Percentage Exponential Distribution

[Desolrit]

I have 15 values (non-numerical), each assigned a percentage of probability;
-1st is 80%
-15th is 0.1%

How would I assign the remaining 19.9% to the other 13 values, the chance exponentially decreasing from 80% and 0.1%?

Thanks in advance.

 

[Ishuda]

I have 15 values (non-numerical), each assigned a percentage of probability;
-1st is 80%
-15th is 0.1%

How would I assign the remaining 19.9% to the other 13 values, the chance exponentially decreasing from 80% and 0.1%?

Thanks in advance.

You may, or may not, be able to get a simple geometric expression. First, lets assume a simple geometric progression so that
a_j = A r^j, j = 0, 1, 2, ..., 14
Since the sum of all the probabilities must be 1, we have
\(\displaystyle A \overset{14}{\underset{j=0}{\Sigma}} r^j = 1\)
or
r = 1 - A (1 - r¹⁵)
The reason I wrote it that way is that iteration scheme, i.e.
r_j+1 = 1 - A (1 - r_j¹⁵)
with r₀=0.5 converges fairly fast for reasonable A. You would then have to check to see if a_j were equal to 0.001. It doesn't in this case

 

[Ishuda]

Just because I got interested. Notice there is some overflow/underflow (accuracy) problems for very small A.

A	a(14)	r
0.010	0.010	1.000
0.012	0.012	1.000
0.014	0.014	1.000
0.017	0.017	1.000
0.021	0.021	1.000
0.025	0.025	1.000
0.030	0.030	1.000
0.036	0.035	0.999
0.043	0.041	0.997
0.052	0.047	0.994
0.062	0.051	0.986
0.074	0.050	0.973
0.089	0.045	0.952
0.107	0.037	0.927
0.128	0.028	0.896
0.154	0.019	0.863
0.185	0.013	0.826
0.222	0.007	0.784
0.266	0.004	0.736
0.319	0.001	0.682
0.383	0.000	0.617
0.460	0.000	0.540
0.552	0.000	0.448

 

[Ishuda]

Exponentially means compounding; changing your range:
decreasing from 800 to 1 (same thing!).
Same as going the other way: 1 to 800 in 15 years;
a deposit of $1 grow to $800 in 14 years:
1(1 + i)^14 = 800 ; solve for i : ~.612 (61.2 %).

1,2,3,5,7,11,18,28,46,73,118,191,308,496,800

Above are rounded; as example, the 5th term "7" is really 6.752....

Example of calculation (last 2 terms): 496 * 1.612 = 800.

What you have said is true but only considers the number as numbers and not as probabilities. That is the numbers need to be normalized so that their sum is 1 if they are to be treated as (exclusive) probabilities. So, letting x = 1+i ~ 1.612, we want
a \(\displaystyle \Sigma_{i=0}^{14}\, x^i\) = 1
or
a = (x-1)/(x¹⁵ - 1) ~ .000475
or that 800 would turn into about 0.38. Please note that in the table provided A=0.383 gives an r of 0.617 or an x = 1/r = 1.621, that is the same to two decimal places which is probably what one could expect toward either end of the table.