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 `$X_{t}$`

, `$t=1\ldots$`

. 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

```
$$
Y_{t}^{m}=\begin{cases}
1 & \text{if }X_{t(m-1)+1}=\cdots=X_{tm}=1,\\
0 & \textrm{otherwise}.
\end{cases}
$$
```

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 (`$m=7\times86400$`

), 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 `$u=\pi^{m'}$`

, where `$m'$`

is another positive integer.
The induced prior is
```
\[
p(u)=(m/m')u^{m/m'-1},
\]
```

or a `$Beta(m/m',1)$`

prior, as can be seen from the following manipulation
```
\begin{eqnarray*}
P(U\leq u) & = & P(\pi^{m\cdot m'/m}\leq u),\\
& = & P(\pi^{m}\leq u^{m/m'}),\\
& = & u^{m/m'},
\end{eqnarray*}
```

with the the derivative being the induced prior
```
\[
p(u)=(m/m')u^{m/m'-1}=(m/m')u^{m/m'-1}.
\]
```

Then it is well known that the posterior `$p(u\mid\{Y_{t}^{m}\})$`

is beta distributed with parameters

```
\begin{eqnarray*}
\alpha & = & S_{n}+m/m',\\
\beta & = & n-S_{n}+1,
\end{eqnarray*}
```

where `$S_{n}$`

are the number of success in your data and `$n$`

is
the number of trials on the `$m'$`

scale.

Since the expectation of the a `$Beta(\alpha,\beta)$`

variable is `$\alpha/(\alpha+\beta)$`

,
it follows that

```
\[
E[Y_{n+1}=1]=\frac{\alpha}{\alpha+\beta}=\frac{S_{n}+m/m'}{n+1+m/m'},
\]
```

where `$m$`

is the original time scaling and `$m'$`

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{eqnarray*}
& & P(\text{rains next day}\mid\text{rained every day last week})\\
& = & \frac{7+1}{7+1+1}=\frac{8}{9}=0.888\ldots.
\end{eqnarray*}
```

Calculating on the hour scale instead, we find that `$m/m'=24$`

, that
`$S_{n}=24\cdot7$`

and `$n=24\cdot7$`

```
\begin{eqnarray*}
& & 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'+r}{n+1+m/m'+r},\\
& = & \prod_{r=0}^{k-1}\frac{24\cdot8+r}{24\cdot8+1+r}.
\end{eqnarray*}
```

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
```
\[
\prod_{r=0}^{k-1}\frac{24\cdot8+r}{24\cdot8+1+r}=\frac{8}{8+1}=0.888...,
\]
```

and the probabilities match.

We can do the same exercise with minutes instead, where `$m/m'=60^{2}\cdot24$`

,
and `$S_{n}=n$`

are equal to `$60^{2}\cdot24\cdot7$`

,
```
\begin{eqnarray*}
& & P(\text{rains next day}\mid\text{rained every day last week})\\
& = & \prod_{r=0}^{k-1}\frac{60^{2}\cdot24\cdot8+r}{60^{2}\cdot24\cdot8+1+r},\\
& \approx & 0.888....
\end{eqnarray*}
```