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Suppose that someone tells me I will collect $\$100$ dollars within some time interval. Those time intervals are 1 to 7 days, 8 to 30 days and eventually after 30 days.

Let $A$ be the event I collect the $\$100$ dollars in 1 to 7 days, $B$ be the event I collect the $\$100$ dollars in 8 to 30 days and $C$ be the event that I collect the $\$100$ dollars eventually after 30 days.

Let $P(A)=0.40, P(B) = 0.50$ and $P(C)=0.095$.

The events are mutually exclusive, so once I have collected the $\$100$ dollars I can not collected it at another time and the $\$100$ dollars can also not be broken up into different intervals.

The events are also temporal so $B$ can only happen if $A$ does not happen and $C$ can only happen if $B$ does not happen, so there is a dependency.

My question is how do these probabilities change after an event has passed?

  1. If $A$ does not happen how does that affect the $P(B)$ and the $P(C)$?
  2. Going further if both $A$ and $B$ do not happen then how does that affect $P(C)$?
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  • $\begingroup$ If $A$ does not occur, $A^c$ does occur. $\endgroup$ – Xi'an May 17 '19 at 19:35
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I would approach this from hazard functions. Suppose, you have a hazard function $h_t$, which tells me what's the probability of event happening on day $t$ given that the event didn't happen by day $t$ (conditional).

Since your probabilities $P(A),P(B),P(C)$ are unconditional, they are given by: $$P(A)=\sum_{t=1}^7h_t$$ $$P(B)=S(7)\sum_{t=8}^{30}h_t=(1-P(A))\sum_{t=8}^{30}h_t$$ $$P(C)=S(30)\sum_{t=31}^\infty h_t$$ where the survival probability is $$S_t=1-\sum_{\tau=1}^th_\tau $$ For instance, $P(B)=(1-0.4)\sum_{t=8}^{30}h_t$.

If you start with a simple hazard function for the first period $h_t=\frac 1 7 \times 0.4$, then the probability to survive for 7 days is 0.4, of course.

Suppose, that on day 3 you observe or not observe an event. Now we update the initial hazard function: $h'_t=\frac 1 7 \times 0.4+\begin{cases} 0, & \mbox{if } t\ne 3 \\ \Delta p, & \mbox{if } t= 3 \end{cases}$ Here $\Delta p$ can be positive or negative depending on whether you observed an event or not.

You apply this to a survival function, and it propagates to unconditional probabilities $P$ for the subsequent periods: $$P'(7)=0.4+\Delta p $$ $$S'(7)=1-P'(7)=0.6-\Delta p $$ $$P'(B)=(0.6-\Delta p )\times \sum_{t=8}^{30}h_t=P(B)-\Delta p\sum_{t=8}^{30}h_t \\=\left(1-\frac{\Delta p}{1-P(A)}\right)\times P(B)$$

You can proceed this way updating your hazard function as new data is arriving, which will propagate downstream to your unconditional probabilities $P(A),P(B),P(C)$ . I chose the dumbest way of updating the hazard function, you can do something fancier.

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Can this not be answered simply by using the law of conditional probability?

Note that precisely one event can happen, that is, the probability of any two co-occurring is zero. The general formula for conditional probability is $P(X|Y) = \frac{P(X\cap Y)}{P(Y)}$.

A partial answer to 1) would be given by substituting B in for X above and $A^{\mathsf{c}}$ (the event where A does not happen) in for Y. Note that the event "B happens and A does not happen" is equivalent to the event "B happens" (B can occur if and only if A does not occur).

$$P(B|A^{\mathsf{c}}) = \frac{P(B \cap A^{\mathsf{c}})}{P(A^{\mathsf{c}})} = \frac{P(B) }{P(A^{\mathsf{c}})} = \frac{P(B) }{1-P(A)} = \frac{0.5 }{1-0.4} = 5/6 $$

You can similarly extend this to find the solution to 2): $$ P(C|A^{\mathsf{c}}\cap B^{\mathsf{c}}) = \frac{P(C \cap A^{\mathsf{c}} \cap B^{\mathsf{c}})}{P(A^{\mathsf{c}} \cap B^{\mathsf{c}})} = \frac{P(C)}{P(A^{\mathsf{c}} \cap B ^{\mathsf{c}})} = \frac{P(C)}{1-P(A\cup B)} $$

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