# Loss functions for specific probability distributions?

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 ($$\frac12\mathbb{E}\left[\left([X-x]^+\right)^2\right]$$)!

• I think this question might be interesting for meta, to discuss where the line between statistics and or should be Jun 1, 2019 at 8:53
• I've never heard about this kind of "Loss functions" in statistics ... Jun 20, 2019 at 9:44
• 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. Jul 31, 2019 at 7:54

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.

References

 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.

 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.

 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.

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

Let $$L_1(r)$$, $$L_c(r)$$ and $$L_2(r)$$ be the first-order, complementary and second-order loss functions respectively. Also, define $$F(x)$$ as the CDF of a distribution defined at $$x$$ with a set of parameters and $$f(x)$$ as the PDF or PMF (continuous vs discrete) defined at $$x$$ with a set of parameters.

Note: Observe that from the answer of @TheSimpliFire, the first-order loss function can be expressed as$$n(x)=\Bbb E(X)-L(x).$$

In this post $$n(x)$$ is given directly. So for example, in the case of the exponential distribution, you can see that

$$n(x)=\frac{1}{\lambda}-\left(\frac{1}{\lambda}(1-e^{-\lambda x})\right)$$, which gives $$n(x) = \frac{e^{-\lambda x}}{\lambda}$$, which is the same as $$L_1(r) = \frac{e^{-\beta r}}{\beta}$$, where $$\lambda=\beta$$.

Normal distribution

$$p = \frac{r-\mu}{\sigma}$$.

Here, $$F(.)$$ and $$f(.)$$ are the CDF and PDF of the standard normal distribution respectively. $$L_1(r) = (\mu-r)[1-F(p)]+\sigma f(p)$$ $$L_c(r) = (r-\mu)F(p)+\sigma f(p)$$ $$L_2(r) = \frac{1}{2}[(r-\mu)^2+\sigma^2][1-F(p)]-\frac{1}{2}\sigma f(p)[r-\mu]$$

Log-normal distribution

$$p_1 = \frac{ln(r)-\mu-2\sigma^2}{\sigma}$$, $$p_2 = \frac{ln(r)-\mu-\sigma^2}{\sigma}$$, $$p_3 = \frac{ln(r)-\mu}{\sigma}$$

$$L_1(r) = e^{\mu+\frac{\sigma^2}{2}}[1-F(p_2)]-r[1-F(p_3)]$$ $$L_c(r) = rF(p_3) - e^{\mu+\frac{\sigma^2}{2}}F(p_2)$$ $$L_2(r) = \frac{r^2}{2}[1-F(p_3)]-re^{\mu+\frac{\sigma^2}{2}}[1-F(p_2)]+\frac{1}{2}e^{2\left(\mu+\frac{\sigma^2}{2}\right)}[1-F(P_1)]$$

Exponential distribution

$$L_1(r) = \frac{e^{-\beta r}}{\beta}$$ $$L_c(r) = r-\left(\frac{1-e^{-\beta r}}{\beta}\right)$$ $$L_2(r) = \frac{e^{-\beta r}}{\beta^2}$$

Gamma distribution

$$L_1(r) = \frac{\alpha}{\beta}[1-F(r; \alpha+1, \beta)]-r[1-F(r; \alpha, \beta)]$$ $$L_c(r) = rF(r; \alpha, \beta)-\frac{\alpha}{\beta}F(r; \alpha+1, \beta)$$ $$L_2(r) = \frac{r^2}{2}[1-F(r; \alpha, \beta)]-\frac{r\alpha}{\beta}[1-F(r; \alpha+1, \beta)]+\frac{\alpha(\alpha+1)}{2\beta^2}[1-F(r;\alpha+2, \beta)]$$

Negative binomial distribution

$$L_1(r) = \frac{np}{1-p}[1-F(r-2; n+1, p)]-r[1-F(r-1; n, p)]$$ $$L_c(r) = rF(r-1; n, p) - \frac{np}{1-p}F(r-2; n+1, p)$$ $$L_2(r) = \left(\frac{r^2+r}{2}\right)[1-F(r-1; n, p)]-\frac{rnp}{(1-p)}[1-F(r-2;n+1, p)]+\frac{(np)^2+np^2}{2(1-p)^2}[1-F(r-3; n+2, p)]$$

Geometric distribution

$$L_1(r) = \frac{(1-p)^r}{p}$$ $$L_c(r) = \frac{(1-p)^r+pr-1}{p}$$ $$L_2(r) = \frac{(1-p)^{r+1}}{p^2}$$

Logarithmic distribution

$$\beta = -\frac{1}{\ln(1-p)}$$

$$L_1(r) = \frac{\beta p^r}{1-p}-r[1-F(r-1)]$$ $$L_c(r) = rF(r) - \beta\left[\frac{1-p^{r+1}}{1-p}-1\right]$$ $$L_2(r) = \frac{1}{2}[r^2+r][1-F(r-1)]+\frac{\beta (2r+1)p^r}{2(p-1)}-\frac{\beta p^r[p(r-1)-r]}{2(1-p)^2}$$

Poisson distribution

$$L_1(r) = -(r-\lambda)[1-F(r)]+\lambda f(r)$$ $$L_c(r) = (r-\lambda)F(r) + \lambda f(r)$$ $$L_2(r) = \frac{1}{2}\left([(r-\lambda)^2 + r][1-F(r)]-\lambda(r-\lambda)f(r)\right)$$