<p>You are adopting a frequentist approach and getting a frequentist result and error bar. The coefficients of a polynomial fit are orthogonal and therefore <em>if</em> you believe the model is an adequate representation, then the usual error propagation formulae should give an appropriate uncertainty in the value of <span class="math-container">$-b/2a$</span>.</p>
<p>It looks to me like your data barely constrains where the peak is and hence the very large uncertainty.</p>
<p>In comments you say</p>
<blockquote>
<p>decreasing the uncertainty on a (that is, saying we have less deviation with regard to what a might be) actually increases the resultant error bar.</p>
</blockquote>
<p>The uncertainty in <span class="math-container">$-b/2a$</span> is given by
<span class="math-container">$$\Delta(-b/2a) = \left(\frac{b}{2a}\right) \left[ \left(\frac{\Delta a}{a}\right)^2 + \left(\frac{\Delta b}{b}\right)^2 \right]^{1/2}$$</span>
so decreasing the uncertainty in <span class="math-container">$a$</span> decreases the uncertainty in the peak position.</p>
<p>If you think there are other (physical) constraints, beyond the displayed data, as to where the peak position should be, then a more correct approach is to use a Bayesian estimation of the posterior probability on the peak position. This would be formed by multiplying the likelihood of the fit (which is what you have now) by your prior information on where the peak should be and using this posterior to estimate the peak position and its uncertainty.</p>

CC BY-SA 4.0
2024-06-06T19:18:49.847Z
It seems that the '400 year' result was primarily an artifact of floating point precision within my shoddy ODR implementation.
Removing the floating point precision by adjusting the year so that the first data point is at year 0 for fitting purposes gave an error within a few years, which is perfectly acceptable.
HobbitJack
CC BY-SA 4.0
2024-06-06T22:23:58.393Z
@HobbitJack this is why you should not have hidden the numerical values of your x-axis on the plot; It's very possible someone would have spotted that right away.
uhoh
CC BY-SA 4.0
2024-06-06T06:56:20.887Z
2024-06-06T07:32:42.063Z
2024-06-06T07:32:42.063Z
ProfRob
<p>One approach that may be used is to simulate multiple results based on the uncertainties in your values, calculate a vertex for each simulated dataset, and plot the distribution of those vertices.</p>
<p>I've done this in a very rough way, by simply applying a normally distributed error to the each point.</p>
<p><a href="https://i.sstatic.net/gYYbdJSI.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/gYYbdJSI.png" alt="enter image description here" /></a></p>
<p>In the diagram, the cloud of points E2:E14 are the points from your data, shifted vertically by a random amount, and the cloud of red dots represent the vertex on each of many simulations.</p>
<p>On this very simple you can see how the coordinates of the vertex are distributed, and that this distribution has a long &quot;tail&quot; pointing up to the right. This gives a visual impression of the region in which the vertex is likely to be with 95% probability.</p>
<p>I think you are doing something more advanced than my simple little toy problem, but the same principle could apply.  You can use the error bars to apply an error to each point and potentially get a graph rather like this one.</p>

CC BY-SA 4.0
2024-06-05T21:17:26.617Z
2024-06-05T21:17:26.617Z
James K
<p>I have a long-term lightcurve for my source, to which I have used orthogonal distance regression to fit a quadratic model [expected flux]=a[year]²+b[year]+c.
Part of our data analysis includes finding the peak time of this lightcurve -- that is, the vertex of the fit-parabola. For each parameter, of the parabola, we have an uncertainty. Calculating the year-value of the vertex is as trivial as year=-b/2a. However, performing uncertainty propagation using the standard uncertainty propogation equation (<span class="math-container">$δf=\sqrt{\sum{(\frac{δf}{δx}*δx})^2}$</span>) on this quickly shows a mathematically unstable result which offers an uncertainty range of nearly 400 years, which is non-physical given that this entire dataset is on the scale of about a decade, and not hundreds of years.</p>
<p>I am wondering what statistical methods one would use to compute a numerically reasonable error bar for the vertex of a parabola given a, b, c, δa, δb, and δc.</p>
<p><a href="https://i.sstatic.net/DdTA1Dy4.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/DdTA1Dy4.png" alt="Long term lightcurve" /></a></p>

