If you cannot directly call histogram to bin your data while visualizing it but need to visualize the output of one or more calls to histcounts (for example if you call histcounts repeatedly with a common set of bin edges and a subset of your data each time then add the bin counts together), call histogram with the 'BinCounts' and 'BinEdges' properties. Compare the two figures created by this example: rng default x = randn(10000, 1); y = randn(10000, 1) + 5; [minCombined, maxCombined] = bounds([x; y]); BE = linspace(minCombined, maxCombined, 100); figure h1 = histogram([x; y], BE); bincounts = histcounts(x, BE); bincounts2 = histcounts(y, BE); figure; h2 = histogram('BinCounts', bincounts + bincounts2, 'BinEdges', BE); While I did concatenate x and y together to identify the minimum and maximum of the combined data to generate a set of bins that would work for both data sets, you may be able to generate a common set of bin edges without needing x and y to be in memory simultaneously. If you're planning on working with data too large to fit in memory regularly, I would like to point out that the histcounts and histogram functions support tall arrays with some limitations given in the Extended Capabilities section on their documentation pages.

Histogramming 1 to 100 inclusive with h = histogram(1:100, 'BinWidth', 50) gives:

Let's see the bin edges:

h.BinEdges

ans =

     0    50   100

From MATLAB's help:

Each bin includes the left edge, but does not include the right edge, except for the last bin which includes both edges

That means that values 1 to 100 are histogrammed in this format:

Bin 1 => edges: [0 50) => Included values: [1, 2, 3, .., 49]         (n = 49)
Bin 2 => edges: [50 100] => Included values: [50, 51, 52, .., 100]   (n = 51)

histcount(X) partitions X in the same manner as histogram(X). Therefore, the results are what you should expect and in fact very reasonable.

For probability, each element in the output is the number of elements in the input that fall into that bin divided by the total number of elements in the input. So if you sum the elements in the output, what you get is the total number of elements in the input that fall into any of the bins divided by the total number. That's why its row in the table in the description of the 'Normalization' name-value argument says "The sum of the bin values is less than or equal to 1." It can be less than 1 if the 'BinLimits' or 'BinEdges' that you specified exclude one or more of the points in the input from being assigned into any of the bins, for example. For pdf, each element in the output is the number of elements in the input that fall into that bin divided by the product of the width of the bin and the total number of elements in the input. If each of your bins were 1 unit wide, the 'pdf' and the 'probability' would be the same. If each of your bins were 0.1 units wide, each element in the output normalized by 'pdf' would be ten times as large as the corresponding element in the output normalized by 'probability' and if I summed the output of 'pdf' normalization I'd expect to get a result of 10. x = randn(1, 1e5); prob_BW1 = histcounts(x, 'BinWidth', 1, 'Normalization', 'probability'); pdf_BW1 = histcounts(x, 'BinWidth', 1, 'Normalization', 'pdf'); prob_BWtenth = histcounts(x, 'BinWidth', 0.1, 'Normalization', 'probability'); pdf_BWtenth = histcounts(x, 'BinWidth', 0.1, 'Normalization', 'pdf'); format longg shouldBeSame = [prob_BW1.', pdf_BW1.'] BWtenth_results = [prob_BWtenth; pdf_BWtenth; pdf_BWtenth./prob_BWtenth].' All the elements in the third column of BWtenth_results are either 10 (or close to it) or NaN (if there's no data in x that fell into that particular bin.) And as I said above, the sum of the probabilities is 1 but the sum of the PDF values is 10 because the bin width was 1/10. [sum(prob_BWtenth), sum(pdf_BWtenth)] All those calculations I did assumed that the bin width was the same for each bin. If your bins had different widths (because you selected a non-uniformly spaced set of BinEdges) then the equivalent of the third column of BWtenth_results for that set of bins would reflect the spacing for each different bin.