- 31 May '15 07:59 / 15 editsI have just a few questions about the cumulative distribution probability of fractiles;

(see:

http://en.wiktionary.org/wiki/fractile

“...The value of a distribution for which some fraction of the sample lies below.

The q-quantile is the same as the (1/q)-fractile.

The median is the .5-fractile.

...“

also see;

http://en.wikipedia.org/wiki/Cumulative_distribution_function

particularly where it says;

" ...random variable X with a given probability distribution will be found to have a value**less than or equal to x**...." )

But, although I make perfect sense of it for continuous random variables, it isn't completely clear to me how that works for discrete random variables.

I explain with an example:

Suppose we have a discrete random variable v that can only be one of 3 values; v=0 or v=1 or v=2.

To keep this simple, lets say this is a uniform distribution i.e. each of these 3 possible values has equal probability and thus each has a probability of 1/3.

Now, lets imagine representing this with a simple bar chart with a bar for each of those 3 v values and the area of each bar representing its probability (with all 3 bars having equal width so its probability is also proportional to its height and all 3 bars thus having the same height )

What is the cumulative distribution probability of the 0.5 fractile of this probability distribution?

The http://en.wiktionary.org/wiki/fractile says;

“The value of a distribution for which some fraction of the sample lies below. “

But where exactly does that mean we should draw our imaginary vertical line for the 0.5 fractile on our bar chart so that the total integral to the left of that vertical line is the cumulative distribution of that 0.5 fractile?

Should we draw that imaginary vertical line exactly between the v=0 bar and the v=1 bar i.e. the left-hand side of the v=1? (in which case that would make the cumulative distribution of 0.5 fractile = 1/3 probability )

Or should we draw imaginary vertical line exactly through the centre of the v=1 bar i.e. to cut it into two equal halves? (in which case that would make the cumulative distribution of 0.5 fractile = 1/2 probability )

Or should we draw imaginary vertical line exactly between v=1 bar and the v=2 bar i.e. the right-hand side of the v=1 bar? (in which case that would make the cumulative distribution of 0.5 fractile = 2/3 probability )

Also; where should we draw this imaginary vertical line for the 0.5 fractile if we had 4 bars for the possible values of v=0 or v=1 or v=2 or v=3 all with equal probability i.e. each of the 4 bars is of equal height and thus has ¼ probability? - 31 May '15 08:55 / 9 editshang on! Not sure but I think I might have got it!

I think wasn't thinking clearly about the definition of "fractile" but, if I am now thinking about this right, the answer is, for my 3 bar example (my first example ), the vertical line should cut straight in the middle of the v=1 bar so that the cumulative probability of the 0.5 fractile is simply 1/2.

And, for my 4 bar example (my second example ), the vertical line should go exactly between the v=1 bar and the v=2 bar so that the cumulative probability of the 0.5 fractile is also simply 1/2.

More generally, and only if I am thinking about it completely correctly now, EVEN always for a*discrete*random variable;

**The cumulative probability of n-fractile ( 0 <= n <= 1 ) is simply n.**

But, because this is critical to the research I am currently doing, I**MUST**make absolutely sure I understood that correctly before moving on.

So;

Anyone:

Have I now got that exactly right?