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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?

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  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:

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

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

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