Calculate recurring probability with break and change in probability

Question

I'm somewhat uncertain on how to calculate recurring probability with two timeframes. Assume an event occurs on a daily basis with the probability of 1/600. If I wanted to calculate how many days it would take for it to occur at least once with the probability of 90% the formula would be

ln(1-0.90)/ln(1-1/600) = 1380.40.

Meaning it would take at least 1380.40 days.

Now if I turn this around and calculate what probability I can achieve in just 1000 days it would be.

ln(1-x)/ln(1-1/600) = 1000

x being 0.81

My two questions are: How would this change if I were to measure this event with a break in between and also how would it change if the probability changed after the break?

So if I measure the event for 500 days, don't take measurements for a few months and then start measuring again for 500 days. Is it save to say, that the probability is still 81%?

How does this change if the probability was 1/600 the first time

ln(1-x)/ln(1-1/600) = 500

and 1/800 the second time?

ln(1-x)/ln(1-1/800) = 500

What would be the overall probability that I recorded the event at least once during both measurements?

Answer 1

According to your description, most likely the model is the discrete Geometric distribution where each day (as a single Bernoulli trial) is treated independent to any other day.

Therefore the a "break" or any number of breaks of any length doesn't matter (as if they don't exist). This is the memoryless property of Geometric distribution, which is also mentioned in the wiki page.

"...start measuring again for $500$ days. Is it save to say, that the probability is still $81\%$?"

The answer is yes.

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Next, when there are different "success" probability $p_1 = \frac1{600}$ and $p_2 = \frac1{800}$ (or whatever numbers you want to plug in), we have in fact two distributions (of the same type) such that the calculation needs to be done taking that into account.

Formally, we have $X_1 \sim \mathrm{Geo}[p_1]$ for before the break and $X_2 \sim \mathrm{Geo}[p_1]$ for afterwards, where $X_1 \perp X_2$ (that they are independent). The task is to find $\Pr\{ X_1 + X_2 \geq 1 \}$.

\begin{align} \Pr \bigl\{ X_1 + X_2 \geq 1 \bigr\} &= \Pr \bigl\{ X_1 \geq 1 \bigr\} \cdot \Pr \bigl\{ X_2 = 0 \bigr\} \\ & \hspace{8pt} + \Pr \bigl\{ X_1 = 0 \bigr\} \cdot \Pr \bigl\{ X_2 \geq 1 \bigr\} \\ & \hspace{8pt} + \Pr \bigl\{ X_1 \geq 1 \bigr\} \cdot \Pr \bigl\{ X_2 \geq 1 \bigr\} \end{align} The plus signs is for decomposing the events into disjoint events, and the multiplying signs are for two connecting two intervals (just like two consecutive days). These apply only when $X_1 \perp X_2$.

Note that $\Pr\{ X_2 \geq 0 \} = 1$ simply means that the number of occurrence after break can be any number (as there's already one before break). Similarly, $\Pr\{ X_1 \geq 0 \} = 1$.

Therefore, with the two lengths of observations $n_1 = 500$ and $n_2 = 500$ (or whatever numbers you want to plug in) \begin{align} \Pr \bigl\{ X_1 + X_2 \geq 1 \bigr\} &= \bigl(1 - (1 - p_1 )^{n_1} \bigr) \cdot (1 - p_2 )^{n_2} \\ & \hspace{8pt} + (1 - p_1 )^{n_1} \cdot \bigl(1 - (1 - p_2 )^{n_2} \bigr) \\ & \hspace{8pt} + \bigl(1 - (1 - p_1 )^{n_1} \bigr) \cdot \bigl(1 - (1 - p_2 )^{n_2} \bigr) \end{align} Plug in the numbers $p_1 = \frac1{600}, p_2 = \frac1{800}$, and $n_1 = n_2 = 500$, we have \begin{align} \Pr \bigl\{ X_1 + X_2 \geq 1 \bigr\} &\approx (1-0.4343) \cdot 0.5351 \\ & \hspace{8pt} + 0.4343 \cdot (1-0.5351) \\ & \hspace{8pt} + (1-0.4343) \cdot (1-0.5351) \\ &\approx 0.302681 + 0.201925 + 0.263023 \\ &\approx 0.767629 \end{align}

This $0.767629$ is the desired probability of "at least one occurrence in the two intervals together as a whol".

Note that you can plug in $p_1 = p _2 =\frac1{600}$ (for $n_1 = n_2 = 500$) to recover the same old $81\%$, which can be directly calculated as $\bigl(1 - (1 - p )^n \bigr)$ with $p = 1/600$ and $n = 1000$.