Confidence interval of an average coefficient calculated from curve fits, with each curve fit using a...












1














I am curve fitting a model to data sets in order to determine a coefficient for each curve fit/data set. I then calculate an average coefficient and the 95% confidence interval for this average. However, each curve fit involves the use of a parameter (in the model equation) which itself has a 95% confidence interval. How do I propagate the 95% confidence interval for this parameter into the 95% confidence interval for the coefficient calculated from curve fits?



Here is an example version of the model that I am curve fitting to data:
$$y=text{erfc}left(frac{x}{A+B}right)$$
$A$ is the curve fit variable and $B$ is the pre-determined parameter with a 95% confidence interval. Changing the value of $B$ changes the calculated value of $A$, so unless I'm mistaken, the confidence interval of $B$ should propagate into that of $A$. How is this done mathematically?



Edit: I found the answer to my question. It's included below.










share|cite|improve this question
























  • Could I calculate $x$ with the upper and lower bounds of $K$, based on its confidence interval, and then state that the confidence interval of $x$ is between the far extremes of the confidence intervals of $x$ calculated for the upper and lower bounds of $K$? For example, calculating $x=10pm10$ for the lower bound of $K$, $x=20pm10$ for the upper bound of $K$, $x=15pm10$ for the average value of $K$, and stating that $x=15pm15$.
    – Cedric Eveleigh
    Nov 9 at 1:13












  • Just using intervals is problematic -- extremes of x are not necessarily found at the extremes of K (you would need to verify that x is a monotonic function of K), and usually x is nonuniformly distributed, but an interval can't express anything about that. My advice for a relatively simple approximation to the full Bayesian taco is to generate a random sample from a distribution over K, and see what resulting distribution of x you get.
    – Robert Dodier
    Nov 11 at 6:22










  • FYI I updated the question with different variable names.
    – Cedric Eveleigh
    Nov 24 at 19:52
















1














I am curve fitting a model to data sets in order to determine a coefficient for each curve fit/data set. I then calculate an average coefficient and the 95% confidence interval for this average. However, each curve fit involves the use of a parameter (in the model equation) which itself has a 95% confidence interval. How do I propagate the 95% confidence interval for this parameter into the 95% confidence interval for the coefficient calculated from curve fits?



Here is an example version of the model that I am curve fitting to data:
$$y=text{erfc}left(frac{x}{A+B}right)$$
$A$ is the curve fit variable and $B$ is the pre-determined parameter with a 95% confidence interval. Changing the value of $B$ changes the calculated value of $A$, so unless I'm mistaken, the confidence interval of $B$ should propagate into that of $A$. How is this done mathematically?



Edit: I found the answer to my question. It's included below.










share|cite|improve this question
























  • Could I calculate $x$ with the upper and lower bounds of $K$, based on its confidence interval, and then state that the confidence interval of $x$ is between the far extremes of the confidence intervals of $x$ calculated for the upper and lower bounds of $K$? For example, calculating $x=10pm10$ for the lower bound of $K$, $x=20pm10$ for the upper bound of $K$, $x=15pm10$ for the average value of $K$, and stating that $x=15pm15$.
    – Cedric Eveleigh
    Nov 9 at 1:13












  • Just using intervals is problematic -- extremes of x are not necessarily found at the extremes of K (you would need to verify that x is a monotonic function of K), and usually x is nonuniformly distributed, but an interval can't express anything about that. My advice for a relatively simple approximation to the full Bayesian taco is to generate a random sample from a distribution over K, and see what resulting distribution of x you get.
    – Robert Dodier
    Nov 11 at 6:22










  • FYI I updated the question with different variable names.
    – Cedric Eveleigh
    Nov 24 at 19:52














1












1








1







I am curve fitting a model to data sets in order to determine a coefficient for each curve fit/data set. I then calculate an average coefficient and the 95% confidence interval for this average. However, each curve fit involves the use of a parameter (in the model equation) which itself has a 95% confidence interval. How do I propagate the 95% confidence interval for this parameter into the 95% confidence interval for the coefficient calculated from curve fits?



