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 $P(C)$?
  • $\begingroup$ If $A$ does not occur, $A^c$ does occur. $\endgroup$
    – Xi'an
    Commented May 17, 2019 at 19:35

2 Answers 2


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.


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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