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suppose we have $n$ Geometric$(p)$ random variables $X_1,\dots,X_n$, and let $Y_t$ be the number of these random variable still alive at time $t$, i.e.

$Y_t = \sum_{i=1}^n \mathbb{1}\{X_i \geq t\}$

I'm interested in upper bounding the expected first time that $Y_t = 0$ using a martingale approach. Here's an example argument that doesn't quite work: we can easily show that

$M_t = \log(Y_t) - \log(n(1-p)^t)$

is a supermartingale. Now we can apply the optional stopping theorem for any stopping time $\tau$ that occurs almost surely to get a bound

$\mathbb{E}[\tau] \leq \frac{1}{\log(\frac{1}{1-p})}(\log(n) - \mathbb{E}[\log(Y_{\tau})])$

However, if we let $\tau$ be the first time that all our geo random variable die, i.e. the first time that $Y_t = 0$, then we can't get anything useful from this bound, since $\mathbb{E}[\log(Y_{\tau})] = -\infty$. I'm wondering if there's a modification of this argument that works? Or if anyone knows alternate approaches? Note that this time is the same as the expected maximum of $n$ geometric random variables. But I'm specifically wondering about a martingale approach (just out of curiosity).

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  • $\begingroup$ Your notation seems to imply that you have a martingale in $t$ and not in $n$. What is the filtration then ? Is the time $n$ or $t$ ? $\endgroup$
    – Synia
    Commented Aug 23, 2017 at 16:59
  • $\begingroup$ Ah I realize the notation is confusing, sorry. The martingale is in $t$, the filtration is $\sigma(Y_{t-1},\dots,Y_0)$ where $Y_0 = n$ (equivalently $\sigma(M_{t-1},\dots,M_0))$. $n$ is a parameter. In words, $n$ is the number of people alive initially, at each ``round'' $t=1,2,\dots$ each person can die with probability $p$. $Y_t$ is the number of people alive at round $t$. $\endgroup$
    – AustinC
    Commented Aug 24, 2017 at 4:49
  • $\begingroup$ I am not sure if a Martingale approach can be used here but I would be interested to know : I came across a similar problem in moais.imag.fr/membres/marc.tchiboukdjian/pub/isaac10.pdf and wanted to use a Martingale approach but it did not work and we add to use a more ad-hoc method. $\endgroup$
    – N. Gast
    Commented Aug 24, 2017 at 6:56

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