# How to relate a binned pdf to a pdf, taking into account how the data is binned?

I have a data set which I represent as a PDF histogram Histogram[Data,{"Raw", NumberOfBins},"PDF"]. I want to know the distribution parameters of my data.

I've seen that we can use FindDistributionParameters[Data, NormalDistribution[mu, sigma]] to determine distribution parameter values. However how one bins data can really change how the distribution manifests. For example data that may appear Gaussian distributed, may have a slight exponential tilt only revealed when binning is fine enough. So is it possible to find distribution parameters in a way that are tied to the way it is binned?

• If you can avoid it, don't bin. If you end up with binned data, then do as @kglr shows with WeightedData (and hopefully with an associated bootstrap to get estimates of standard errors) or use a maximum likelihood approach which uses the CDF function which gives you the probability of getting a particular count in a bin which in turn goes into the log likelihood function that you'll choose the parameters that maximize the log likelihood. You can also get estimates of standard errors. (If someone doesn't write that up, I will after I get some pizza and beer for dinner.)
– JimB
Aug 10, 2019 at 1:52
• But maybe this is a duplicate: mathematica.stackexchange.com/questions/39679/….
– JimB
Aug 10, 2019 at 1:53
• Interesting, maybe I am looking at PDF's in the wrong way. Previously my method was to produce histogram, defining the number of bins however. Then extract the centre of each bin and associated count/PDF value. Then fit my distribution to this. With this method clearly the way you bin data will affect the resultant fit/PDF parameters. Enjoy your dinner. Aug 10, 2019 at 1:57
• You must have learned that technique a long time ago or had a non-statistics professor that learned it a long time ago.
– JimB
Aug 10, 2019 at 5:13

Don't bin if you don't have to bin. But if that's all you have....

If you have binned data with $$n$$ bins, boundaries $$x_ 1< x_ 2< \cdots < x_ {n + 1}$$, and counts $$c_ 1, c_ 2, \ldots, c_n$$ for a proposed distribution with cumulative distribution function (CDF) $$F$$, then maximum likelihood estimators are the values of the parameters that maximum the likelihood. Usually the log of the likelihood is maximized as that can be more numerically stable when iteration is needed and sometimes results in simple closed-form estimators. We have

$$log (L) = \sum_ {i = 1}^n c_i \log (F (x_ {i + 1}) - F (x_ {i}))$$

Here is some code when the distribution is normal with unknown mean and variance:

(* Random sample from a known distribution *)
SeedRandom[12345];
n = 10000;
data = RandomVariate[NormalDistribution[5, 3], n];

(* Create a histogam *)
nBins = 20;
h = HistogramList[data, nBins];
(* Bin boundaries *)
x = h[[1]]
(* {-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18} *)
(* Frequency counts *)
c = h[[2]]
(* {4,7,27,65,136,244,443,656,949,1234,1299,1292,1148,932,690,420,250,122,53,17,9,0,2,1} *)

(* Find the log of the likelihood for the binned data *)
logL = Total[Table[c[[i]] Log[CDF[NormalDistribution[μ, σ], x[[i + 1]]] -
CDF[NormalDistribution[μ, σ], x[[i]]]], {i, nBins}]];

(* Find values of μ and σ that maximize the log of the likelihood *)
(* Initial values *)
(μ0 =  Sum[c[[i]] (x[[i + 1]] + x[[i]])/2, {i, nBins}]/Total[c]) // N
(* 4.9439 *)
(σ0 = (Sum[c[[i]] ((x[[i + 1]] + x[[i]])/2 - μ0)^2, {i, nBins}]/Total[c])^(1/2)) // N
(* 2.9738228281705013 *)
(* Maximim likelihood estimates *)
mle = FindMaximum[{logL, σ > 0}, {{μ, μ0}, {σ, σ0}}]
(* {-25063.7, {μ -> 4.94984, σ -> 2.96156}} *)

(* Now get estimates of the associated standard errors *)
(covMat = -Inverse[D[logL, {{μ, σ}, 2}] /. mle[[2]]]) // MatrixForm
seμ = covMat[[1, 1]]^0.5
(* 0.029773837258604677 *)
seσ = covMat[[2, 2]]^0.5
(* 0.021152624920503942 *)

(* Display histogram and estimated density *)
Show[Histogram[data, nBins, "PDF"],
Plot[PDF[NormalDistribution[μ, σ] /. mle[[2]], z], {z, x[[1]], x[[nBins + 1]]}]]

For example data that may appear Gaussian distributed, may have a slight exponential tilt only revealed when binning is fine enough.

is true but has nothing to do with fitting a specific distribution. The fit is conditional on assuming the form of the distribution (i.e., the form known but not necessarily all of the parameters). If you suspect deviations from a particular distribution then you need to try different forms of distributions or fit a nonparametric density estimate (using SmoothHistogram or SmoothKernelDistribution) but that requires un-binned data.

