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A standard geometric distribution can be interpreted as the number of Bernoulli trials required to get one success. However, what if the probability of success if each trial diminishes by some factor with each failure?

Let $p$ be probability of success of the first trial and $d$ be the diminishing factor of each failure such that the probability of success of trial $n$ is $pd^{n-1}$. I have been trying to calculate the expected value of this distribution without success. My main stumbling block is the probability of failing $n$ times:

$$ P(n\text{ failures}) = \prod_{k=1}^n (1-pd^{k-1}) $$

Given that, the expected value would be

$$ E(X)=\sum_{n=1}^{\infty} npd^{n-1}F(n-1) $$

Is there a way to get a closed form for this?

I also suspect that there may be something screwy going on with the variance of this distribution. Since the probability of success tends to 0 exponentially, empirical trials either finish very quickly or rocket off to infinity as the probabilities become infinitesimal.

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The random variable $X$ is not well-defined. For any $d\lt 1$, there is a non-zero probability that there will never be a success.

To show this, first note that it is clear that for any $n$, there is a non-zero probability that the game will last at least $n$ trials. We can choose $n$ large enough that $pd^{n-1}$ is as small as we wish. Given that there have been $n$ trials without a success, the probability that there will never be a success is $\prod_{k=n+1}^\infty(1-pd^{k-1})$. For small $pd^{k-1}$, we can lower bound the logarithm of this. For example, it is $\gt -\sum_{n+1}^\infty 2pd^{k-1}$.

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  • $\begingroup$ I have been suspecting as much. Is it possible to add the "no success" outcome and thus make it well-defined? $\endgroup$ – Max Apr 30 '14 at 17:55
  • $\begingroup$ A random variable takes on real number values. If your random variable $X$ is to measure the length of the game, there seems to be no sensible value to assign to it if the game does not terminate. We could condition on the game terminating. Basically what this comes down to is that if $e$ is the probability the game does not terminate, we multiply all of the commonsense probabilities for $X$ by $\frac{1}{1-e}$. $\endgroup$ – André Nicolas Apr 30 '14 at 18:06

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