\Phi\left(\frac{C}{\sigma}\right)\right).$$

  • $\begingroup$ @Did thanks, I'm going to fix it. $\endgroup$ – the_candyman Feb 20 '17 at 17:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.