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Consider the following sequence of Bernoulli trials with the following varying probability of success at each trial:

  • At the first trial: the probability of success is $p_1 = 1 / n_1$, with $n_1 = 1$.
  • At the second trial: the probability of success is $p_2 = 1 / n_2$, with $ n_2 = \begin{cases} n_1 + 1 & \text{if the first trial was a success,} \\ n_1 & \text{if the first trial was a failure.} \\ \end{cases}$
  • $\ldots$
  • At the $i$th trial: the probability of success is $p_i = 1 / n_i$, with $ n_i = \begin{cases} n_{i-1} + 1 & \text{if the $(i-1)$th trial was a success,} \\ n_{i-1} & \text{if the $(i-1)$th trial was a failure.} \\ \end{cases}$
  • Etc.

I am trying to get my teeth into the probability mass function of this sequence and expected value of $n_i$ for a certain $i$, but I would be interested in any insightful information I could attain.

What I have so far:

  • Binary tree diagram showing a 5-trial progression of the process
  • After the $i$th trial happens:
    • There are ${i - 1 \choose k - 1}$ ways of getting $n_i = k$, with $k \in \{1, \ldots, i\}$ .

Any hint on how to find a compact expression (if any)? My main issue is generalizing the products of probabilities (elegantly) without keeping track of every single one of them "manually" on the expressions.

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  • $\begingroup$ Oh my... I overlooked that. Edited. Thanks, @Henry $\endgroup$ – DIVVS IVLIVS Mar 13 '18 at 12:25
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The generating functions $$g_k(s)=E(s^{n_k})$$ solve the recursion $g_1(s)=s$ and $$g_{k+1}(s)=g_k(s)-(1-s)\int_0^s\frac1tg_k(t)dt$$ To characterize the sequence $(g_k)$, a common trick is to consider the series $$G_x(s)=\sum_{k=1}^\infty g_k(s)x^k$$ then the recursion on the sequence $(g_k)$ yields $$G_x(s)=xs+xG_x(s)-x(1-s)\int_0^s\frac1tG_x(t)dt$$ From there, one can show that each function $G_x$ solves a first order linear equation, hence some "explicit" formulas for $G_x$ are available. However, to extract precise informations from such formulas is not always simple...

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  • $\begingroup$ Thanks for your comment @Did. I lack proper knowledge about generating functions, so I guess it is time to teach myself about them. Thanks again! $\endgroup$ – DIVVS IVLIVS Mar 13 '18 at 13:50

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