Exponential Decay fit and printed parameters with integer data [closed]

Question

Problem:

I'm trying to plot an exponential decay curve on a set of integer data, and print the various parameters and their uncertainties on the plot itself. Having adapted code from another question, this worked successfully with non-integer values with uncertainties in the data set. Here the data set is integer only and does not contain uncertainties and my attempt at adapting it has been unsuccessful.

The target output is something like this:

Code: (contains data)

dataHist5 = {{18, 74}, {36, 64}, {54, 62}, {72, 54}, {90, 47}, {108, 
    39}, {126, 40}, {144, 35}, {162, 34}, {180, 29}, {198, 34}, {216, 
    30}, {234, 31}, {252, 22}, {270, 14}, {288, 14}, {306, 13}, {324, 
    25}, {342, 18}, {360, 11}, {378, 13}, {396, 16}, {414, 13}, {432, 
    12}, {450, 10}, {468, 12}, {486, 11}, {504, 13}, {522, 9}, {540, 
    8}, {558, 7}, {576, 5}, {594, 5}, {612, 4}, {630, 5}, {648, 
    2}, {666, 1}, {684, 3}, {702, 2}, {720, 2}, {738, 1}, {756, 
    1}, {774, 1}, {792, 0}, {810, 0}, {828, 1}, {846, 2}, {864, 
    0}, {882, 1}, {900, 1}};
fitData = dataHist5;

Clear[A, k]
uncertainties = sqrt[dataHist5[[2]]];
fit = NonlinearModelFit[fitData, A Exp[-k t], {A, k}, t, 
  Weights -> 1/dataHist5[[2]]]

{A, k} = {A, k} /. fit["BestFitParameters"];
{\[Sigma]A, \[Sigma]k} = fit["ParameterErrors"];
hLife = Log[2]/Around[k, \[Sigma]k];
halfLife = hLife[[1]];
seA = Around[A, \[Sigma]A];
sehalfLife = hLife[[2]];

Show[Plot[fit[x], {x, 0, 900}, PlotRange -> All, 
  PlotTheme -> "Detailed", PlotStyle -> Red, Axes -> False, 
  Frame -> {{True, False}, {True, False}}, 
  FrameLabel -> {"Time /s", 
    "Counts Recorded in the Previous 15 seconds"}, 
  ImageSize -> Large], ListPlot[dataHist5, ImageSize -> Large], 
 Graphics[Inset[
   Framed[Column[{Style["Run 0", Bold], 
      Row[{"Data Points = ", Length[dataHist5], "/50"}], 
      Row[{Subscript[t, Style["1/2", FontSize -> 10]], " = ", 
        PlusMinus[NumberForm[halfLife, 4], 
         NumberForm[sehalfLife, 3]]}], 
      Row[{"A = " PlusMinus[Round[A], Round[\[Sigma]A]]}],
      Row[{"\[Lambda] = " PlusMinus[NumberForm[k, 3], 
          NumberForm[\[Sigma]k, 2]]}], 
      Row[{Superscript[\[Chi], 2], "= ", 
        NumberForm[fit["ANOVATableSumsOfSquares"][[2]], 4]}], 
      Row[{"Reduced " Superscript[\[Chi], 2], "= ", 
        NumberForm[fit["ANOVATableMeanSquares"][[2]], 3]}]}], 
    Background -> White, RoundingRadius -> 5], {Right, Top}, 
   Scaled[{1.1, 1.2}]]], 
 PlotLabel -> 
  Style["Decay Curve of Phosphorus-30 by \[Beta]+ Emission", Bold]]

Current Output:

Among many errors are 'NonlinearModelFit::wtsln: The number of weights 2 specified by Weights->{0.0277778,0.015625} is not the same as the number of data points 50.' and 'Set::shape: Lists {A,k} and {A,k}/. NonlinearModelFit[<<1>>][BestFitParameters] are not the same shape.'(though there are other errors as well)

Attempted Solution (Following JimB's answer):