Here is an example version of the model that I am curve fitting to data:
$$y=text{erfc}left(frac{x}{A+B}right)$$
$A$ is the curve fit variable and $B$ is the pre-determined parameter with a 95% confidence interval. Changing the value of $B$ changes the calculated value of $A$, so unless I'm mistaken, the confidence interval of $B$ should propagate into that of $A$. How is this done mathematically?



Edit: I found the answer to my question. It's included below.










share|cite|improve this question















I am curve fitting a model to data sets in order to determine a coefficient for each curve fit/data set. I then calculate an average coefficient and the 95% confidence interval for this average. However, each curve fit involves the use of a parameter (in the model equation) which itself has a 95% confidence interval. How do I propagate the 95% confidence interval for this parameter into the 95% confidence interval for the coefficient calculated from curve fits?



Here is an example version of the model that I am curve fitting to data:
$$y=text{erfc}left(frac{x}{A+B}right)$$
$A$ is the curve fit variable and $B$ is the pre-determined parameter with a 95% confidence interval. Changing the value of $B$ changes the calculated value of $A$, so unless I'm mistaken, the confidence interval of $B$ should propagate into that of $A$. How is this done mathematically?



Edit: I found the answer to my question. It's included below.







data-analysis confidence-interval error-propagation






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share|cite|improve this question













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edited Nov 24 at 19:45

























asked Nov 5 at 22:38









Cedric Eveleigh

113




113












  • Could I calculate $x$ with the upper and lower bounds of $K$, based on its confidence interval, and then state that the confidence interval of $x$ is between the far extremes of the confidence intervals of $x$ calculated for the upper and lower bounds of $K$? For example, calculating $x=10pm10$ for the lower bound of $K$, $x=20pm10$ for the upper bound of $K$, $x=15pm10$ for the average value of $K$, and stating that $x=15pm15$.
    – Cedric Eveleigh
    Nov 9 at 1:13












  • Just using intervals is problematic -- extremes of x are not necessarily found at the extremes of K (you would need to verify that x is a monotonic function of K), and usually x is nonuniformly distributed, but an interval can't express anything about that. My advice for a relatively simple approximation to the full Bayesian taco is to generate a random sample from a distribution over K, and see what resulting distribution of x you get.
    – Robert Dodier
    Nov 11 at 6:22










  • FYI I updated the question with different variable names.
    – Cedric Eveleigh
    Nov 24 at 19:52


















  • Could I calculate $x$ with the upper and lower bounds of $K$, based on its confidence interval, and then state that the confidence interval of $x$ is between the far extremes of the confidence intervals of $x$ calculated for the upper and lower bounds of $K$? For example, calculating $x=10pm10$ for the lower bound of $K$, $x=20pm10$ for the upper bound of $K$, $x=15pm10$ for the average value of $K$, and stating that $x=15pm15$.
    – Cedric Eveleigh
    Nov 9 at 1:13












  • Just using intervals is problematic -- extremes of x are not necessarily found at the extremes of K (you would need to verify that x is a monotonic function of K), and usually x is nonuniformly distributed, but an interval can't express anything about that. My advice for a relatively simple approximation to the full Bayesian taco is to generate a random sample from a distribution over K, and see what resulting distribution of x you get.
    – Robert Dodier
    Nov 11 at 6:22










  • FYI I updated the question with different variable names.
    – Cedric Eveleigh
    Nov 24 at 19:52
















Could I calculate $x$ with the upper and lower bounds of $K$, based on its confidence interval, and then state that the confidence interval of $x$ is between the far extremes of the confidence intervals of $x$ calculated for the upper and lower bounds of $K$? For example, calculating $x=10pm10$ for the lower bound of $K$, $x=20pm10$ for the upper bound of $K$, $x=15pm10$ for the average value of $K$, and stating that $x=15pm15$.
– Cedric Eveleigh
Nov 9 at 1:13






