Suppose we know, for each year $k$, that your chance of dying from all causes is $d_k$: during your first year of life your chance of dying before you turn 2 is $d_1$; during the second year, your chance not to make it to year 3 is $d_2$, etc. We want to know the probability $D_n$ that you will die exactly during year $n$. My guess would be this:

$$D_n = (1 - d_1)(1 - d_2)\dots(1 - d_{n-1})d_n$$

Does this make sense? Intuitively I would say so, but I'm not convinced. My confusion stems from the fact I am not sure whether we're talking about $n$ different, dependent events each with its own probability, or one single event with different probabilities of happening at different times. Thanks!

  • 1
    $\begingroup$ "...assuming that you didn't die until now..." On what do you base this interpretation? $P(E_k)$ is said to be the probability of the event that you die in year $k$. So unconditional. $\endgroup$
    – drhab
    Jul 10, 2018 at 7:48
  • $\begingroup$ It seems that you are making confusion between the events "you die exactly on year $k$" and "you die on some year $\le k$". $\endgroup$
    – Crostul
    Jul 10, 2018 at 7:57
  • $\begingroup$ @drhab $P(E_k)$ should be the probability of death that a person has in general when they hit age $k$. If you don't make it to age $k$, though, $P(E_k)$ does not make sense for you. That's my interpretation. $\endgroup$
    – Nicola
    Jul 10, 2018 at 8:17
  • $\begingroup$ @Crostul Can you elaborate? I'm not even trying to compute anything about the event "you die on some year $\leq k$", so I'm not sure what you mean $\endgroup$
    – Nicola
    Jul 10, 2018 at 8:19
  • $\begingroup$ Fine, but then it is simply wrong and confusing to name it as "the event that your death occurs in year $k$". You should repair that. $\endgroup$
    – drhab
    Jul 10, 2018 at 8:24

1 Answer 1


Yes, it does make sense.

Let $F_n$ denote the event that you die in your $n$-th year.

Let $E_{n}$ denote the event that you live longer than $n$ years.

Then $P(F_n\mid E_{n-1})=d_n$ and $P(F_n^{\complement}\mid E_{n-1})=1-d_n$.

And $E_n=F_1^{\complement}\cap\cdots\cap F_n^{\complement}$ so that: $$P(E_n)=P(F_n^{\complement}\mid E_{n-1})P(E_{n-1})=(1-d_n)P(E_{n-1})$$leading to: $$P(E_n)=(1-d_1)(1-d_2)\cdots(1-d_n)$$so that:

$$P(F_n)=P(F_n\cap E_{n-1})=P(E_{n-1})P(F_n\mid E_{n-1})=(1-d_1)(1-d_2)\cdots(1-d_{n-1})d_n$$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .