# Can I estimate the frequency of an event based on random samplings of its occurrence?

This question is just for fun, so if it isn't fun then please feel free to ignore it. I already get a lot of help from this site so I don't want to bite the hand that feeds me. It's based on a real life example and it's just something I've wondered about a lot.

I visit my local dojo to train on an essentially random basis Monday-Friday. Let's assume I visit twice a week. This means I visit exactly twice, every week, with only the two days varying. There is one individual who is nearly always there whenever I am there. If he visits on the same day as me then I will see him. Let's assume he's there 90% of the time when I'm there. I want to know two things:

1) how often he trains

2) whether he comes on a random basis or on set days of the week.

I'm guessing perhaps we have to assume one to guess the other? I've really got nowhere with this at all. I just think about it in the warm-up every week and am baffled anew. Even if somebody gave me a way in to think about the problem I would be most grateful.

Cheers!

• @Chris, well you need to start by defining your model. When you say you visit twice a week at random, that can potentially mean many things. For example, you could go every week exactly two times, chosen as a random combination of two elements of the set $\{\text{Mon},\ldots,\text{Fri}\}$, or you could go on average twice a week where, say, you flip a biased coin with probability of heads of 2/5 and you go every day you see a head. These aren't the only options. – cardinal Mar 2 '11 at 12:59
• Also are you assuming that you'll always see him if you visit the dojo on the same day as him? If not, i think we'd need to know something about the length of your sessions and the length of his sessions compared to the length of time each day the dojo is open. – onestop Mar 2 '11 at 13:11
• @Chris, @onestop, this question reminds me of, and is related to, a technique used to sample people that might be reluctant to answer a question truthfully, often due to the social stigma of responding affirmatively. You introduce a random element to the sampling such that with fairly high probability the respondent responds affirmatively (more embarrassing answer) even if in truth they would have responded negatively. If the probability of a randomly determined "yes" is high enough, the "embarrassment bias" is reduced. Of course, one has to sample more people, too. – cardinal Mar 2 '11 at 13:22
• @Chris You would have to make some more assumptions. As it is now there are a multiplicity of valid explanations. Here’s a silly one: Are the individual’s visits independent of yours? If not, maybe he will only visit when you visit (he looks for your car outside everyday), but tosses a coin (with probability 0.9) before deciding whether to go inside. – vqv Mar 2 '11 at 13:38
• Simple solution: ask him :-). – whuber Mar 2 '11 at 16:49

The model is this: represent this individual's attendance as a sequence of indicator (0/1) variables $(q_i)$, $i=1, 2, \ldots$. You randomly observe a two-element subset out of each weekly block $(q_{5k+1}, q_{5k+2}, \ldots, q_{5k+5})$. (This is a form of systematic sampling.)
1. How often does he train? You want to estimate the weekly mean of the $q_i$. The statistics you gather tell you the mean observation is 0.9. Let's suppose this was collected over $w$ weeks. Then the Horvitz-Thompson estimator of the total number of the individual's visits is $\sum{\frac{q_i}{\pi_i}}$ = ${5\over2} \sum{q_i}$ = ${5\over2} (2 w) 0.9$ = $4.5 w$ (where $\pi_i$ is the chance of observing $q_i$ and the sum is over your actual observations.) That is, you should estimate he trains 4.5 days per week. See the reference for how to compute the standard error of this estimate. As an extremely good approximation you can use the usual (Binomial) formulas.