Could I calculate $x$ with the upper and lower bounds of $K$, based on its confidence interval, and then state that the confidence interval of $x$ is between the far extremes of the confidence intervals of $x$ calculated for the upper and lower bounds of $K$? For example, calculating $x=10pm10$ for the lower bound of $K$, $x=20pm10$ for the upper bound of $K$, $x=15pm10$ for the average value of $K$, and stating that $x=15pm15$.
– Cedric Eveleigh
Nov 9 at 1:13














Just using intervals is problematic -- extremes of x are not necessarily found at the extremes of K (you would need to verify that x is a monotonic function of K), and usually x is nonuniformly distributed, but an interval can't express anything about that. My advice for a relatively simple approximation to the full Bayesian taco is to generate a random sample from a distribution over K, and see what resulting distribution of x you get.
– Robert Dodier
Nov 11 at 6:22




Just using intervals is problematic -- extremes of x are not necessarily found at the extremes of K (you would need to verify that x is a monotonic function of K), and usually x is nonuniformly distributed, but an interval can't express anything about that. My advice for a relatively simple approximation to the full Bayesian taco is to generate a random sample from a distribution over K, and see what resulting distribution of x you get.
– Robert Dodier
Nov 11 at 6:22












FYI I updated the question with different variable names.
– Cedric Eveleigh
Nov 24 at 19:52




FYI I updated the question with different variable names.
– Cedric Eveleigh
Nov 24 at 19:52










3 Answers
3






active

oldest

votes


















0














Your question is a bit confusing -- if x is the adjustable parameter to be optimised, and K is an input (from a disribution), what is the independent variable? Nevertheless, if I assume for the sake of argument that it's there somewhere, conceptually you should fit by optimisation of a single parameter



$y = text{erfc}(theta)$,



then rearrange the basic error propagation (eq 3.20 from Bevington) to extract the uncertainty in $x$ (i.e. solve for $sigma_x$ from the following):



$sigma_theta^2 = sigma_x^2 + sigma_K^2 pm 2sigma_{xK}^2$



where the correlation coefficient of the covariance term would be 1.



This assumes a normal distribution for all quantities, of course, which may not be the case. Monte Carlo sampling would be the more robust way of solving this, if you've got any computational nous.






share|cite|improve this answer





























    0














    This is a good question, and unfortunately there is no easy answer. The right way to go about problems like this is via Bayesian inference. In this context, that amounts to assuming that every uncertain variable has a distribution, conditional or unconditional, and then deriving the distribution over what you consider interesting, given any data at hand, and marginalizing (i.e. integrating) over the other variables you consider uninteresting. My advice is to try to state the whole Bayesian model, and then make simplifying assumptions (possibly drastic) to arrive at something tractable.



    There are probably many non-Bayesian methods, which amount to specific simplifications of the Bayesian approach, making use of particular assumptions, possibly unstated. My advice is, don't resort immediately to simplifications, instead work out the whole model and consider what are reasonable and workable assumptions for your problem.






    share|cite|improve this answer





















    • I'm curious what you think about what I suggested in a comment to the question.
      – Cedric Eveleigh
      Nov 11 at 1:13



















    0














    I found a way to accomplish the propagation of uncertainty: Combine $A$ and $B$ into a single term—say, $C$—and curve fit to determine $C$. Then, solve $C = A + B$ for $A$ and calculate the uncertainty of $A$ with the uncertainties of $C$ and $B$.






    share|cite|improve this answer





















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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      0














      Your question is a bit confusing -- if x is the adjustable parameter to be optimised, and K is an input (from a disribution), what is the independent variable? Nevertheless, if I assume for the sake of argument that it's there somewhere, conceptually you should fit by optimisation of a single parameter



      $y = text{erfc}(theta)$,



      then rearrange the basic error propagation (eq 3.20 from Bevington) to extract the uncertainty in $x$ (i.e. solve for $sigma_x$ from the following):



      $sigma_theta^2 = sigma_x^2 + sigma_K^2 pm 2sigma_{xK}^2$



      where the correlation coefficient of the covariance term would be 1.



