Berkeley CSUA MOTD:Entry 53378
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2022/06/25 [General] UID:1000 Activity:popular
 6/25

 2009/9/18-29 [Computer/Theory] UID:53378 Activity:nil ```9/18 I forgot my math. Say the probability of a bug happening is the unknown fixed value p in [0,1]. I attempt to reproduce the bug until it happens once. Then it happens at my n-th trial, so I stop. Now, what is the expected value of p? Is it E(p) = 1/n? Thx. \_ Did a quick program to test. Looks like E(p) = 1/n (given assumption n(p) = sum x = 1 to inf of x*(1-p)^(x-1)*p where n(p) = expected n for a given p) \_ Consider the case n=1 to see why E(p) = 1/n can't possibly be right. You need to specify a prior distribution and then apply Bayes' theorem. If your prior distribution is the uniform distribution on [0,1] (i.e. before running the experiment you think all values for p between 0 and 1 are equally likely), then if you see your bug for the first time in the nth trial, you'll get a posterior distribution of (n)(n+1)(p)(1-p)^(n-1). The expected value of this is the root r of the equation (1+nr)(1-r)^n = 1/2. For n = 1, E(p)=1/sqrt(2). For n = 2, E(p) = 1/2. For n = 3, it's around .3857. This sequence probably has some name, but I don't know it. [In practice that might be a stupid prior to use for your problem. But you have to have some prior, otherwise the problem is not well-specified mathematically.] \_ Doy, wow I'm an idiot. I'm wrong, this person is right. -pp \_ That is a fine answer above. The part about Bayes Rule and having to make some assumptions about priors/distribution properties is hard to explain. All I have to add is OP may enjoy reading about the NEGATIVE BINOMIAL DISTRIBUTIION. (while you are at it, you might read up on POISSON DISTRIBUTION/ PROCESS ... which can be quite useful in a lot of engineering contexts).```