• Okay, I'm going to accept your answer, as you also demonstrated why my approach is flawed and I shouldn't be using histograms to extract distribution parameters -- rather just use them as an illustration and diagnostics. Aug 10, 2019 at 12:27

You can (1) use HistogramDistribution with the same bin specification to get hd, (2) use the properties "PDFValues" and "BinDelimiters" of hd to construct a WeightedData object wd, (3) use FindDistributionParameters with wd as the first argument:

SeedRandom[1]
Data = RandomVariate[NormalDistribution[5, 3], 100];

FindDistributionParameters[Data, NormalDistribution[mu, sigma]]

{mu -> 4.97099, sigma -> 3.02726}

NumberOfBins = 5;
hd = HistogramDistribution[Data, {"Raw", NumberOfBins}];

hd["PDFValues"]

{0.0104376, 0.0782821, 0.13047, 0.0365317, 0.00521881}

hd["BinDelimiters"]

{-3.83229, 0., 3.83229, 7.66458, 11.4969, 15.3292}

wd = WeightedData[MovingAverage[hd["BinDelimiters"], 2], hd["PDFValues"]];
FindDistributionParameters[wd, NormalDistribution[mu, sigma]]

{mu -> 4.98198, sigma -> 3.06583}

NumberOfBins = 10;
hd = HistogramDistribution[Data, {"Raw", NumberOfBins}];

hd["PDFValues"]

{0.00587116, 0.0176135, 0.0880674, 0.0469693, 0.135037, 0.129166, 0 .105681, 0.035227, 0.0117423, 0.0117423}

hd["BinDelimiters"]

{-3.40648, -1.70324, 0., 1.70324, 3.40648, 5.10972, 6.81296, 8.51621, 10 .2194, 11.9227, 13.6259}

wd = WeightedData[MovingAverage[hd["BinDelimiters"], 2], hd["PDFValues"]];
FindDistributionParameters[wd, NormalDistribution[mu, sigma]]

{mu -> 4.9905, sigma -> 3.05878}

To see that Histogram with "PDF" as height specification and Plot of the PDF if hd give the same picture:

histogram = Histogram[Data, {"Raw", NumberOfBins}, "PDF", ChartStyle -> Blue,
ImageSize -> 300, Frame -> True, Axes -> False, AspectRatio -> 1];

pdfhd = ParametricPlot[{x, v PDF[hd, x]}, {x, -5, 15}, {v, 0, 1},
MeshFunctions -> {# + 50 #2 &}, Mesh -> 50, MeshStyle -> Thick,
MeshShading -> {Red, Opacity[0]}, PlotRange -> All,
AspectRatio -> 1, Axes -> False, ImageSize -> 300];

Row[{histogram, pdfhd, Show[histogram, pdfhd]}, Spacer[10]]

• Nice, let me play around with your solution to see if I understand it and it solves my problem. Thanks. Aug 10, 2019 at 1:59

If the objective is what you state below

For example data that may appear Gaussian distributed, may have a slight exponential tilt only revealed when binning is fine enough.

and the raw data is available, then departures from normality (or particular distributions of interest) can many times be best displayed as a nonparametric density estimate. Now that we have computers, histograms as so old-school.

Consider a mixture of two normal distributions. We can take a sample, estimate the probability density function, and compare that to a normal distribution with the same mean and variance. That way one can potentially see where there might be departures from a normal distribution: bimodality, skewness, etc.

(* Random sample from a known distribution *)
n = 500;
d = MixtureDistribution[{0.6, 0.4}, {NormalDistribution[5, 3], NormalDistribution[8, 1]}];
SeedRandom[12345];
data = RandomVariate[d, n];

(* Nonparametric density estimate *)
skd = SmoothKernelDistribution[data, "LeastSquaresCrossValidation"];

(* Mean and standard deviation of data *)
mean = Mean[data];
sd = StandardDeviation[data];

Plot[{PDF[d, x], PDF[NormalDistribution[mean, sd], x], PDF[skd, x]}, {x, Min[data], Max[data]},
PlotLegends -> {"True distribution", "Normal with same mean and sd",
"Nonparmetric density estimate"},
PlotStyle -> {Green, Blue, Red}]

• Thanks for taking the time to give an additional answer -- I'll have a good think about this. You've taught me a fair amount. What is the best way to choose a smoothing parameter in the SmoothKernelDistribution if your data isn't Gaussian. I've seen the rules of thumb here: en.wikipedia.org/wiki/… but these obviously apply for Gaussian data. Aug 11, 2019 at 16:18
• Probably the "LeastSquaresCrossValidation" option is overall best. Also, one needs to use the "Bounded" option if the domain is bounded in any way (i.e., say all values are between $0$ and $2\pi$). Looking at a QuantilePlot can also be helpful to look for departures from normality (or any other specified distribution).
– JimB
Aug 11, 2019 at 18:39