23 ms·

Okay, I'm stumped. Isn't the gaussian function a probability density function, which means it should have an area of 1 by definition? Are you taking it as f(x)

by ulucs 6y ago

Okay, I'm stumped. Isn't the gaussian function a probability density function, which means it should have an area of 1 by definition? Are you taking it as f(x) = exp(-x^2)? To keep f(0) equal to 1?

em500 6y ago
You're stumped because most of the "statistics primer" section in that post doesn't make any sense. The connection between the Gaussian density and the sqrt(pi) heuristic is mostly imaginary. The original heuristic (pi, sqrt(pi), pi^2) works pretty much the same with 3 instead of pi, so you can view the pi versions as numerology or charitably a nice mnemonic.
- ToJans 6y ago
  Could you elaborate? Assume we convert all our happy path estimates to minutes. What I'm saying is that each "estimated minute" is more likely to have a gaussian distribution than a uniform standard distribution,because normal distribution is more likely to occur in nature. while I can understand that this is a controversial assumption, I'm not the first one to make it. Referring to numerology seems a bit odd? I'm really looking for a proper way here, I'm quite a rational person, so numerology is not really my cup of tea TBH.
  - ToJans 6y ago
    Addendum: this is what I was referring to: https://en.wikipedia.org/wiki/Gaussian_integral https://en.wikipedia.org/wiki/Gaussian_integral
  - em500 6y ago
    Your blog post has so many errors that I don't even know where to start. As another poster mentioned, areas under non-degenerate probability density functions are 1 by definition, whether they're uniform, Gaussian, standardized or not. What you described as a "standard uniform distribution" is really a degenerate distribution[1], meaning that you assume no uncertainty at all (stdev=0). There's nothing "uniform" about that, you might just as well start with a Gaussian with stdev=0. "converting from a standard uniform distribution to a Gaussian distribution" as you described does not make any sense at all. If you replace an initial assumption of a degenerate distribution with a Gaussian, as you seem to be doing, you replace a no-uncertainty (stdev=0) assumption with some uncertainty (so the uncertainty blow-up is infinite), but it doesn't affect point estimates such as the mean or median, unless you make separate assumptions about that. There is nothing in your story that leads to multiplying some initial time estimate by sqrt(pi). The only tenuous connection with sqrt(pi) in the whole story is that the Gaussian integral happens to be sqrt(pi). There are some deep mathematical reasons for that, which has to do with polar coordinates transformations. But it has nothing to do with adjusting uncertainties or best estimates. [1] https://en.wikipedia.org/wiki/Degenerate_distribution https://en.wikipedia.org/wiki/Degenerate_distribution
    - ToJans 6y ago
      Thank you for your valuable feedback; it will take some time to process, and I will adjust the blog post as my insights grow (potentially discarding the whole idea, but for me it's a learning process.)
  - yorwba 6y ago
    > What I'm saying is that each "estimated minute" is more likely to have a gaussian distribution If that were your actual assumption, you should measure the variance of the difference between your estimate and the actual time taken, use that to determine a confidence interval (e.g. 95% of the time, the additional delay is less than x) and then add it to your estimate.
ToJans 6y ago
My apologies, I'm by no means well versed in math, so I might use the wrong terminology. Here is a good example video of what I am alluding to: https://www.youtube.com/watch?v=9CgOthUUdw4 https://www.youtube.com/watch?v=9CgOthUUdw4 Maybe I should refine my blog post or include the video in the explanation?
- sumtechguy 6y ago
  You are the second person I have come across that uses PI to get software estimates. "I am not sure why but it works, except when it doesnt then I know exactly why" What I found was most people over/under estimate things. They also tend to do it consistently at the same rate. Typically they have a scaling factor you can use. Around 3 seems to be the sweet spot for most people. You could just as easily use 3.1 or 3.2 and get a similar answer. I usually go for 3 because I can do that in my head without a calculator.

by ulucs 6y ago

Okay, I'm stumped. Isn't the gaussian function a probability density function, which means it should have an area of 1 by definition? Are you taking it as f(x) = exp(-x^2)? To keep f(0) equal to 1?

em500 6y ago

You're stumped because most of the "statistics primer" section in that post doesn't make any sense. The connection between the Gaussian density and the sqrt(pi) heuristic is mostly imaginary. The original heuristic (pi, sqrt(pi), pi^2) works pretty much the same with 3 instead of pi, so you can view the pi versions as numerology or charitably a nice mnemonic.

ToJans 6y ago
Could you elaborate? Assume we convert all our happy path estimates to minutes. What I'm saying is that each "estimated minute" is more likely to have a gaussian distribution than a uniform standard distribution,because normal distribution is more likely to occur in nature. while I can understand that this is a controversial assumption, I'm not the first one to make it. Referring to numerology seems a bit odd? I'm really looking for a proper way here, I'm quite a rational person, so numerology is not really my cup of tea TBH.
- ToJans 6y ago
  Addendum: this is what I was referring to: https://en.wikipedia.org/wiki/Gaussian_integral https://en.wikipedia.org/wiki/Gaussian_integral
- em500 6y ago
  Your blog post has so many errors that I don't even know where to start. As another poster mentioned, areas under non-degenerate probability density functions are 1 by definition, whether they're uniform, Gaussian, standardized or not. What you described as a "standard uniform distribution" is really a degenerate distribution[1], meaning that you assume no uncertainty at all (stdev=0). There's nothing "uniform" about that, you might just as well start with a Gaussian with stdev=0. "converting from a standard uniform distribution to a Gaussian distribution" as you described does not make any sense at all. If you replace an initial assumption of a degenerate distribution with a Gaussian, as you seem to be doing, you replace a no-uncertainty (stdev=0) assumption with some uncertainty (so the uncertainty blow-up is infinite), but it doesn't affect point estimates such as the mean or median, unless you make separate assumptions about that. There is nothing in your story that leads to multiplying some initial time estimate by sqrt(pi). The only tenuous connection with sqrt(pi) in the whole story is that the Gaussian integral happens to be sqrt(pi). There are some deep mathematical reasons for that, which has to do with polar coordinates transformations. But it has nothing to do with adjusting uncertainties or best estimates. [1] https://en.wikipedia.org/wiki/Degenerate_distribution https://en.wikipedia.org/wiki/Degenerate_distribution
  - ToJans 6y ago
    Thank you for your valuable feedback; it will take some time to process, and I will adjust the blog post as my insights grow (potentially discarding the whole idea, but for me it's a learning process.)
- yorwba 6y ago
  > What I'm saying is that each "estimated minute" is more likely to have a gaussian distribution If that were your actual assumption, you should measure the variance of the difference between your estimate and the actual time taken, use that to determine a confidence interval (e.g. 95% of the time, the additional delay is less than x) and then add it to your estimate.

ToJans 6y ago

My apologies, I'm by no means well versed in math, so I might use the wrong terminology. Here is a good example video of what I am alluding to: https://www.youtube.com/watch?v=9CgOthUUdw4 https://www.youtube.com/watch?v=9CgOthUUdw4 Maybe I should refine my blog post or include the video in the explanation?

sumtechguy 6y ago
You are the second person I have come across that uses PI to get software estimates. "I am not sure why but it works, except when it doesnt then I know exactly why" What I found was most people over/under estimate things. They also tend to do it consistently at the same rate. Typically they have a scaling factor you can use. Around 3 seems to be the sweet spot for most people. You could just as easily use 3.1 or 3.2 and get a similar answer. I usually go for 3 because I can do that in my head without a calculator.