      This assumes a normal distribution for all quantities, of course, which may not be the case. Monte Carlo sampling would be the more robust way of solving this, if you've got any computational nous.






      share|cite|improve this answer


























        0














        Your question is a bit confusing -- if x is the adjustable parameter to be optimised, and K is an input (from a disribution), what is the independent variable? Nevertheless, if I assume for the sake of argument that it's there somewhere, conceptually you should fit by optimisation of a single parameter



        $y = text{erfc}(theta)$,



        then rearrange the basic error propagation (eq 3.20 from Bevington) to extract the uncertainty in $x$ (i.e. solve for $sigma_x$ from the following):



        $sigma_theta^2 = sigma_x^2 + sigma_K^2 pm 2sigma_{xK}^2$



        where the correlation coefficient of the covariance term would be 1.



        This assumes a normal distribution for all quantities, of course, which may not be the case. Monte Carlo sampling would be the more robust way of solving this, if you've got any computational nous.






        share|cite|improve this answer
























          0












          0








          0






          Your question is a bit confusing -- if x is the adjustable parameter to be optimised, and K is an input (from a disribution), what is the independent variable? Nevertheless, if I assume for the sake of argument that it's there somewhere, conceptually you should fit by optimisation of a single parameter



          $y = text{erfc}(theta)$,



          then rearrange the basic error propagation (eq 3.20 from Bevington) to extract the uncertainty in $x$ (i.e. solve for $sigma_x$ from the following):



          $sigma_theta^2 = sigma_x^2 + sigma_K^2 pm 2sigma_{xK}^2$



          where the correlation coefficient of the covariance term would be 1.



          This assumes a normal distribution for all quantities, of course, which may not be the case. Monte Carlo sampling would be the more robust way of solving this, if you've got any computational nous.






          share|cite|improve this answer












          Your question is a bit confusing -- if x is the adjustable parameter to be optimised, and K is an input (from a disribution), what is the independent variable? Nevertheless, if I assume for the sake of argument that it's there somewhere, conceptually you should fit by optimisation of a single parameter



          $y = text{erfc}(theta)$,



          then rearrange the basic error propagation (eq 3.20 from Bevington) to extract the uncertainty in $x$ (i.e. solve for $sigma_x$ from the following):



          $sigma_theta^2 = sigma_x^2 + sigma_K^2 pm 2sigma_{xK}^2$



          where the correlation coefficient of the covariance term would be 1.



          This assumes a normal distribution for all quantities, of course, which may not be the case. Monte Carlo sampling would be the more robust way of solving this, if you've got any computational nous.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 8 at 5:12









          GoneAsync

          1011




          1011























              0














              This is a good question, and unfortunately there is no easy answer. The right way to go about problems like this is via Bayesian inference. In this context, that amounts to assuming that every uncertain variable has a distribution, conditional or unconditional, and then deriving the distribution over what you consider interesting, given any data at hand, and marginalizing (i.e. integrating) over the other variables you consider uninteresting. My advice is to try to state the whole Bayesian model, and then make simplifying assumptions (possibly drastic) to arrive at something tractable.



              There are probably many non-Bayesian methods, which amount to specific simplifications of the Bayesian approach, making use of particular assumptions, possibly unstated. My advice is, don't resort immediately to simplifications, instead work out the whole model and consider what are reasonable and workable assumptions for your problem.






              share|cite|improve this answer





















              • I'm curious what you think about what I suggested in a comment to the question.
                – Cedric Eveleigh
                Nov 11 at 1:13
















              0














              This is a good question, and unfortunately there is no easy answer. The right way to go about problems like this is via Bayesian inference. In this context, that amounts to assuming that every uncertain variable has a distribution, conditional or unconditional, and then deriving the distribution over what you consider interesting, given any data at hand, and marginalizing (i.e. integrating) over the other variables you consider uninteresting. My advice is to try to state the whole Bayesian model, and then make simplifying assumptions (possibly drastic) to arrive at something tractable.



