For a random variable $X$ with pdf $f(x)$, the loss function* is defined as $$n(x) = \mathbb{E}[(X-x)^+] = \int_{x}^\infty (y-x)f(y)dy,$$ where $a^+ = \max\{a,0\}$. Or, for a discrete distribution, $$n(x) = \mathbb{E}[(X-x)^+] = \sum_{y=x}^\infty (y-x)f(y).$$ Loss functions are used frequently in inventory theory and other fields.

*This is different from the "loss function" used in machine learning.

For some well known probability distributions, there are explicit forms for the loss function, typically using the pdf/pmf and cdf. For example, if $X$ has a standard normal distribution, then $$n(x) = \phi(x) - x(1-\Phi(x)),$$ where $\phi(\cdot)$ and $\Phi(\cdot)$ are the standard normal pdf and cdf. And if $X$ has a Poisson($\lambda$) distribution, then $$n(x) = -(x-\lambda)(1-F(x)) + \lambda f(x).$$ These explicit forms are nice because they can be calculated without performing numerical integration or computing long sums, using pdf/pmf and cdf functions that are built into nearly every programming language and mathematical software package.

I have seen explicit forms for loss functions for a handful of distributions, but typically somewhat scattershot in the appendix of an inventory-theory textbook (e.g., Zipkin 2000). I've never found them nicely collated anywhere.

Do you know of a resource to find explicit-form loss functions for more probability distributions?

Bonus points if the resource also has complementary loss functions ($\mathbb{E}[(X-x)^-]$) and second-order loss functions ($\frac12\mathbb{E}\left[\left([X-x]^+\right)^2\right]$)!

  • 1
    $\begingroup$ I think this question might be interesting for meta, to discuss where the line between statistics and or should be $\endgroup$ Jun 1, 2019 at 8:53
  • $\begingroup$ I've never heard about this kind of "Loss functions" in statistics ... $\endgroup$ Jun 20, 2019 at 9:44
  • $\begingroup$ Revisiting this, the second-order loss function is equivalent to calculating $\frac12\left(\Bbb V[(X-x)^+]+n(x)^2\right)$ but the closest literature I can find for the variance is through James, G. M. under squared error loss. $\endgroup$
    – TheSimpliFire
    Jul 31, 2019 at 7:54

1 Answer 1


There is indeed a paper titled Loss Distributions that provides the limited expected value functions $L(x)$ for several probability distributions (on page 15). It is directly related to the first-order loss function $n(x)$ through $$n(x)=\Bbb E(X)-L(x)\tag1$$ and notice that the loss function can also be written as $$n(x)=\int_x^\infty yf(y)\,dy-x(1-F(x))\tag2$$ after splitting the term $(y-x)$. The expressions for $L(x)$ and $\Bbb E(X)$ are tabulated below.

\begin{array}{c|c}\small\sf{Distribution}&L(x)&\Bbb E(X)\\\hline\small\sf{Log-Normal}&e^{\mu+\frac{\sigma^2}2}\Phi\left(\frac{\ln x-\mu-\sigma^2}{\sigma}\right)+x\left[1-\Phi\left(\frac{\ln x-\mu}{\sigma}\right)\right]&e^{\mu+\frac{\sigma^2}2}\\\hline\small\sf{Exponential}&\frac1\lambda\left(1-e^{-\lambda x}\right)&\frac1\lambda\\\hline\small\sf{Pareto}&\frac{\beta-\beta^\alpha(x+\beta)^{1-\alpha}}{\alpha-1}&\frac{\alpha\beta}{\alpha-1}\\\hline \small\sf{Burr}&\small\frac{\lambda^{1/\tau}\Gamma\left(\alpha-\frac1\tau\right)\Gamma\left(1+\frac1\tau\right)}{\Gamma(\alpha)}{\rm B}\left(1+\frac1\tau,\alpha-\frac1\tau;\frac{x^\tau}{\lambda+x^\tau}\right)+x\left(\frac\lambda{\lambda+x^\tau}\right)^\alpha&\frac{\lambda^{1/\tau}\Gamma\left(\alpha-\frac1\tau\right)\Gamma\left(1+\frac1\tau\right)}{\Gamma(\alpha)}\\\hline\small\sf{Weibull}&\frac{\Gamma\left(1+\frac1\tau\right)}{\beta^{1/\tau}}\Gamma\left(1+\frac1\tau,\beta x^\alpha\right)+xe^{-\beta x^\alpha}&\frac{\Gamma\left(1+\frac1\tau\right)}{\beta^{1/\tau}}\\\hline\small\sf{Gamma}&\frac\alpha\beta F(x,\alpha+1,\beta)+x(1-F(x,\alpha,\beta))&\frac\alpha\beta\end{array}

Notice how in the majority of cases, $\Bbb E(X)$ is the same as the starting coefficient of $L(x)$. Of course, $n(x)$ can be found using $(1)$.

In particular, for extensive details on the first-order loss function (and its complementary function) for the normal distribution, I highly recommend Piecewise linear approximations of the standard normal first order loss function.

For a more generic and heuristic approach on any type of distribution, there is a follow-up paper on Piecewise linearisation of the first order loss function for families of arbitrarily distributed random variables.


[1] Burnecki, K., Misiorek, A., Weron, R. (2010). Loss Distributions. MPRA Paper No. 22163. Available from: https://mpra.ub.uni-muenchen.de/22163/2/MPRA_paper_22163.pdf.

[2] Rossi, R., Tarim, S.A., Prestwich, S., Hnich, B. (2013). Piecewise linear approximations of the standard normal first order loss function. Available from: https://arxiv.org/pdf/1307.1708.pdf.

[3] Rossi, R., Hendrix, E.M.T. (2014). Piecewise linearisation of the first order loss function for families of arbitrarily distributed random variables. Proceedings of MAGO. pp. 1-4. Available from: https://gwr3n.github.io/chapters/Rossi_et_al_MAGO_2014_2.pdf.

  • 3
    $\begingroup$ Wow, this is fantastic. I use these functions all the time but had no idea about paper [1]. (I think I knew of [2] and [3].) Thanks also for that Herculean use of MathJax!! $\endgroup$ Jun 1, 2019 at 11:26
  • 1
    $\begingroup$ This is pure art $\endgroup$ Jun 1, 2019 at 13:06
  • $\begingroup$ @TheSimpliFire I’m just waiting a bit to see whether other answers get posted... :) $\endgroup$ Jun 1, 2019 at 13:09
  • $\begingroup$ Obv I upvoted. I’d upvote it twice if I could. :) $\endgroup$ Jun 1, 2019 at 13:12
  • $\begingroup$ @LarrySnyder610 Herculean is one allusion, Ceasarean is another. Reference. Reducing the font size (see here small?) and using 2 lines. $\endgroup$
    – Rob
    Jun 14, 2019 at 1:00

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.