Issue with power functions of p-boxes that straddle

[leslieDLcy]

The current algorithm for the power function of p-box objects are not correct for cases that straddle 0.

Below are two simple examples to illustrate the problem. For the first Gaussian p-box, it seems an additional envelope to 0 will solve the issue. But this solution seems inadequate for the second example of a Uniform case.

The function verify_power_calculations_plot(a, 2) means that we are calculating $a^2$. reference suggests a discretisation-interval propagation-stacking interval solution; while direct suggests the current algorithm which naively takes takes the power of the edges.

[ScottFerson]

L: [I can view the issue, but I'm not sure how to respond there.] Ugh! Powers are so tricky. And with uncertainty, they are really tricky. I remember I had dozens of pages of checks I was using for the Risk Calc implementation (and it's *still* got holes and bugs). Zero isn't the only point of worry; so is one. Exponents that are intervals and p-boxes are a nightmare. This is a known bug in Risk Calc [see https://sites.google.com/site/riskcalcblog/corrections/square-of-straddling-p-box]. The blog offers a *work-around*, which seems to work with both of your examples. (I superimposed them over the calculations you made with verify_power_calculations_plot.) Nevertheless, I wouldn't trust it as far as I can throw it. The Risk Calc code is below. The graphs beneath the code are for your second example, which I call "b". What is the algorithm you used for verify_power_calculations_plot? I've been stymied by trying to implement powers in the Kaucher code too. There I'm I'm not even sure I can tell what the correct should be! Maybe we can talk about this a bit. I presume that these calculations are needed for something Simon wants to do so that just returning an error or a "not yet" message is not an option. Sc // Risk Calc code *func square() return env($1^2,rightside(abs(env(rightside(min($1,0)),leftside(max($1,0)))) ^ 2))* a = N([-1,2],0.1) a2 = a^2 a2 ~(range=[0.588849,5.09668], mean=[1.01,4.02], var=[0,3.08]) sa = square(a) sa ~(range=[0,5.09668], mean=[0,4.1], var=[0,3.08]) clear; show sa in blue; show a2 in red; show a2 on '1111' in red b =U([-3,-2], [7,8]) b2 = b ^ 2 b2 ~(range=[1.2326e-32,64], mean=[12.1,18.4], var=[164,378]) sb = square(b) sb ~(range=[0,64], mean=[11.4,18.4], var=[164,378]) clear; show sb in blue; show b2 in red; show b2 on '1111' in red [image: image.png] [image: image.png]

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On Wed, Nov 19, 2025 at 9:11 PM (Leslie) Yu Chen ***@***.***> wrote: *leslieDLcy* created an issue (leslieDLcy/PyUncertainNumber#126) <#126> The current algorithm for the power function of p-box objects are not correct for cases that straddle 0. Screenshot.2025-11-19.at.21.06.53.png (view on web) <https://github.com/user-attachments/assets/3b01977b-ec3c-4bed-8f4d-b4c081e0930f> Screenshot.2025-11-19.at.21.06.02.png (view on web) <https://github.com/user-attachments/assets/fe227c90-84b7-4b1a-956c-f60960dbad42> — Reply to this email directly, view it on GitHub <#126>, or unsubscribe <https://github.com/notifications/unsubscribe-auth/AL3U54FNL6VXQPLDJC66VLT35TMGZAVCNFSM6AAAAACMT3MRV6VHI2DSMVQWIX3LMV43ASLTON2WKOZTGY2DINBTGY2DINA> . You are receiving this because you were assigned.Message ID: ***@***.***>