              There are probably many non-Bayesian methods, which amount to specific simplifications of the Bayesian approach, making use of particular assumptions, possibly unstated. My advice is, don't resort immediately to simplifications, instead work out the whole model and consider what are reasonable and workable assumptions for your problem.






              share|cite|improve this answer





















              • I'm curious what you think about what I suggested in a comment to the question.
                – Cedric Eveleigh
                Nov 11 at 1:13














              0












              0








              0






              This is a good question, and unfortunately there is no easy answer. The right way to go about problems like this is via Bayesian inference. In this context, that amounts to assuming that every uncertain variable has a distribution, conditional or unconditional, and then deriving the distribution over what you consider interesting, given any data at hand, and marginalizing (i.e. integrating) over the other variables you consider uninteresting. My advice is to try to state the whole Bayesian model, and then make simplifying assumptions (possibly drastic) to arrive at something tractable.



              There are probably many non-Bayesian methods, which amount to specific simplifications of the Bayesian approach, making use of particular assumptions, possibly unstated. My advice is, don't resort immediately to simplifications, instead work out the whole model and consider what are reasonable and workable assumptions for your problem.






              share|cite|improve this answer












              This is a good question, and unfortunately there is no easy answer. The right way to go about problems like this is via Bayesian inference. In this context, that amounts to assuming that every uncertain variable has a distribution, conditional or unconditional, and then deriving the distribution over what you consider interesting, given any data at hand, and marginalizing (i.e. integrating) over the other variables you consider uninteresting. My advice is to try to state the whole Bayesian model, and then make simplifying assumptions (possibly drastic) to arrive at something tractable.



              There are probably many non-Bayesian methods, which amount to specific simplifications of the Bayesian approach, making use of particular assumptions, possibly unstated. My advice is, don't resort immediately to simplifications, instead work out the whole model and consider what are reasonable and workable assumptions for your problem.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Nov 8 at 17:59









              Robert Dodier

              1968




              1968












              • I'm curious what you think about what I suggested in a comment to the question.
                – Cedric Eveleigh
                Nov 11 at 1:13


















              • I'm curious what you think about what I suggested in a comment to the question.
                – Cedric Eveleigh
                Nov 11 at 1:13
















              I'm curious what you think about what I suggested in a comment to the question.
              – Cedric Eveleigh
              Nov 11 at 1:13




              I'm curious what you think about what I suggested in a comment to the question.
              – Cedric Eveleigh
              Nov 11 at 1:13











              0














              I found a way to accomplish the propagation of uncertainty: Combine $A$ and $B$ into a single term—say, $C$—and curve fit to determine $C$. Then, solve $C = A + B$ for $A$ and calculate the uncertainty of $A$ with the uncertainties of $C$ and $B$.






              share|cite|improve this answer


























                0














                I found a way to accomplish the propagation of uncertainty: Combine $A$ and $B$ into a single term—say, $C$—and curve fit to determine $C$. Then, solve $C = A + B$ for $A$ and calculate the uncertainty of $A$ with the uncertainties of $C$ and $B$.






                share|cite|improve this answer
























                  0












                  0








                  0






                  I found a way to accomplish the propagation of uncertainty: Combine $A$ and $B$ into a single term—say, $C$—and curve fit to determine $C$. Then, solve $C = A + B$ for $A$ and calculate the uncertainty of $A$ with the uncertainties of $C$ and $B$.






                  share|cite|improve this answer












                  I found a way to accomplish the propagation of uncertainty: Combine $A$ and $B$ into a single term—say, $C$—and curve fit to determine $C$. Then, solve $C = A + B$ for $A$ and calculate the uncertainty of $A$ with the uncertainties of $C$ and $B$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 24 at 19:49









                  Cedric Eveleigh

                  113




                  113






























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