4 ms·

Exactly, that's not the question you're answering, but it is the question that is being asked in this context: what is the probability of rapid unplanned disass

by pps43 8y ago

Exactly, that's not the question you're answering, but it is the question that is being asked in this context: what is the probability of rapid unplanned disassembly of SpaceX rockets.

joshuamorton 8y ago
Hence the "but you can integrate to answer that". If your rocket has a 50% chance of being explosion proof and a 50% chance of blowing up half the time (you could imagine this as any number of situations, like two otherwise identical rockets which you use), any given launch has a 25% chance of rapid unplanned disassembly. You can do the same thing over a continuous random variable.
- pps43 8y ago
  But you know neither the chance of the new rocket being explosion proof, nor the chance of the rocket blowing up half the time. You built the rocket, launched it once, and it did not blow up. What should be integrated to get the probability of rapid unplanned disassembly?
  - joshuamorton 8y ago
    Enter the Beta(variate) distribution. The beta distribution[1] is a cool statistical distribution defined by BetaDist{a,b} (or alpha, beta, but that's too much work), where a is the number of successes and b is the number of failures you've sampled. It has a number of cool properties, chief among them that given X = BetaDist{a,b}, then cdf(X, x) = the probability that the mean of the distribution you are approximating is less than x. It has a bunch of other nice properties too (like E[X] = a / (a + b), which should be obvious), but those aren't as relevant here. So let's say that you assume a uniform prior. This is defined as BetaDist{1,1} [2]. this is probably the wrong prior. So you might have a better idea. If, for example, you believe there is a 10% chance of your rocket exploding based on some calculations you've done, you might use a differently tuned beta distribution, like BetaDist{9,1} or BetaDist{4.5,.5} if you were feeling uncertain (but in general it would likely be better to use {8,2} in that situation iirc). But let's assume {1,1} for now. So you launch your rocket. Everything goes great. You update your distribution. Its a success. So you get BetaDist{2,1} [3]. So what is the chance your rocket explodes? Well the cdf of your beta distribution is the probability that the mean is less than x. So the pdf of the beta distribution is the probability that the mean is exactly x. So then The integral from 0 -> 1 of `(1 - x) * pdf(X, x) dx` is the estimated probability that your rocket explodes on its next launch, since that's "for every value x, the likelyhood of the distribution being that one multiplied by the chance your rocket explodes given that distribution". For the one rocket case, this happens to be equal to E[X] = a / (a + b), so it's 1/3. For the two rocket case, you apply reinforcement learning/k-armed bandit strategies like UBC1[4] or Thompson Sampling[5]. These are algorithms that will result in you picking the best rocket with as few unnecessary explosions as possible, provably. You can see some related discussion I've had on HN about these algorithms [6]. [1]: https://en.wikipedia.org/wiki/Beta_distribution https://en.wikipedia.org/wiki/Beta_distribution [2]: http://www.wolframalpha.com/input/?i=beta+distribution+(1,1) http://www.wolframalpha.com/input/?i=beta+distribution+(1,1) [3]: http://www.wolframalpha.com/input/?i=beta+distribution+(2,+1) http://www.wolframalpha.com/input/?i=beta+distribution+(2,+1... [4]: http://banditalgs.com/2016/09/18/the-upper-confidence-bound-algorithm/ http://banditalgs.com/2016/09/18/the-upper-confidence-bound-... [5]: https://en.wikipedia.org/wiki/Thompson_sampling https://en.wikipedia.org/wiki/Thompson_sampling [6]: https://news.ycombinator.com/item?id=17014232 https://news.ycombinator.com/item?id=17014232
    - pps43 8y ago
      Wouldn't it be easier to just add one success and one failure and calculate (events+1)/(trials+2) instead of events/trials?
      - joshuamorton 8y ago
        I'm not sure what you mean.
        pps43 8y ago
        I'm talking about pseudocounts (add one success and one failure) [1] or maybe Agresti-Coull estimator (add two successes and two failures) [2]. [1] https://en.wikipedia.org/wiki/Rule_of_succession https://en.wikipedia.org/wiki/Rule_of_succession [2] http://users.stat.ufl.edu/~aa/articles/agresti_coull_1998.pdf http://users.stat.ufl.edu/~aa/articles/agresti_coull_1998.pd...

