Jensen's inequality; Hazard, Likelihood and Prior?

Primer:

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?

Summary:

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.

• Are you assuming the bias $\theta$ is the same for all coins, i.e., at all times? – Xi'an Mar 22 '17 at 8:05
• If the outcome $D_t$ is binary, why bother with all these notations when $\theta=\mathbb{P}(D_t=1|\theta)$ would suffice? – Xi'an Mar 22 '17 at 8:09
• 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. – tmo Mar 22 '17 at 9:59
• 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! – tmo Mar 22 '17 at 10:12
• If $\theta$ changes after each time, there is no learning from past experiments. – Xi'an Mar 22 '17 at 11:07