Timeline for Estimation of the error for zero degree of freedom fit with error bar
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Oct 29, 2018 at 3:24 | history | edited | JimB | CC BY-SA 4.0 |
Added in confidence intervals for the predicted mean.
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| Oct 28, 2018 at 16:03 | history | edited | JimB | CC BY-SA 4.0 |
Added more details about the explicit use of the Poisson distribution in the estimation process.
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| Oct 28, 2018 at 2:18 | history | edited | JimB | CC BY-SA 4.0 |
Change answer to match the more specific additional information recently added by the OP.
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| Oct 28, 2018 at 2:14 | comment | added | JimB | You learned from a different set of statistics classes than I did: I don't know what a "condition for the convergence toward Poisson distribution" means. $k_1$ is almost certainly not an integer as suggested by the comment about the binomial distribution in your previous comment and (at least conceptually) $x_i$ would follow a Poisson distribution which means that you do really require Poisson regression. I will rewrite my answer. | |
| Oct 27, 2018 at 17:52 | comment | added | Dalnor | Oh yes you are right. As I am dealing with large number ($x_i > 1000$) I expect the estimator to be very very close to the real variance. So I think in practice the error on the error would not be that large. Also I do not think Poisson regression is ok since my data does not always fulfill the condition for the convergence toward Poisson distribution. However the condition for convergence toward Normal Distribution are fulfill ($x_1>1000$ and $p = exp(-t_i/k1) \in [0.05,0.90]$). I will edit my inital post for adding details | |
| Oct 27, 2018 at 17:09 | comment | added | JimB | But $\sqrt{x_i}$ is not the known variance: that is a known "estimate" of the variance. There's a big difference. I think what you have is Poisson regression with two data points. I'll have to delete or at least modify my answer given this new information. | |
| Oct 27, 2018 at 10:53 | comment | added | Dalnor | In fact we do know the variance of $x_i$ since it follow binomial distribution $\binom{a}{\exp(-t/k_1)}$ ( we are speaking about radioactive decay). This can be approximate with Poisson law for short $t$ and so variance is $\sqrt{x_i}$. For long $t$ we need indeed an estimation of $a$ and $k_1$ but with one or two recurrence step this is ok. | |
| Oct 26, 2018 at 22:16 | comment | added | JimB | I would not conclude that there is no built-in in Mathematica tool to do this but it does seem unlikely. But if one doesn't know the fixed coefficients, it's usually unlikely (and therefore, unbelievable) that one would "know" the variance. So still color me skeptical. | |
| Oct 26, 2018 at 22:07 | comment | added | Dalnor | Thank you for your help. From your answer I conclude there is no tool for this. I am surprise to have to use random variable to have a result. Now I think the answer is quite simple, error popagation : $ da = \sqrt{ (\frac{da}{dy1})^2 + \frac{da}{dy2})^2 }$ You sould indeed be skeptical. But on one side my systematic errors are way below the statistical error (and well estimate ) so I can do an almost perfect fit with only pure statistical error. One can show that for a define time of data acquisition less data point is more sensitive (smaller error)...so 2 points is the best. | |
| Oct 26, 2018 at 19:39 | history | answered | JimB | CC BY-SA 4.0 |