Dear Mats Lindqvist,
I've found several possible explanations, so please check if anyone results to be your case:
1) As it is, g is not a probability distribution, because the convolution using conv simply multiplies elements from the vectors without taking into consideration the size of the bin. In other words, you have to multiply/divide by the bin size (in your example, 0.01) each time you convolve (in your example, twice), depending on how you define the uniform distributions (see item 3).
2) To get the integral of g equal to unity, you might need to consider a bin size smaller than the precision of the limits of the uniform distributions. In your example, the precision is 0.001 (take a look at Di_h_max, for example). Otherwise, you will encounter some border effects depending on how you define the uniform distribution (i.e. the integral will slightly differ from 1).
3) You might need to be careful about how you define the uniform distributions in general (as samples from a continuous distribution or as a discrete distribution) and at the edges of the intervals where the distribution is different from zero, especially in your particular discretization.
In your example, if you take x=0.0005:.001:.4 (so that each bin is defined in the centre and the size of the bin matches the precision of the borders of the uniform distributions), you define the uniform distributions as if they were continuous (i.e., assuming that a<b are the edges, then for each x, the value of unifpdf(x) is 1/(b-a)) and you multiply each convolution for the size of the bin (the differential in the convolution integration) when you obtain a probability g that, when integrated (i.e. sum all the terms and multiply by the size of the bin) is unity, and thus it is a proper piece-wise constant continuous probability distribution.
Does this solve your problem? Or did I misunderstand your question?
Best regards, Hugo
