I have a series of probabilities that event $E$ occurs at iteration $i$ with $i=1,2,....$. These are denoted $p_1,p_2,....$. These probabilities are that $E$ occurs on iteration $i$, but do not consider whether $E$ occurred before $i$.

I want to transform these $p$s into a new series such that $\hat{p}^i$ = probability of first occurrence of $E$ on $i$. My thought is that this is:

$$\hat{p}^i = (1-p_1)(1-p_2)....(1-p_{i-1})p_i$$

Assuming $p_i$ is independent of $p_j$ for $i>j$.

But how do I transform if they are not independent?

In my problem, the probability $E$ occurs over time approaches a stationary distribution and is increasing up to that point. E.g, after some iteration $T$ $p_{t} = X \forall t>T$. "An event waiting to happen"

  • $\begingroup$ +1 agree with your answer, and yes, it certainly assumes independence $\endgroup$ – MikeP May 19 '17 at 14:21
  • $\begingroup$ @MikeP I edited my question with new information asking about that. $\endgroup$ – Tommy May 19 '17 at 14:24
  • $\begingroup$ You would normally speak of independent events, rather than probabilities. What do you mean by "agnostic"? If the probability of an event at $j$ is the same whether or not an event occured at $i$, then the events are independent. $\endgroup$ – GeoMatt22 May 19 '17 at 15:00

They could still be independent. I can come up with an example, although very contrived.

Suppose that p-primes are 0.1, 0.2, 0.3, 0.4 (have to sum to 1)

We can work backwards and come up with ps assuming independence to match using:

$$p'(i)=\prod_{j=1}^{i-1}(1-p(j))\cdot p(i)$$


Using Matlab ps are 0.1000, 0.2222, 0.4286, 1.0000

  • $\begingroup$ from summing my $p$s I have learned that I must not have independence. Thus I've edited my question to ask how to transform it not independent. $\endgroup$ – Tommy May 19 '17 at 18:51

You would normally talk about independent events rather than probabilities.

To be more precise, let $E_i$ be a binary indicator such that $$E_i=\begin{cases} 1 & \text{an event occurs at iteration }i \\ 0 & \text{otherwise} \end{cases}$$

Then the marginal probability of $E_i=1$, with respect to an event at some other iteration $j$, will be \begin{align}\Pr\big[E_i=1\big] &= \Pr\big[E_i=1\mid E_j=1\big]\times\Pr\big[E_j=1\big] \\ &+\Pr\big[E_i=1\mid E_j=0\big]\times\Pr\big[E_j=0\big]\end{align} If the event $E_i$ is independent of the event $E_j$, then you will have $$\Pr\big[E_i=1\mid E_j=1\big] = \Pr\big[E_i=1\mid E_j=0\big] = p_i$$ This intuitively would seem to correspond to your statement about "agnostic".

Mathematically, since you are specifying the probabilities $p_i$ without reference to the sequence of values $E_{j<i}$, your model is assuming independent events. (Otherwise you would need to marginalize over possible "prefix sequences" to determine the $p_i$.)

  • $\begingroup$ this answer helps. because my $p$s do not seem to add to 1, so I guess I am violating independence somewhere. $\endgroup$ – Tommy May 19 '17 at 18:34
  • $\begingroup$ can I re-ask, how to transorm if they are not independent? I edited my question $\endgroup$ – Tommy May 19 '17 at 18:39
  • $\begingroup$ Where do your $p_i$ values come from? My understanding was that multiple events can occur, so the sum of the $p$'s would not sum to $1$. (As distinguished from the $\hat{p}$'s; e.g. you have a variable-$p$ version of this.) $\endgroup$ – GeoMatt22 May 19 '17 at 18:53
  • $\begingroup$ more information: 1) only one event can occur. 2) i've updated the question with information about $p$s approaching a stationary distribution. $\endgroup$ – Tommy May 19 '17 at 18:58
  • $\begingroup$ yes it looks like a variable p version of that; is there a straitforward answer to that? $\endgroup$ – Tommy May 19 '17 at 19:06

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