The ParameterTable elements are seemingly being pulled, but not the DevianceTable ones, and I'm not super sure I've worked out the uncertainty in A and t correctly. Also the axes labels aren't appearing.

glm = GeneralizedLinearModelFit[dataHist5, t, t, 
  ExponentialFamily -> "Poisson"]

halfLife = (Log[E, 2]/glm["ParameterTableEntries"][[2, 1]])
sehalfLife = (((glm["ParameterTableEntries"][[2, 2]])/(glm[
    "ParameterTableEntries"][[2, 1]])))*halfLife
k = glm["ParameterTableEntries"][[2, 1]]
\[Sigma]k = glm["ParameterTableEntries"][[2, 2]]
edp = glm["DevianceTableEntries"][[4, 2]]
redp = (glm["DevianceTableEntries"][[4, 2]])/(glm[
     "DevianceTableEntries"][[3, 2]])
A = E^(glm["ParameterTableEntries"][[1, 1]])
\[Sigma]A = (A*(glm["ParameterTableEntries"][[1, 1]])/E)

Show[ListPlot[dataHist5], 
 Plot[glm[t], {t, 0, 900}, PlotRange -> All, PlotTheme -> "Detailed", 
  PlotStyle -> Red, Axes -> False, 
  Frame -> {{True, False}, {True, False}}, 
  FrameLabel -> {"Time /s", 
    "Counts Recorded in the Previous 15 seconds"}, 
  ImageSize -> Large], 
 Graphics[Inset[
   Framed[Column[{Style["Run 0", Bold], 
      Row[{"Data Points = ", Length[dataHist5], "/50"}], 
      Row[{Subscript[t, Style["1/2", FontSize -> 10]], " = ", 
        PlusMinus[NumberForm[halfLife, 4], 
         NumberForm[sehalfLife, 3]]}], 
      Row[{"A = " PlusMinus[Round[A], Round[\[Sigma]A]]}],
      Row[{"\[Lambda] = " PlusMinus[NumberForm[k, 3], 
          NumberForm[\[Sigma]k, 2]]}], 
      Row[{Superscript[\[Chi], 2], "= ", NumberForm[edp, 4]}], 
      Row[{"Reduced " Superscript[\[Chi], 2], "= ", 
        NumberForm[redp, 3]}]}], Background -> White, 
    RoundingRadius -> 5], {Right, Top}, Scaled[{1.1, 1.2}]]], 
 PlotLabel -> 
  Style["Decay Curve of Phosphorus-30 by \[Beta]+ Emission", Bold]]

Answer 1

If the "integer data" consists of counts that might be expected to follow a Poisson distribution given the predictor value, then a Poisson regression should be considered. That can be accomplished using GeneralizedLinearModelFit.

glm = GeneralizedLinearModelFit[dataHist5, t, t, ExponentialFamily -> "Poisson"]
Show[ListPlot[dataHist5], Plot[glm[t], {t, 0, 900}]]

One of the characteristics of a lack-of-fit is that the variability about the fitted curve is larger (or at least different) from what one would expect when there is Poisson variability. A parameter that characterizes that is the "dispersion" parameter. It should be around 1.0 if the observed variability is what is expected with a Poisson distribution.

GeneralizedLinearModelFit has an option for "EstimatedDispersion" but it appears to always have the value 1.0 so I don't know what it is doing. A more reasonable estimator is residual deviance divided by the residual degrees of freedom:

glm["DevianceTable"]

The estimated dispersion parameter is 42.4657/48 = 0.884702 which is a bit less than one suggesting that there is no additional variation in the data to be explained after the Poisson model is considered.

So what model is actually being fit?

$$Y|t \sim \text{Poisson}(e^{a+b t})$$

We have the coefficients of the fit:

glm["ParameterTable"]

So the estimate of the intercept ($a$) is found in the row labeled 1 and the estimate of the slope ($b$) is found in the row labeled t.

In short in this case there is no need to use weights.