Suppose we are flipping coins starting at some time $t$. At time $t$ the probability we obtain heads is $\frac{1}{\sqrt{t}}$. If the coin lands tails, at time $t+1$ the probability of heads is now $\frac{1}{\sqrt{t+1}}$ and so on... What is the expected number of flips until the coin lands heads? In particular I would be interested in an upper bound on the expectation.. If this is difficult suppose we also make the assumption that it is bounded by $T$, e.g. there at most $T > t$ coins.

If the probabilities are constant at $p$ at each round, the expected value is $1/p$ which is like one coupon being collected in the coupon collector problem. Since the probability of heads is less than $1/\sqrt{t}$ at each round, $\sqrt{t}$ is a lower bound for the expectation. On the other hand, the probability is greater than $1/\sqrt{T}$ in the bounded case, and so $\sqrt{T}$ is an upper bound. The question is where in this regime $[\sqrt{t}, \sqrt{T}]$ does the expectation fall..

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    $\begingroup$ In the bounded case, do you assume a success if you run out of coins? $\endgroup$ – Clement C. Oct 18 '18 at 18:24

Lemma. In the unbounded case, the expected number of flips is at most $\sqrt t + 3/2$.

Proof. Let r.v. $F$ be the number of flips until a head. Then the expected number of flips is

\begin{align} E[F] & = \sum_{i=1}^\infty \Pr[F \ge i] && (1)\\ &= \sum_{i=0}^\infty \Pr[\text{first $i$ flips are tails}] && (2)\\ &= \sum_{i=0}^\infty \prod_{j=t}^{t+i-1} (1-1/\sqrt{j}) && (3)\\ &\le \sum_{i=0}^\infty \exp(\textstyle-\sum_{j=t}^{t+i-1} 1/\sqrt j) & \text{as } 1+z\le e^z~~ & (4)\\ &\le \sum_{i=0}^\infty \textstyle\exp(-\int_{t}^{t+i} 1/\sqrt x ~dx) & \text{as } \textstyle \sum_{j=a}^{b-1} f(j) \ge \int_{a}^b f(x)\, dx~~&(5)\\[-10ex] &&~~\text{for $f$ decreasing}~~\\ &= 1+\sum_{i=1}^\infty \textstyle\exp(2\sqrt{t}-2\sqrt{t+i}) & \text{as } \textstyle\int 1/\sqrt x ~dx = 2\sqrt x~~&(6)\\ &= 1+e^{2\sqrt{t}} \sum_{i=t+1}^\infty e^{-2\sqrt i}&&(7)\\ &\le 1+e^{2\sqrt{t}} \int_{t}^\infty e^{-2\sqrt x}\,dx &\text{as } \textstyle \sum_{i=a}^{\infty} f(i) \le \int_{a-1}^\infty f(x)\, dx ~~&(8)\\ &&\text{for monotonic $f$}~~\\ \\ &= 1 + e^{2\sqrt{t}} \,e^{-2\sqrt t}(\sqrt t + 1/2) & \text{as } \textstyle\int_t^\infty e^{-2\sqrt x}\,dx = e^{-2\sqrt t}(\sqrt t + 1/2)~~&(9)\\\\ &= \sqrt {t} + 3/2&&~~\Box \end{align}

Of course the same bound holds in the bounded case as well.

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  • $\begingroup$ Just as a remark, eyeballing the argument: the same analysis goes through for $1/t^{\alpha}$, for any $0<\alpha < 1$ (and for $\alpha=1$, one gets an infinite expectation). $\endgroup$ – Clement C. Oct 19 '18 at 16:11
  • $\begingroup$ FWIW, my guess based on calculations and computations is that the true expectation is $\sqrt t + 1/2 - \Theta(1/\sqrt t)$. $\endgroup$ – Neal Young Oct 19 '18 at 19:56
  • $\begingroup$ @ClementC., for $1/t^\alpha$ with $\alpha\ne 1/2$, what integral do you get for in Step (6) in the (generalized) proof that you have in mind? $\endgroup$ – Neal Young Oct 19 '18 at 21:26
  • $\begingroup$ I am most likely missing something here. Why doesn't $$\frac{x^{1-\alpha}}{1-\alpha}\Bigg|_{t}^{t+i}$$ work? $\endgroup$ – Clement C. Oct 19 '18 at 21:29
  • $\begingroup$ @ClementC. -- Sorry, not Step (6), I meant to say Step (9). E.g. for $\alpha=0.4$? $\endgroup$ – Neal Young Oct 19 '18 at 21:50

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