A simple coin game: You sequentially flip a coin in each trial that can either land heads or tails $(H,T)$. If you flip $T$ you are allowed to play on, if you flip $H$ you lose. Importantly, for each new round you are handed a coin with a (unknown) bias to either side.

  1. Can we describe the probability of being allowed to continue (i.e. only getting $T$) in a Bayesian non-parametric sense?
  2. Under this formulation (given it is correct) Jensen's inequality yields a bound on the survival function. Is there a intuition for this?

Setup and definitions:

Assume that some unknown stochastic process or model $m$ with parameters $\theta\in\Omega$ is producing binary data, over discrete time $t_1<t_2<...t_k$ such that data at time $t$, $D(t)\in{0,1}$. ($D(t)$ is not really a function.)

Given that all prior n observations were one and the parameter of the model is described by $\theta$, the immediate (or event) probability of an entry of one to the dataset at time $t_k$, is given by the hazard function

$$ \lambda(t_k\vert\theta)=\mathbb{P}(D(t_k)=1\vert D(t_{k-n}=1),\theta). $$

Alternatively, let $T=t_k$ denote the time of the event $D(t_k)=0$.Treating $T$ as a random discrete variable, the hazard function can be written as

$$ \lambda(t_k\vert\theta)=\mathbb{P}(T=t_k\vert T\geq t_k,\theta) = \frac{f(t_k\vert\theta)}{1-F(t_k\vert\theta)}. $$

This says that the hazard function is the conditioned probability of an event (the entry being zero), given that the event has not happened yet. In discrete time this is a probability, and not a rate.

Think of playing a round of "elimination coin". You flip a coin every time $t_1,t_2..t_k$ and if you realise $H$ on the $t_k$ trial, you lose. The coin may not be fair and is described by the parameter $\theta\in\Omega$. If the event time $T$ is exponentially distributed with a fixed $\theta$, the hazard is constant at $\beta=\frac{1}{\lambda}$, the scaling of the exponential distribution.

Let $\phi(t_k\vert\theta)=(1-\lambda(t_k\vert\theta))$ clearly, this is the immediate event probability (of survival) given $\theta$.

Define the marginalized event probability as

$$ \phi(t_k\vert\theta) = \sum_{\theta\in\Omega}\phi(t_k,\theta) = \sum_{\theta\in\Omega}\phi(t_k\vert\theta)p(\theta). $$

Here, $p(\theta)$ plays the role of a "mixing" distribution or prior. Colloquially; If we expect $\theta$ to be some specific parameter value, $\theta^{'}$, half of the time, then the above equation is a weighted average over different parameters.

Again, think of "elimination coin", but this time $\theta$ is a random variable with the density $p(\theta)$. If there is just two states; a fair coin and a very biased one - then the probability of $H$ is a weighted average over the probability (or rate) with which these different parameters are realized.

The definition of the Survival function $S(\cdot)$ is

$$ S(t_{k}\vert\theta)=\mathbb{P}(T\geq t_{k}\vert\theta) $$

and can be written as the simple recursive relation of the hazard as

$$ S(t_{k}\vert\theta)=(1-\lambda_{1})(1-\lambda_{2})...(1-\lambda_{k-1}) $$

where $\theta$ is removed for notational simplicity. With a bit of simple algebra it is easy to show that

$$ S(t_{k}\vert\theta) = \prod_{1}^{k-1}\sum_{\theta\in\Omega}\phi(t_k\vert\theta)\cdot p(\theta).$$

Is this result correct / uncontroversial?

Jensen's inequality:

To find the time-average lifespan from $t_k=1$ to $\tau$

$$ \left\langle S(t_k\vert\theta)\right\rangle _{\tau} = \frac{1}{\tau}\prod_{1}^\tau\sum_{\theta\in\Omega}\phi(t_k\vert\theta))\cdot p(\theta). $$

which is a product (over time) of a sum (over parameters) over products (over likelihoods times priors), and does (to me) not afford any meaningful simplification. Taking the log, we realize Gibbs version of Jensen's inequality:

$$ \ln\left\langle S(t_k\vert\theta)\right\rangle _{\tau} \leq \frac{1}{\tau}\sum_{1}^\tau\ln\sum_{\theta\in\Omega}\phi(t_k\vert\theta))\cdot p(\theta). $$

In this setup, does this bound have any intuition? Alternatively, is there anyway to derive "interesting" quantities from the above, like Shannon entropy or surprise?


Is this expression equal (or come close to) time-average (Shannon) surprise? (entropy)

$$ \left\langle S(t_k\vert\theta)\right\rangle _{\tau} = \frac{1}{\tau}\prod_{1}^\tau\sum_{\theta\in\Omega}\phi(t_k\vert\theta))\cdot p(\theta). $$

Does this boundary bear any intuition to a survival/reliability/first passage time?

$$ \ln\left\langle S(t_k\vert\theta)\right\rangle _{\tau} \leq \frac{1}{\tau}\sum_{1}^\tau\ln\sum_{\theta\in\Omega}\phi(t_k\vert\theta))\cdot p(\theta). $$

I realize this is quite lengthy. I am not a mathematician, so notation might not be correct. But I hope that the intuition is correct. Any calls for clarification is appreciated.

  • $\begingroup$ Are you assuming the bias $\theta$ is the same for all coins, i.e., at all times? $\endgroup$ – Xi'an Mar 22 '17 at 8:05
  • $\begingroup$ If the outcome $D_t$ is binary, why bother with all these notations when $\theta=\mathbb{P}(D_t=1|\theta)$ would suffice? $\endgroup$ – Xi'an Mar 22 '17 at 8:09
  • $\begingroup$ Hi Xi'an. 1) No. At each time step $t_1<t_2...$ the value of the bias $\theta$ is drawn from $p(\theta)$. 2) Because, implicit in $\theta = \mathbb{P}(D_t=1\vert\theta)$ is the important assumption that all prior data points must be equal to one as well. This is expressed in $\phi(\cdot)$ and $S(\cdot)$ above. $\mathbb{P}(D_t=1\vert\theta)$ is (unless otherwise stated) a unconditional probability. Hope this makes sense. $\endgroup$ – tmo Mar 22 '17 at 9:59
  • $\begingroup$ Despite the (perhaps too complicated) notation, do you have any insight into: 1) If the equations are correct? 2) If the bound is meaningful or yields any intuition? 3) If I can derive time-average surprise (entropy) from the second last equation. Thanks! $\endgroup$ – tmo Mar 22 '17 at 10:12
  • 1
    $\begingroup$ If $\theta$ changes after each time, there is no learning from past experiments. $\endgroup$ – Xi'an Mar 22 '17 at 11:07

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