We begin by throwing some darts
at a unit circle dartboard with GeoGebra.
The dots mark the places where the darts land.
We'll keep track of a random variable R , which measures the distance the dart lands from the center.
Two Questions:
For now let's return to the simple experiment.
With the darts falling at random anywhere in the circle it should seem reasonable that:
We generalize and define the probability distribution function F for the random variable R by
In the case of the dart variable R,
{ 
0 when A `<=` 0 

F(A) = 
A^{2} when 0< A <1 

1 when A `>=` 1 
The probability that a dart falls in any particular band (called an annulus) formed by concentric circles is also easy to calculate from the areas.
The probability that `A < R <= B` is just `F(B)  F(A)`.
With this analysis it should be clear that the probability that R = A is zero since the circle of radius A is a region in the plane with area zero.
This result can be interpreted as saying that the likelihood of the dart landing on a circle of a given radius is very small. And in an experiment, any specified number A from 0 to 1 is equally likely to occur as the value of R.
Yet the formula above also suggests that the probability that the value of R will lie between `1/8` and `1/4` is not as large as the probability that R will lie between `3/4` and `7/8`.
This leads to the concepts of average probability density and point probability density.
The average probability density for an interval [A,B] is the ratio of the probability that R will fall in a certain interval [A,B] to the length of that interval, BA. That is,
The densities for the intervals `[1/4,3/8]` and `[3/4,7/8]` illustrate why larger values of R are more likely by measuring the average density of comparable length intervals that contain them.
For the interval `[1/4,3/8]` we have the average density is
while for the intervals `[3/4,7/8]` the average density is
The point probability density of the random variable
R at the point A, dF(A), is the limit as `B rarr A` of the average probability
densities for intervals with endpoints A and B. So
dF(A) = 



This is the key relation between the distribution function
F and the probability density function of a random variable.
REMARKS on the DENSITY FUNCTION.
So to find the probability that a random variable is between A and B we need only integrate its density function from A to B.
Density and darts.
One way to estimate the definite integral
of the density function over an interval [A,B] is to THROW DARTS at the
entire region under the graph of the density function and above the X 
axis for the range of the random variable.
Then .. determine the proportion of darts that have fallen with first coordinate between A and B.
MEDIAN of a random variable.
Let's return to our darts. One problem is to find a value for A so that the probability that `R <= A` is the same as the probability that R > A. It should seem reasonable that this probability is 1/2. This number is called the MEDIAN of R.
So... we want `F(A) = A^2 = 1/2`.
Not too hard for algebra... the MEDIAN of R is `A = {sqrt{2}}/2` .
Let's check by running the darts on X(PLORE).
Finally , let's return to the first question of the average or what is called the MEAN of the random variable:
First let's note that when `B~~ A, F(B)  F(A)~~dF(A)*(BA)`.
Let's cut the interval from 0 to 1 into N pieces of equal length. For example if N = 5 we would have the intervals with length 1/5.
Now when N is large the length of these intervals will get small and there won't be much variation of the value of R in that interval.
To estimate the average from the theory, it would seem
sensible
to choose one number to represent the numbers from each interval  call it
"`r_k` ". Now estimate the probability that a dart would fall in that interval,
call that "`p_k`". Then, multiply the representative number by that probability
and finaly add those numbers up to find an estimate for the average, i.e.,
If we choose the right hand endpoint of each interval we would find an overestimate:
Can you
see why?
For example when N = 5 we would find
overest `= r_1*p_1 + r_2*p_2 + r_3*p_3 + r_4*p_4 + r_5*p_5`
`= 1/5*1/25 + 2/5*3/25 + 3/5*5/25 + 4/5 * 7/25 + 5/5 * 9/25 `
`= {1+6+15+28+45}/125 = 95/125 = 19/25 = .76`
This would be an overestimate for the average.
For an underestimate we would use the left hand endpoints....
underest `= r_0*p_0 + r_1*p_1 + r_2*p_2 + r_3*p_3 + r_4*p_4 `
`= 0 + 1/5*3/25 + 2/5*5/25 + 3/5*7/25 + 4/5 * 9/25 `These two estimates have an average averest = (overest +underest)/2 = `{.76+.56}/2`= .66
which is still an underestimate. Can you tell why?
THE MEAN MEETS THE RIEMANN INTEGRAL:
Now remember that `p_k = F(A_k)  F(A_{k1}) ~~ dF(A_{k1}) *1/N`.
So to estimate the average value of R theoretically we could consider
`sum_{k=1}^{k=N} r_k*p_k ~~ sum_{k=1}^{k=N} r_k *dF(A_{k1})*1/N`.
AHahhh! this last expression is precisely a Riemann Sum that estimates the definite integral
`int_0^1 x * dF(x) dx`
Thus the MEAN of the random variable R must be
`int_0^1 x * 2x = 2{x^3}/3 _0^1 = 2/3`.