Darts:
An Introduction to Probability with Calculus

We begin by throwing some darts at a unit circle dartboard with GeoGebra

 Use the reset to throw the darts again. This is a Java Applet created using GeoGebra from www.geogebra.org - it looks like you don't have Java installed, please go to www.java.com

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:

• How many darts fall within a distance of 1/2 for the center?

• What do you think the average value of R will be?

For now let's return to the simple experiment.

With the darts falling at random anywhere in the circle it should seem reasonable that:

The probability the dart falls into any particular region R inside the circle is proportional to ratio of the area A of that region to the area of the unit circle, i.e., `A/pi` With a concentric circle of radius 1/2, the ratio of the area of the circle with radius 1/2 to the unit circle is   `{pi/4} / pi = 1/4`.

We generalize and define the probability distribution function F for the random variable R by

F(A) = probability that  R `<=` A

In the case of the dart variable R,

 0   when A `<=` 0 F(A) = A2 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, B-A. That is,

`barF(A,B) = {F(B)-F(A)}/ {B-A}`.

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

`barF(1/4,3/8) = {9/64 - 4/64}/{3/8 - 1/4} = 5/8`;

while for the intervals `[3/4,7/8]` the average density is

`barF(3/4,7/8) = {49/64 - 36/64}/{7/8 - 3/4} = 13/8`

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) = `lim` `{F(B) - F(A)}/{B - A}` `B rarr A`
= F '(A) = the derivative of the function F at A.

Thus in the case of the darts,
dF(A)= 2A  for `0<=A<=1`,
and dF(A) = 0 for all other 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.

1. `dF(A) >= 0` for all A provided it exists.

2. DENSITY AND THE FUNDAMENTAL THEOREM OF CALCULUS
`int_A^B dF(x) dx = F(B) - F(A)`

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 GeoGebra.

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)*(B-A)`.

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 finally add those numbers up to find an estimate for the average, i.e.

Average value of R ` ~~ sum_{k=1}^{k=N} r_k*p_k`.

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 `
`=  {3+10 + 21+36}/125 = 70/125 = 14/25 = .56`

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_{k-1})  ~~  dF(A_{k-1}) *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_{k-1})*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`.

Final Comment:  For any  random  variable, X, where X has values between A and B,  F  is used to denote the distribution  function  and f  is used to denote the density function.

When F and f  are continuous functions, the mean  of the random variable  is `int_A^B x f(x) dx `.

In particular when the variable X is uniform in the sense that the likelihood of  X falling within any interval is the same for intervals of equal length, then `f(x) = 1/{B-A}`.

If the random variable is given by evaluating a function P at X, so R = P(X), then the average value for R is

`int_A^B P(x) f(x) dx = 1/{B-A} int_A^B P(x) dx` .

This is usually described as the average value of the function P over the interval [A,B].