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Assume that we have an experiment $A$ which has the outcome "success" or "failure". Each time we try $A$ the probability of getting a success is different but always less than or equal to a given constant $p~(0<p<1$). The probability of success for every trial does not depent on previous trials.

Let $X$ be the random variable representing the number of times we have to repeat $A$ until we get the first success. Let $Y$ be a random variable that follows the geometric distribution with parameter $p$, i.e. $\Pr (Y=k) = (1-p)^{k-1} \cdot p$.

Does $X \geq_{\text{st}} Y$ hold? Why?

Remarks

  1. According to this link independent trials with non-constant probability are called "Poisson Trials".
  2. $\geq_{\text{st}}$ is the stochastic ordering. See the wikipedia page on Stochastic Ordering.
  3. The straightforward way to approach the problem is as follows:
    If the probability of success in the $i$-th trial is $p_i \leq p$ it should hold that (for $k \geq 1$): $$ \Pr (X=k) = (1-p_1) \cdot \dots \cdot (1-p_{k-1}) \cdot p_k~. $$ Now we have that \begin{align*} & X \geq_{\text{st}} Y & \Leftrightarrow\\ & (\forall a) \Pr (X \geq a) \geq \Pr (Y \geq a) & \Leftrightarrow\\ & (\forall a) \Pr (X < a) \leq \Pr (Y < a)\\ \end{align*} Since the probability of getting a success in $A$ is always at most equal to $p$, the fact that $(\forall a) \Pr (X < a) \leq \Pr (Y < a)$ seems "reasoble", but I am unable to prove it.
Ioannis Kokkinis
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1 Answers1

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The event $\{X \ge a\}$ is simply the event that the first $a-1$ trials are failures, so it has probability $$P(X \ge a) = \prod_{k=1}^{a-1} (1-p_k).$$ It's clear that this is greater than $P(Y \ge a) = (1-p)^{a-1}$.

To see it another way, let's for simplicity write $q_k = 1-p_k$, and let $Q_n = \prod_{k=1}^n q_n$. Then we have $$\begin{align*}P(X \ge a) &= \sum_{n=a}^\infty P(X=n) \\ &= \sum_{n=a}^\infty \prod_{k=1}^{n-1} q_k (1-q_n) \\ &= \sum_{n=a}^\infty Q_{n-1} (1-q_n) \\ &= \sum_{n=a}^\infty (Q_{n-1} - Q_n).\end{align*}$$ This is a telescoping sum (since $Q_n \to 0$) so $P(X \ge a) = Q_{a-1}$.

Nate Eldredge
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