# Finding a distribution to fit truncated data

I have certain experimental data with the following PDF.

As you may note the data does not take values lower than 0.

Using FindDistribution I get the following fit:

Is there any good way to fit this kind of "truncated" data with any of the built in distributions?

I tried to use some non negative distributions in Target distributions:

 TargetFunctions -> {LogNormalDistribution, NormalDistribution,
ExponentialDistribution, InverseGaussianDistribution}


But didn't get better results.

The data for the Histogram using HistogramList is the following:

{{0., 0.005, 0.01, 0.015, 0.02, 0.025, 0.03, 0.035, 0.04, 0.045, 0.05,
0.055, 0.06, 0.065, 0.07, 0.075, 0.08, 0.085, 0.09, 0.095, 0.1,
0.105, 0.11, 0.115, 0.12, 0.125, 0.13, 0.135, 0.14, 0.145, 0.15,
0.155, 0.16, 0.165, 0.17, 0.175, 0.18, 0.185, 0.19, 0.195, 0.2,
0.205, 0.21, 0.215, 0.22, 0.225, 0.23, 0.235, 0.24, 0.245, 0.25,
0.255, 0.26, 0.265, 0.27, 0.275, 0.28, 0.285, 0.29, 0.295, 0.3,
0.305, 0.31, 0.315, 0.32, 0.325, 0.33, 0.335, 0.34}, {8.35598,
8.30599, 6.7505, 6.00847, 5.41484, 4.65615, 3.66651, 3.48715,
3.47286, 3.45858, 3.66571, 3.7673, 3.7546, 3.71889, 3.66492,
3.69746, 4.05458, 4.31727, 3.91094, 4.22997, 4.57044, 5.01803,
4.90375, 4.78233, 4.99502, 4.71249, 4.45139, 4.43949, 4.48314,
4.47123, 4.37917, 4.46488, 4.55615, 4.44187, 4.29664, 3.98316,
3.95221, 3.73317, 3.01257, 2.63084, 2.44751, 2.25228, 2.25149,
2.01023, 2.04991, 1.86659, 1.69675, 1.44597, 1.16582, 1.02694,
0.637274, 0.569023, 0.463472, 0.367444, 0.295225, 0.233323,
0.192849, 0.126979, 0.0880914, 0.0253957, 0.0261893, 0.0047617,
0.0198404, 0.0222213, 0.0253957, 0.0388872, 0.0111106, 0.00238085}}


Yo can see the data (around 250.00 points) in the following link:

• Can you provide the input data you used and put it e.g. on pastebin? – halirutan May 12 '17 at 3:25
• @halirutan The original data has around 250.000 values. I'll make it available trough mi google drive – BPinto May 12 '17 at 3:36
• Why does your data contain a lot of negative values? Assuming I have loaded your csv file into data can you please add how you achieved your first histogram? – halirutan May 12 '17 at 4:02
• @halirutan I'm sorry, I'll correct the file. The actual data is just the square of each value. – BPinto May 12 '17 at 4:06
• After some sleep, I reworked my answer to include all 4 peaks in your original data. Please see my edits. – halirutan May 12 '17 at 12:21

As you said in the comments, the data for your histograms did contain negatives and you squared all of them. I suggest to consider looking at your values before you square them

Although I have no insight into the underlying process that created the data, one could assume that you have 4 mixed normal distributions here. It seem the left peak is again divide into two distributions. Let us use this for a start and define a mixture distribution:

dist = MixtureDistribution[{a, b, c,
d}, {NormalDistribution[μ1, σ1],
NormalDistribution[μ2, σ2],
NormalDistribution[μ3, σ3],
NormalDistribution[μ4, σ4]}]


Now we can estimate the parameters by using FindDistributionParameters. I have roughly estimated the initial the initial conditions by just looking where the peaks are, and how high and wide they are:

sol =
FindDistributionParameters[data,
dist, {{a,
1}, {μ1, -.4}, {σ1, .1}, {b, .3}, {μ2, .1}, {\
σ2, .05}, {c, .5}, {μ3, .4}, {σ3, .1}, {d, .4}, {\
μ4, -.2}, {σ4, .1}}]
(* {a -> 0.376702, b -> 0.125485, c -> 0.275036,
d -> 0.222777, μ1 -> -0.395739, σ1 -> 0.0586256, μ2 ->
0.103749, σ2 -> 0.0496838, μ3 -> 0.337538, σ3 ->
0.0866439, μ4 -> -0.217508, σ4 -> 0.0911891} *)


With 250.000 data values, Mathematica has no problem finding a good fit. Looking at it reveals:

pdf1 = PDF[dist, x] /. sol;

Show[
Histogram[data, {0.03}, "PDF"],
Plot[PDF[dist, x] /. sol, {x, -.6, .6}]
]


Now, since you squared all your values, we need to transform the found mixture distribution as well and you will end up with a distribution that has only values for positive x and fits your data very nicely:

pdf = PDF[TransformedDistribution[u^2, u \[Distributed] dist], x] /. sol;
Show[
Histogram[data^2, {0.007}, "PDF"],
Plot[pdf, {x, 0, .6}]
]


Because your distribution is (at least) bimodal and because the probability density is greater than zero at zero, you might consider using nonparametric density estimation rather than maybe a poor approximation with a mixture of a small number of distributions. (However, if it works with a mixture distribution, then that's great.)

You've got lots and lots of data and you don't need to force a description with a mixture of standard distributions.

Many times just using SmoothKernelDistribution on the data gets you everything you need. In this case because the density is definitely not zero at the border of zero, one needs to do a bit of (commonly used) trickery but one still gets a legitimate and appropriate estimate of the probability density. (See Silverman, Page 20).

First one creates a dataset that includes a reflection of the data:

data2 = Flatten[{data, -data}];


Then the SmoothKernelDistribution function is used followed by truncating the resulting distribution to be between zero and $\infty$.

skd2 = SmoothKernelDistribution[data2];
skd = TruncatedDistribution[{0, ∞}, skd2];


One then has access to the functions associated with a probability distribution: PDF, CDF, Expectation, etc.

Plot[PDF[skd, x], {x, 0, 0.4}]


To repeat: when you've got lots of data you are not restricted to standard distributions. Really.

• Would it be possible to use the TransformedDistribution function on a distribution found by SmoothKernelDistribution? Or has this disadvantages? – halirutan May 12 '17 at 14:55
• Good idea. It would be nice if that worked and it would be more straightforward than what I did. But with this data I just get the command echoed back when attempting to use PDF and CDF with that transformed distribution. However, Expectation seems to work with it. – JimB May 12 '17 at 16:31
• @JimBaldwin I found that SmoothKernelDistribution[ data, Automatic, {"Bounded", {0, [Infinity]}, "Gaussian"}] also works and seem to make a better fit. Thanks for pointing out this solution. – BPinto May 13 '17 at 3:34
• The "Bounded" option works great. Thanks for pointing that out. One doesn't need to go through the "data reflection" trick that I included. But I would have to believe that's what's done behind the curtain. The fits seem similar to me but using the "Bounded" option does result in smaller bandwidths which means a "bumpier" curve. – JimB May 13 '17 at 15:51
• This seems very relevant: mathematica.stackexchange.com/questions/88908/…. – JimB May 14 '17 at 1:48