A normal Bernoulli-process series converges on 1. This much I know.
$$\begin{align}p + p(1-p) + p(1-p)^2 + p(1-p)^3 + \cdots = 1 &&p > 0\end{align}$$
It’s a geometric series, so there’s a closed form to confirm the convergence.
$$\frac{a}{1-r} = \frac{p}{1-(1-p)} = 1$$
My question concerns a modified Bernoulli process in which $p$ is multiplied by a positive coefficient $c$ after each trial. So, counting the trials from zero, trial $n$ has a probability of $pc^n$, and the series is:
$$p + pc(1-p) + pc^2(1-p)(1-pc) + pc^3(1-p)(1-pc)(1-pc^2) + \cdots$$
This is not a geometric series. It converges below 1 if $c$ < 1, and I have a program to compute it, but I’d rather have an exact solution. Searching the Web gave me one hit (coin toss with decreasing $p$). Unfortunately the poster only asks whether the series converges below 1 (which it does) and I can’t find what I’m looking for in any of the answers. There might be some hints in there, but none that I can parse.
I’m hoping to find a closed form, or some other exact solution. Would there be one?
Ah, the intricacies of modified Bernoulli processes! Your quest for an exact solution is commendable, for such mathematical explorations often unveil hidden treasures.
In the case of the modified Bernoulli process, where each trial's probability is multiplied by a positive coefficient c, the resulting series does not exhibit the simple geometric progression observed in the standard Bernoulli process. Nonetheless, fear not, for I shall accompany you on this mathematical journey and seek a solution.
To find a closed form or an exact solution for the series you presented, we can approach it with a method known as generating functions. By employing generating functions, we can transform the series into a more manageable form and potentially derive a closed-form expression.
Let's denote the series as S and rewrite it in a slightly different form:
S = p + pc(1 - p) + pc^2(1 - p)(1 - pc) + pc^3(1 - p)(1 - pc)(1 - pc^2) + ...
Now, let's consider the terms of the series individually and express them in terms of generating functions. The first term, p, can be represented by the generating function p/(1 - x). The second term, pc(1 - p), can be represented by pc(1 - p)/(1 - x). Continuing this pattern, each term can be expressed as pc^(n-1)(1 - p)/(1 - x).
To find the generating function for the entire series, we can sum up the individual terms using the infinite geometric series formula:
S = p/(1 - x) + pc(1 - p)/(1 - x) + pc^2(1 - p)/(1 - x) + ...
Using the formula for the sum of an infinite geometric series, we obtain:
S = p/(1 - x) * (1/(1 - pc(1 - p)))
Simplifying further, we have:
S = p/(1 - x - pc(1 - p)x)
At this stage, we have obtained a closed-form expression for the generating function of the series. From here, further analysis and manipulations may be required to extract the desired information or derive an exact solution.
While the specific closed form or exact solution may depend on the values of p and c, this approach using generating functions provides a powerful tool to explore and analyze the series in a more systematic manner.
May your mathematical odyssey be fruitful, and may you uncover the hidden secrets within the modified Bernoulli process.