What does standby time mean?

What is a non-iterative way to compute the mean time until a random event within a limited window?

• Consider that the probability of me getting a speeding ticket on any particular day is p.  Imagine that I only care about my next speeding ticket because I lose my driving license after that.  I know that I can calculate the expected time until I get a speeding ticket as follows: With probability p, T=1. (The event happens tomorrow.) If it doesn't happen tomorrow, then we expect to wait the same amount of time starting then. This gives us:  E(T) = p + (1-p)(E(T)+1) And we can solve for E(T) to get:  E(T) = 1/p However, what if I will be selling my car in N days, so, to put it in computer science terms, the depth of our recursion is limited to N? Of course, we could calculate this by manually "unrolling" N times, so for N=3, our E(T) is (I've capitalized the inner-most P):  E(T) = p + (1-p)(p + (1-p)(P+1)+1) The problem is that my actual goal is to come up with a way to estimate p given an actual duration of time until a speeding ticket is received.  With no time limit this is easy, we just reverse  E(T) = 1/p , to get (where d is the observed duration until a ticket): p=1/d. But can I do this in the scenario where I have a limited time window? I know I can do it computationally by calculating E(T) manually for a fixed N and a variety of values of  p, and then creating a reverse-lookup table, but I'm hoping there is a better way. I then have a follow-on problem if you'll indulge me: My eventual goal is to come up with an overall estimate P by taking the mean of a variety of values  p_1, p_2, ...  I get for a number of independent observations of the type described above, each of which results in measurements  d_1, d_2, ... . Now I realize it may be preferable to take the mean of  d_1, d_2, ...  and calculate P=1/D from that, but some external constraints prevent me from taking that approach, I have to take the mean of the probabilities. Now the question: If I don't get a speeding ticket within the  N day time window, what  p_n value do I use to update my mean, to account for this observation?  Could it be  p=0?

• Answer:

  The discussion for question is relevant to this answer. I'm not sure I understand all the aspects of your question, but here is my first try at an answer. I take your expression [math]E(T)[/math] to be the probability of the event happening at day [math]T[/math] without it occurring before. The probability that the event occurs on day [math]d[/math] without it occurring before is given by [math]1 - (1-\lambda)^d[/math] Where I use [math]\lambda[/math] to denote the probability of it occurring in a particular day. The probability of it not occurring after [math]d[/math] days is [math](1-\lambda)^d[/math] I assume that your second question is the following: having not observed the event after [math]d[/math] days, what range of values should I believe are possible for the unknown "true" probability of daily occurrence [math]\lambda[/math]? These calculations are the province of the celebrated Bayes' theorem. See David MacKay's book "Information Theory, Inference, and Learning Algorithms" for a great discussion on the topic. Here is one answer to your second question assuming that we are equally likely to expect any value in the unit range for [math]\lambda[/math]: [math]p(\lambda | d ) = (1+d)(1-\lambda)^d[/math] As you suggest, the value [math]\lambda=0[/math] is the maximum of the distribution (this is a probability density so don't expect it to be bounded at one!). If you plot the results you get, for any [math]d[/math], a sharp drop-off for values of [math]\lambda > 1/d[/math]. But nonetheless, there is always a finite, non-zero probability that the daily occurrence probability is say, 99%, no matter how many days you have not seen the event. Of course, given the "subjective" nature of Bayesian reasoning, the formula above is not the only answer as it depends on my prior expectation for the unknown [math]\lambda[/math] parameter. I took that prior to be the "uninformative" one - equally uncertain about all possible values.