How Assign Exponentialy increasing probablities to a random number

Question

Suppose we have a random variables with values from 1 to 1000 and we have to assign exponential increasing values to them (1 has low probability and 1000 has highest but increase in probabilities is exponential). How can i write equation for this.

Second how can i randomly pick any one of the a random variables with chances that highest probability random variable is most likely to be returned.

Let me rephrase the problem, It is same as in https://stackoverflow.com/questions/10942318/random-numbers-based-on-a-probability only difference i explain here

considering S is random number generated to map to the numbers following exponential growth rate (exponential probability increase) for n=1 S=1 or 2 for n=2 S=3 or 4 or 5 or 6 (S=2pow(n) ) and so. How i can solve now with arithmetic series formula ( i think it is must ) as shown in link above for exponential distribution to get n .

Answer 1

For some $0 \le \alpha \le 1$, You can assign a probability of $\alpha$ to 1000, $\alpha^2$ to 999, ..., $\alpha^{1000}$ to 1. To make this a legitimate probability distribution (a probability mass function, strictly speaking), the probabilities must add up to 1. This implies:

$\sum_{n=1}^{1000} \alpha^n = \frac{\alpha(1-\alpha^{1000})}{1-\alpha} = 1$. You can solve numerically in a program like MATLAB or Python to calculate the value of $\alpha$ which satisfied this equation. Since the summation converges to $\frac{1}{1-\alpha}$, your answer will be very close to $\alpha = 0.5$.

Answer 2

In general you have an exponentially increasing probability on the integers $1$ to $n$ as long as the probability of each number $k$ ($k \neq 1$) is greater than the previous number $k-1$ and the ratio of those two numbers is the same fixed ratio for every $k.$

(Note: In the specific question asked above, $n=1000.$)

In other words, you need to set some fixed ratio $r$ such that $P(2)=r\cdot P(1),$ $P(3)=r^2\cdot P(1),$ $P(4)=r^3\cdot P(1),$ and so forth.

If you do this up to $P(n)=r^{n-1}\cdot P(1)$ and then say $n$ is the largest possible outcome and there are no others, by the law of total probability you must have \begin{multline} P(1)+P(2)+P(3)+P(4)+\cdots+P(n)=\\ P(1) + r\cdot P(1)+ r^2\cdot P(1) + r^3\cdot P(1) + \cdots + r^{n-1}\cdot P(1) = 1. \end{multline}

This implies that $$ P(1)=\frac{1}{1+r+r^2+r^3+\cdots+ r^{n-1}} =\frac{r-1}{r^{n}-1}. $$

So you can set $r$ to any number you like greater than $1$, such as $r=2$ or $r=10$ or even $r=1.005,$ then set $P(1)$ according to the equation above, and then set all the other probabilities as described.

To simulate your distribution in software, a standard method for any probability distribution over the numbers $1$ to $n$ is to compute the cumulative distribution function for every possible outcome. That is, for each possible value $k$, you compute the sum of probabilities $P(1)+P(2)+P(3)+P(4)+\cdots+P(k).$ Since you have an exponentially increasing distribution, however, it will often be more accurate to compute $1-P(n)-P(n-1)-\cdots-P(k+1),$ which is the same thing mathematically but reduces some numerical errors in the computer. In any case the cumulative distribution function is always exactly $1$ at the largest possible random value, $n.$

Continuing the method, you put all the sums of probabilities into an array in increasing sequence. You then use a pseudorandom function such as rand() to generate a pseudorandom floating-point number (not an integer!) in the range $0.0$ to $1.0.$ Next, use binary search to find the smallest number in the array that is still greater than or equal to the output of rand(). Binary search is recommended because it is generally much faster than looking at each number in sequence within the array. Take the position of that number within the array; in most programming languages this will be an integer from $0$ to $n-1.$ Add $1$ to that integer to get your final “random number” in the range $1$ to $n$.