LChrisman/Possible MetaLog application examples

This is a brainstorming ideas page. A place to list ideas for possible examples (either practical or didactic) of the MetaLog distribution. In particular, I'm looking for examples that leverage the results from https://papers.ssrn.com/sol3/papers.cfm?abstract_id=5280979 Baucells, Chrisman, Keelin & Xu (2025)].

Newsboy problem

Q = quantity to order (Decision)
D = Demand (a MetaLog distribution)
Q-D = also a MetaLog for any Q.

[math]\displaystyle{ Profit = p Min( D,Q ) - c Q = p [ Min(D-Q,0) + Q ] - c Q = p Min(D-Q,0) + (p-c) Q }[/math]

[math]\displaystyle{ E[Profit] = p E[ D-Q | D-Q \lt 0 ] + (p-c) Q }[/math]

Closed form for any Q.

Finding optimal Q? The closed form for profit doesn't eliminate the need for optimization.

Solve [math]\displaystyle{ P(D \le Q) = {{p-c}\over p} }[/math] to get optimal Q:

[math]\displaystyle{ Q^* = M_D\left({{p-c}\over p}\right) }[/math]

This is closed-form. (Wikipedia shows this solution).

This example doesn't require any of our new results, but it does show off an advantage of MetaLog.

Options pricing

X = market price at expiration. (A MetaLog)
K = strike price

Note paper: Valentyn Khokhlov (2021), "Conditional Value at Risk and Partial Moments for the Metalog Distributions", arXiv 2102.10999.

Real options

Given a forecasted MetaLog distribution for the price at expiration, the partial expectation: [math]\displaystyle{ E[ X-K | X\gt K] }[/math] gives the option's value at expiration. Uses our closed-form for partial expectation.

Financial options

In an efficient market, financial option valuation incorporates a no-arbitrage constraint.

Consider a European call option on a stock.

No-arbitrage implies:

[math]\displaystyle{ S_0 = e^{-r T} E[x] }[/math]

where

[math]\displaystyle{ S_0 }[/math] is the current stock price.
[math]\displaystyle{ r }[/math] is the risk-free interest rate.
[math]\displaystyle{ T }[/math] is the time to maturity.

The example could:

Each option bid/ask price on the market determines one quantile value.
The no-arbitrage gives us a mean-constraint, i.e., [math]\displaystyle{ E[x] = S_0 e^{r T} }[/math].

Given:

[math]\displaystyle{ C(S, K_i, T) }[/math] for [math]\displaystyle{ i = 1..n }[/math] strike prices.

How does each option imply the quantile?

Breeden–Litzenberger

The Breeden–Litzenberger formula says that the second derivative of price wrt strike price equals the risk-neutral density.

[math]\displaystyle{ {{\partial^2 C(K,T)}\over{\partial K^2}} = e^{r T} f(K) }[/math]

[math]\displaystyle{ f(x) }[/math]: Risk-neutral density function at expiration.
[math]\displaystyle{ C(K,T) }[/math]: Call price

Given the value of an option (say from its current bid/ask), we have K, C. We actually need some way to estimate the second derivative. This could likely be done by using 3 consecutive strike prices at the same expiration.

The formula is obtained by just taking the of both sides twice of:

[math]\displaystyle{ C(S,K) = e^{-rt} \int_K^\infty (x-K) f(x) dx }[/math]

It doesn't matter whether it has a no-arbitrage or not, nor on what model is being used to value the options. It should hold regardless since [math]\displaystyle{ f(x) }[/math] is the price density. Thus, it can also be used for exotic options.

The problem is that [math]\displaystyle{ {{\partial^2 C(K,T)}\over{\partial K^2}} }[/math] has to be estimated, and if estimated from [math]\displaystyle{ C(K_i,T) }[/math] points, is going to be really noisy, and you'll burn the two most extreme strike price points where the number of strike prices is already low. In practice, people fit a smooth curve to the [math]\displaystyle{ C(K_i,T) }[/math] points and use that. It would be nice if one fit was used for that and obtaining the MetaLog coefficients, but that doesn't look possible.

Maintenance scheduling, reliability analysis

Time‑to‑failure distribution, to plan maintenance.
Multiple modes -- e.g,. burn-in failures, wear-out failures.
Closed-form mean time to failure and higher‑order moments for variability.
Mean residual life [math]\displaystyle{ E[ T-t | T \gt t] }[/math].
Compare optimal replacement time w/MetaLog to Weibull fit. (to minimize cost/time).

Queuing models

How about an M/G/1 queue with MetaLog distributed service times. The distribution of issue difficulty might lead to interesting shapes to service times.

Closed form partial expectations and moments for a variety of things might be interesting.

Notes and exploration of this idea are at LChrisman/Queuing theory application of MetaLog.