1
$\begingroup$

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)) $

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

Your Answer

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

Browse other questions tagged or ask your own question.