CC BY-SA 4.0
2024-06-05T02:21:28.250Z
*Welcome!* Some comments and questions - the more details you put into a question, usually the more helpful the answers and the faster they come: 1) How exactly (or with what method or software) did you do "the uncertainty propagation"? 2) There are no numbers on your x-axis. The units are "Year", but is that a 7 year span or 1 year or 20 years? 3) Have you considered asking in another Stack Exchange site, like [Cross-Validated SE](https://stats.stackexchange.com/)? You would have to add all your details before asking there, and we don't cross-post so you'd have to delete here afterwards.
uhoh
CC BY-SA 4.0
2024-06-05T02:32:00.620Z
I chose Astronomy because I wasn't quite sure where to post (I'm doing astrophysics research, after all), but it seems that Cross-Validated would have been a better place to post. Thanks for the directon.
HobbitJack
CC BY-SA 4.0
2024-06-05T02:33:21.653Z
OK it might get an answer here too. I'd say sit tight here for a day or so before considering moving.
uhoh
CC BY-SA 4.0
2024-06-05T05:13:17.650Z
I'm not covinced that it is "non-physical".  Eye-balling the data suggests at least the possibility that a curve could be fit that is almost linear in this region, and extrapolates to the right as an increasing function, with a maximum way off to the right.   It wouldn't be the "best fit" but such a nearly linear curve could be in the 95% confidence interval, so based solely on the data, you can't tell that a maximum exists in the range of years....  But the cross validated people may have better ideas about locating a maxium.
James K
CC BY-SA 4.0
2024-06-05T05:31:23.767Z
The main reason we were concerned the error propagation result is nonphysical answer is that *decreasing* the uncertainty on a (that is, saying we have less deviation with regard to what a might be) actually *increases* the resultant error bar. The other concern is that if the data really were not able to constrain the peak we would expect a linear fit to perform about as well as the quadratic fit, though χ²red is about twice as good for the quadratic fit. The other reason that we believe a constraint of 400 years is non-physical is that the 95% confidence intervals show good convergence.
HobbitJack
CC BY-SA 4.0
2024-06-05T05:35:33.453Z
Pt. 2 Didn't have room in the above comment but I should also add we have other reasons (the object we're studying, other similar trends) to believe that a quadratic, and not linear, fit is best for this data. Those trends have a variety of reasons why they're not as well-suited to perform this analysis, and due to the mathematics of error propogation would suffer from the same mathematical unstability anyways.
HobbitJack
<astrophysics><data-analysis>
Computing error bar for vertex of parabola-fit lightcurve
HobbitJack
# Computing error bar for vertex of parabola-fit lightcurve

(Asked by: username_0 on 2024-06-05)

I have a long-term lightcurve for my source, to which I have used orthogonal distance regression to fit a quadratic model [expected flux]=a[year]²+b[year]+c.
Part of our data analysis includes finding the peak time of this lightcurve -- that is, the vertex of the fit-parabola. For each parameter, of the parabola, we have an uncertainty. Calculating the year-value of the vertex is as trivial as year=-b/2a. However, performing uncertainty propagation using the standard uncertainty propogation equation ($δf=\sqrt{\sum{(\frac{δf}{δx}\*δx})^2}$) on this quickly shows a mathematically unstable result which offers an uncertainty range of nearly 400 years, which is non-physical given that this entire dataset is on the scale of about a decade, and not hundreds of years.

I am wondering what statistical methods one would use to compute a numerically reasonable error bar for the vertex of a parabola given a, b, c, δa, δb, and δc.

[![Long term lightcurve](https://i.sstatic.net/DdTA1Dy4.png)](https://i.sstatic.net/DdTA1Dy4.png)

