Histogram with different sample probabilities Assume we are given a list of samples $L_1,L_2,\ldots,L_n$ of some random variable $L$. By classing them into bins we can easily create a standard histogram.
But now suppose that we associate a probability $p_i$ with each $L_i$. Of course, $\sum_i^n p_i=1$. In the standard case case we have $p_i=1/n$. But if we have very different probabilities,then how would one have to change the plot of a historgram? Is it even possible?
For example, let $n=10$ and $p_1=\ldots=p_{5}=0.05$ and $p_6=\ldots=p_{10}=0.15$, i.e. the samples $L_6,\ldots,L_{10}$ are more probable than the other five samples. But how would this be reflected in the histogram?
 A: You are confusing a probability distribution with the distribution of a sample. The sample is what it is, you do not weight them. If, in fact, a certain point is more probable than another, then it should, on average, come up more often in a sample. For a simple example, consider a die where the sides are $\{1,1,1,2,3,4\}$
If I rolled the die $n$ times, I'd expect to see more $1$'s than any other number. Thus, when you bin by value, this increased probability will automatically be incorporated. Note that it is possible, but less probable, that you will get more $2$'s than $1$'s, in which case your sample will be misleading....this is totally normal and an expected result, especially with small samples.
So, bottom line is that a histogram is a descriptive statistical tool and hence relies only on the sample data -- probability doesn't enter into its construction. On the opposite end, a probability distribution defines the probability of each point, and doesn't rely on a sample.
you are trying to do something inconsistent. If you know the probability, the you don't need the sample, if you don't, then how are you coming up with your probabilities? I would think that the ECDF of the data would be the best estimate of your distribution.
