3 ms·
I compute the standard deviation of n(1) as sqrt(N p (1-p)) where p = 1/10. It's N iid realizations of a biased coin toss. You left out the 1-p part, I think.
by mturmon 15y ago
I compute the standard deviation of n(1) as sqrt(N p (1-p)) where p = 1/10. It's N iid realizations of a biased coin toss.
You left out the 1-p part, I think.
I totally agree with your sentiment that just eyeballing the plot is not the best route here. It's all too easy to convince yourself that the deviation isn't that bad, if you're just looking at the plot without error bars.
- foob 15y agoThanks, good catch. I'm used to thinking about the limit where you have so many bins that the Binomial distribution can be approximated as a Poisson distribution without thinking twice. 10 is sort of borderline but I probably should have been more explicit. If there are only 10 bins then this will introduce a roughly 5% overestimation of the standard deviation in the distribution of n(i). This in turn would be an 11% overestimation of Q and an 11% overestimation of the number of standard deviations you're off by. The statistical error on your estimation of this sigma is generally going to be significantly larger than the errors introduced by making this approximation so it really doesn't matter all that much. As you make the number of bins larger they matter less and less (mturmon probably knows this, I'm just making it clear to anyone reading our comments).