Time in Laplace's rule in the context of a "for all" problem

We are working with a sequence of independent binary events on a discrete time-scale, say seconds, with success probability $\pi$. Call this sequence of events $X_t$, $t=1\dots$. For instance, $X_t=1$ if it rains at second $t$ and $0$ otherwise. For each positive integer $m$, define the “for all”''"-type variable

$$Y_t^m=\{\begin{array}{cc}1& \text{if}{X}_{t(m-1)+1}=\cdots =X_{tm}=1,\\ 0& \text{otherwise}.\end{array}$$

For instance, the variable $Y_t^{60}=1$ if it rained during the entirety of the the $t$th minute. We can do the same thing for days, with $m=86400$, weeks ($m=7\times 86400$), and so on.

Each variable $Y_t^m$ has its own success probability ${\pi }^m$, with $P(X_t=1)=\pi$ being the most refined probability. Suppose you choose an $m$, such as $m=86400$ for daily data, and decide on a uniform prior for ${\pi }^m$. Then you can deduce the implied prior for $u={\pi }^{m^{\prime }}$, where $m^{\prime }$ is another positive integer. The induced prior is $$p\left(u\right)=\left(m/m^{\prime }\right)u^{m/m^{\prime }-1},$$ or a $Beta(m/m^{\prime },1)$ prior, as can be seen from the following manipulation $$\begin{array}{ccc}P(U\le u)& =& P({\pi }^{m\cdot m^{\prime }/m}\le u),\\ & =& P({\pi }^m\le u^{m/m^{\prime }}),\\ & =& u^{m/m^{\prime }},\end{array}$$ with the the derivative being the induced prior $$p\left(u\right)=\left(m/m^{\prime }\right)u^{m/m^{\prime }-1}=\left(m/m^{\prime }\right)u^{m/m^{\prime }-1}.$$ Then it is well known that the posterior $p(u\mid \{Y_t^m\left\}\right)$ is beta distributed with parameters

$$\begin{array}{ccc}\alpha & =& S_n+m/m^{\prime },\\ \beta & =& n-S_n+1,\end{array}$$ where $S_n$ are the number of success in your data and $n$ is the number of trials on the $m^{\prime }$ scale.

Since the expectation of the a $Beta(\alpha ,\beta )$ variable is $\alpha /(\alpha +\beta )$, it follows that

$$E[Y_{n+1}=1]=\frac{\alpha }{\alpha +\beta }=\frac{S_n+m/m^{\prime }}{n+1+m/m^{\prime }},$$ where $m$ is the original time scaling and $m^{\prime }$ is the new time scaling.

Rain example

Suppose our discrete time unit is seconds, and suppose it has rained every second for the last week. Assume a uniform prior on the probability for raining for a whole day. Then $$\begin{array}{ccc}& & P(\text{rains next day}\mid \text{rained every day last week})\\ & =& \frac{7+1}{7+1+1}=\frac{8}{9}=0.888\dots .\end{array}$$ Calculating on the hour scale instead, we find that $m/m^{\prime }=24$, that $S_n=24\cdot 7$ and $n=24\cdot 7$ $$\begin{array}{ccc}& & P(\text{rains next day}\mid \text{rained every day last week}),\\ & =& \int u^{24}f(u\mid S_n+24,n-S_n+1)du,\\ & =& \prod _{r=0}^{k-1}\frac{S_n+m/m^{\prime }+r}{n+1+m/m^{\prime }+r},\\ & =& \prod _{r=0}^{k-1}\frac{24\cdot 8+r}{24\cdot 8+1+r}.\end{array}$$ By taking logarithms, one can show that $$\prod _{r=0}^{n-1}\frac{nm+r}{nm+1+r}=\frac{m}{m+1},$$ for all $n,m$, hence $$\prod _{r=0}^{k-1}\frac{24\cdot 8+r}{24\cdot 8+1+r}=\frac{8}{8+1}=\mathrm{0.888...},$$ and the probabilities match.

We can do the same exercise with minutes instead, where $m/m^{\prime }={60}^2\cdot 24$, and $S_n=n$ are equal to ${60}^2\cdot 24\cdot 7$, $$\begin{array}{ccc}& & P(\text{rains next day}\mid \text{rained every day last week})\\ & =& \prod _{r=0}^{k-1}\frac{{60}^2\cdot 24\cdot 8+r}{{60}^2\cdot 24\cdot 8+1+r},\\ & \approx & \mathrm{0.888....}\end{array}$$