2
$\begingroup$

I'm reading this paper about comparing histograms ["Comprehensive Survey on Distance/Similarity Measures between Probability Density Functions" by Sung-Hyuk Cha] that writes in the introduction:

Here only the nominal type histogram where each level or bin is independent from other levels or bins is considered

and

All measures appearing in this paper have the shuffling invariant property and thus naturally imply the level independency.

I tried Googling these terms, but couldn't find a definition. What are they?

$\endgroup$
0
1
$\begingroup$

They simply mean this: any measure discussed in a paper is a measure between two quantitative variables $X$ and $Y$ which values are proportions or probabilities. Each case $i$ (= observation, data point) - and all the cases are independent of each other - is thus a pair $(X_i,Y_i)$. Sum of values across all cases (cases in the dataset) in $X$ is 1, likewise as in $Y$. Therefore the speach is about aggregated data: each case is a class or category and value is a proportion of the event occured. The order of classes (cases) in the data is arbitrary. The measures discussed in the article are insensitive to the order of the classes. If the classes have to be bins of a discretized continuous scale then the question how to discretize it beforehand is out of the scope of the article.

Now, they simply consider graphically each case/class to be a unique bin on a histogram which in so doing they equate with frequency barchart of a categorical variable (each category = unique data case). So, you're comparing - if to speak graphically - two barcharts; you are in right to reorder bars (coherently in both) as you like, because you are in right to reorder cases in the dataset.

That (very nice) article's name actually misses, for me, the bracketed: "Comprehensive Survey on Distance/Similarity Measures between [two discretized, categorized] Probability Density Functions". That would be more clear.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.