by pps43 8y ago

Exactly, that's not the question you're answering, but it is the question that is being asked in this context: what is the probability of rapid unplanned disassembly of SpaceX rockets.

joshuamorton 8y ago

Hence the "but you can integrate to answer that". If your rocket has a 50% chance of being explosion proof and a 50% chance of blowing up half the time (you could imagine this as any number of situations, like two otherwise identical rockets which you use), any given launch has a 25% chance of rapid unplanned disassembly. You can do the same thing over a continuous random variable.

pps43 8y ago
But you know neither the chance of the new rocket being explosion proof, nor the chance of the rocket blowing up half the time. You built the rocket, launched it once, and it did not blow up. What should be integrated to get the probability of rapid unplanned disassembly?
- joshuamorton 8y ago
  Enter the Beta(variate) distribution. The beta distribution[1] is a cool statistical distribution defined by BetaDist{a,b} (or alpha, beta, but that's too much work), where a is the number of successes and b is the number of failures you've sampled. It has a number of cool properties, chief among them that given X = BetaDist{a,b}, then cdf(X, x) = the probability that the mean of the distribution you are approximating is less than x. It has a bunch of other nice properties too (like E[X] = a / (a + b), which should be obvious), but those aren't as relevant here. So let's say that you assume a uniform prior. This is defined as BetaDist{1,1} [2]. this is probably the wrong prior. So you might have a better idea. If, for example, you believe there is a 10% chance of your rocket exploding based on some calculations you've done, you might use a differently tuned beta distribution, like BetaDist{9,1} or BetaDist{4.5,.5} if you were feeling uncertain (but in general it would likely be better to use {8,2} in that situation iirc). But let's assume {1,1} for now. So you launch your rocket. Everything goes great. You update your distribution. Its a success. So you get BetaDist{2,1} [3]. So what is the chance your rocket explodes? Well the cdf of your beta distribution is the probability that the mean is less than x. So the pdf of the beta distribution is the probability that the mean is exactly x. So then The integral from 0 -> 1 of `(1 - x) * pdf(X, x) dx` is the estimated probability that your rocket explodes on its next launch, since that's "for every value x, the likelyhood of the distribution being that one multiplied by the chance your rocket explodes given that distribution". For the one rocket case, this happens to be equal to E[X] = a / (a + b), so it's 1/3. For the two rocket case, you apply reinforcement learning/k-armed bandit strategies like UBC1[4] or Thompson Sampling[5]. These are algorithms that will result in you picking the best rocket with as few unnecessary explosions as possible, provably. You can see some related discussion I've had on HN about these algorithms [6]. [1]: https://en.wikipedia.org/wiki/Beta_distribution https://en.wikipedia.org/wiki/Beta_distribution [2]: http://www.wolframalpha.com/input/?i=beta+distribution+(1,1) http://www.wolframalpha.com/input/?i=beta+distribution+(1,1) [3]: http://www.wolframalpha.com/input/?i=beta+distribution+(2,+1) http://www.wolframalpha.com/input/?i=beta+distribution+(2,+1... [4]: http://banditalgs.com/2016/09/18/the-upper-confidence-bound-algorithm/ http://banditalgs.com/2016/09/18/the-upper-confidence-bound-... [5]: https://en.wikipedia.org/wiki/Thompson_sampling https://en.wikipedia.org/wiki/Thompson_sampling [6]: https://news.ycombinator.com/item?id=17014232 https://news.ycombinator.com/item?id=17014232
  - pps43 8y ago
    Wouldn't it be easier to just add one success and one failure and calculate (events+1)/(trials+2) instead of events/trials?
    - joshuamorton 8y ago
      I'm not sure what you mean.
      - pps43 8y ago
        I'm talking about pseudocounts (add one success and one failure) [1] or maybe Agresti-Coull estimator (add two successes and two failures) [2]. [1] https://en.wikipedia.org/wiki/Rule_of_succession https://en.wikipedia.org/wiki/Rule_of_succession [2] http://users.stat.ufl.edu/~aa/articles/agresti_coull_1998.pdf http://users.stat.ufl.edu/~aa/articles/agresti_coull_1998.pd...