Timeline for Determining the mean and standard deviation in real time
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 20, 2012 at 13:06 | comment | added | Jason S | OK, fair enough. I won't downvote. | |
| Jan 20, 2012 at 6:50 | history | edited | Jason R | CC BY-SA 3.0 |
noted numeric instability of this method in floating point
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| Jan 20, 2012 at 6:45 | comment | added | Jason R | @JasonS: I would disagree that the technique is inherently flawed, although I agree with your point that it is not a numerically robust method when implemented in floating point. I should have been more clear that I've used this successfully in an application that used integer arithmetic. Integer (or fixed-point implementations of fractional numbers) arithmetic doesn't suffer from the issue you pointed out that causes loss of precision. In that context, it is a suitable method that requires fewer operations per sample. | |
| Jan 20, 2012 at 6:43 | comment | added | Jason R | @Dilip: Thanks for noticing that; I've replaced $\tilde\mu$ with $\tilde\mu^2$. | |
| Jan 20, 2012 at 6:43 | history | edited | Jason R | CC BY-SA 3.0 |
fixed missing exponents in standard deviation estimate
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| Jan 20, 2012 at 6:42 | comment | added | Jason R | Thanks for dusting off this old answer that obviously needs some attention. | |
| Jan 20, 2012 at 4:43 | comment | added | schnarf | This is a really good answer, even though there are numerical problems when using floating-point. This is a problem with running sums in general. Floating-point rounding errors accumulate, and it will become unstable. One solution is to recalculate the sum every so often. Another is to stabilize the filter: a running sum is a filter with poles on the unit circle, and shifting the poles in just a little bit, say, so their radius is 0.999 will make the filter more stable. Another solution is to use integers instead of floats. | |
| Jan 20, 2012 at 1:33 | comment | added | Dilip Sarwate | There seem to be a couple of typos here. Why is the mean being subtracted under the square root sign for $\sigma$? it should be $\mu^2$ to match the displayed equation $\sigma^2 = E(X^2) - (E(X))^2$, no? Also, though I won't vote down this answer, I agree with Jason S that there can be numerical problems in this approach. | |
| Jan 20, 2012 at 0:23 | comment | added | Jason S | -1 when I have enough rep to do so: this has numerical problems. See Knuth vol. 2 | |
| Dec 7, 2011 at 2:07 | vote | accept | jonsca | ||
| Jan 20, 2012 at 0:45 | |||||
| Dec 6, 2011 at 16:12 | comment | added | Jason R | Yes, that's better. I tried to rewrite to make the recursive implementation more clear. | |
| Dec 6, 2011 at 16:10 | history | edited | Jason R | CC BY-SA 3.0 |
incorporated yoda's suggestion
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| Dec 6, 2011 at 15:18 | comment | added | Lorem Ipsum | Maybe rewriting it as $A_{x,i}=x[i]+A_{x,i-1},\ A_{x,0}=x[0]$ will make it clear that you need to add only two numbers at each step, lest someone implement it as summing all $i$ elements of $x$ at each step. | |
| Dec 6, 2011 at 13:34 | history | answered | Jason R | CC BY-SA 3.0 |