**Tags:** astrophysics, data-analysis
> **Comments:**
> **username_1**: *Welcome!* Some comments and questions - the more details you put into a question, usually the more helpful the answers and the faster they come: 1) How exactly (or with what method or software) did you do "the uncertainty propagation"? 2) There are no numbers on your x-axis. The units are "Year", but is that a 7 year span or 1 year or 20 years? 3) Have you considered asking in another Stack Exchange site, like [Cross-Validated SE](https://stats.stackexchange.com/)? You would have to add all your details before asking there, and we don't cross-post so you'd have to delete here afterwards.  
> **username_0**: I chose Astronomy because I wasn't quite sure where to post (I'm doing astrophysics research, after all), but it seems that Cross-Validated would have been a better place to post. Thanks for the directon.  
> **username_1**: OK it might get an answer here too. I'd say sit tight here for a day or so before considering moving.  
> **username_2**: I'm not covinced that it is "non-physical".  Eye-balling the data suggests at least the possibility that a curve could be fit that is almost linear in this region, and extrapolates to the right as an increasing function, with a maximum way off to the right.   It wouldn't be the "best fit" but such a nearly linear curve could be in the 95% confidence interval, so based solely on the data, you can't tell that a maximum exists in the range of years....  But the cross validated people may have better ideas about locating a maxium.  
> **username_0**: The main reason we were concerned the error propagation result is nonphysical answer is that *decreasing* the uncertainty on a (that is, saying we have less deviation with regard to what a might be) actually *increases* the resultant error bar. The other concern is that if the data really were not able to constrain the peak we would expect a linear fit to perform about as well as the quadratic fit, though χ²red is about twice as good for the quadratic fit. The other reason that we believe a constraint of 400 years is non-physical is that the 95% confidence intervals show good convergence.  
> **username_0**: Pt. 2 Didn't have room in the above comment but I should also add we have other reasons (the object we're studying, other similar trends) to believe that a quadratic, and not linear, fit is best for this data. Those trends have a variety of reasons why they're not as well-suited to perform this analysis, and due to the mathematics of error propogation would suffer from the same mathematical unstability anyways.  

---

## 2 answers

### Answer by username_3 on 2024-06-06. Score: 4 | Accepted answer

You are adopting a frequentist approach and getting a frequentist result and error bar. The coefficients of a polynomial fit are orthogonal and therefore *if* you believe the model is an adequate representation, then the usual error propagation formulae should give an appropriate uncertainty in the value of $-b/2a$.

It looks to me like your data barely constrains where the peak is and hence the very large uncertainty.

In comments you say

> decreasing the uncertainty on a (that is, saying we have less deviation with regard to what a might be) actually increases the resultant error bar.

The uncertainty in $-b/2a$ is given by
$$\Delta(-b/2a) = \left(\frac{b}{2a}\right) \left[ \left(\frac{\Delta a}{a}\right)^2 + \left(\frac{\Delta b}{b}\right)^2 \right]^{1/2}$$
so decreasing the uncertainty in $a$ decreases the uncertainty in the peak position.

If you think there are other (physical) constraints, beyond the displayed data, as to where the peak position should be, then a more correct approach is to use a Bayesian estimation of the posterior probability on the peak position. This would be formed by multiplying the likelihood of the fit (which is what you have now) by your prior information on where the peak should be and using this posterior to estimate the peak position and its uncertainty.

> **Comments:**
> **username_0**: It seems that the '400 year' result was primarily an artifact of floating point precision within my shoddy ODR implementation. Removing the floating point precision by adjusting the year so that the first data point is at year 0 for fitting purposes gave an error within a few years, which is perfectly acceptable.  
> **username_1**: @HobbitJack this is why you should not have hidden the numerical values of your x-axis on the plot; It's very possible someone would have spotted that right away.  

---

### Answer by username_2 on 2024-06-05. Score: 4

One approach that may be used is to simulate multiple results based on the uncertainties in your values, calculate a vertex for each simulated dataset, and plot the distribution of those vertices.

I've done this in a very rough way, by simply applying a normally distributed error to the each point.

[![enter image description here](https://i.sstatic.net/gYYbdJSI.png)](https://i.sstatic.net/gYYbdJSI.png)

In the diagram, the cloud of points E2:E14 are the points from your data, shifted vertically by a random amount, and the cloud of red dots represent the vertex on each of many simulations.

On this very simple you can see how the coordinates of the vertex are distributed, and that this distribution has a long "tail" pointing up to the right. This gives a visual impression of the region in which the vertex is likely to be with 95% probability.

I think you are doing something more advanced than my simple little toy problem, but the same principle could apply. You can use the error bars to apply an error to each point and potentially get a graph rather like this one.


---

