3
$\begingroup$

Let's say that an event happens every $\Delta x$ seconds and we sample the time $x$ with some error. The events are not sampled reliably so some are missing in the sample set. However, we do know the ordinal number $k_i$ of every sampled time $x_i$ so we always know exactly how many samples were skipped in between.

I'd like to estimate $\Delta x$ as accurately as possible and I'd like to calculate error bounds on the resulting value.

My intuitive solution is to calculate the difference between every sample and the first one. Then I'd sum up all of these differences and divide them by the total number of $\Delta x$ included in the sum. In other words:

$$ \Delta x \approx \frac {\sum_{i=1}^n x_i-x_0}{\sum_{i=1}^n k_i-k_0} $$

But I don't know if my intuition is correct and don't have enough knowledge to confirm or disprove it. Also, I have no idea how to calculate the error bounds.

EDIT:

Assume normal distribution for sample time errors. Should be good enough for my use case.

$\endgroup$
1
  • $\begingroup$ I know little about statistics so my notation is probably off. Feel free to edit the question or post comments on what to fix. Also, I had no idea which tags to use so feel free to edit those as well. $\endgroup$ Commented Dec 2, 2019 at 14:32

1 Answer 1

2
$\begingroup$

A question about model specification: is the error really iid normal? If so, then the $x_i$ are not necessarily monotic (i.e., it's possible that $x_2 < x_1$), but for many measurement systems that type of error is impossible. But let's ignore that for now.

For practical purposes, probably $$ \Delta x \approx \frac{x_n - x_0}{k_n-k_0}$$ would be fine, since the variance of the error in that estimate is dropping like $k_n-k_0$.

If you want to do more and you really think the error is normal, then this problem maps exactly to linear regression (you're trying to predict $x_i$ from $k_i$); the slope of the fit will give you your $\Delta$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.