# Solve loss function for a normal distribution by integration

I want to compute the loss function for a normal distribution with mean $$\mu$$ and standard deviation $$\sigma$$. I need this for an inventory optimization model as basically I want to know "If I purchase Q units, what is the expected amount of loss sales if the demand follows $$\mathcal{N}(\mu,\sigma)$$.

I know that for a unit normal distribution function $$\mathcal{N}(0,1)$$ we have $$L(Q) = \int_{x=Q}^{\infty}(x-Q)\phi(x)dx =\phi(Q) - Q(1-\Phi(x))$$

but I want to solve this for $$\mathcal{N}(\mu,\sigma)$$

Starting from $$L(Q)$$ definition, I have,

$$L(Q) = \int_{x=Q}^{\infty}(x-Q)\phi(x)dx = \int_{x=Q}^{\infty}x\phi(x)dx - Q\int_{x=Q}^{\infty}\phi(x)dx$$

solving $$Q\int_{x=Q}^{\infty}\phi(x)$$ is fine.

The issue I face is to integrate $$\int_{x=Q}^{\infty}x\phi(x)$$.

This is what I have so far:

Step #1 - use the definition of $$\phi(x)$$ for a normal distribution:

$$\phi(x) = \frac{1}{\sqrt{2\pi\sigma^2}} e^{-\frac{(x - \mu)^2}{2 \sigma^2}}$$

So that

$$\int_{x=Q}^{\infty}x\phi(x)dx = \frac{1}{\sqrt{2\pi\sigma^2}} \int_{x=Q}^{\infty}xe^{-\frac{(x - \mu)^2}{2 \sigma^2}}dx$$

Steph #2 - I guess I have now to define something like

$$u = \frac{(x - \mu)^2}{2 \sigma^2} \text{ and } dx = \frac{x - \mu}{\sigma^2} dd$$

but I am unsure. I also guess, that the result could be something like $$\sigma^2\phi(Q) + (\mu-Q)*(1-\Phi(Q))$$

• Put $u=\frac {x -\mu} {\sigma}$. – Kavi Rama Murthy Mar 14 at 8:50
• YES! I did it. It takes 3 pages of latex to proof it, but it worked. THANK YOU! – Nicolas Mar 14 at 10:34