Accounting for missing samples when calculating average over time deltas I need to calculate the average over a series of data points where each represents the time difference from its predecessor. Only the time delta is relevant here, not the data itself:
08 sec
10 sec
12 sec
06 sec

Samples are collected into a buffer and once either the buffer is filled OR a specific time has passed, the average of the buffer is taken and used for further calculations. 
My problem is that I don't know how to account for the case where time is up and the buffer has not been filled completely or not at all, as this would significantly distort the result.
The example dataset above would net the average of 9sec which would be divided by some pre-determined ideal-average and used to adjust a parameter that controls the influx of new samples. In short, something like a self-tuning rate limiter.
newModifier = oldModifier * actualSamplesPerSec / idealSamplesPerSec

 A: First I'll just describe the problem in the language of estimating the mean of a random variable.
For an exponentially-distributed random variable, given a full buffer of $n$ observations $x_1,\dots,x_n$ the maximum likelihood estimate for the mean is just the sample mean,
$$\frac{1}{n}\sum_{i=1}^n x_i.$$
If the last observation is right-truncated so we only know $x_n\geq T-x_1-\dots-x_{n-1}$ where $T$ is the maximum time, then the maximum likelihood estimate is
$$\frac{T}{n-1}$$
where for $n=1$ there is no maximum; effectively the estimate is $+\infty$.

You mention the output will be compared against an ideal value. To regularize the output for small $n$, one simple approach would be to pre-fill the buffer with one sample of this ideal value.
More generally you can add a prior belief expressed as a pair $\alpha,\beta$. This can be thought of as adding $\alpha$ extra dummy values to the buffer with total $\beta$. Using $\alpha=1$ and $\beta=\text{ideal value}$ is like adding the ideal value as an extra sample. But if you ever need the extra flexibility you could for example use $\alpha=\tfrac 1 2$ and $\beta=\tfrac 1 2(\text{ideal value})$ to give less influence on the output.
For the right-truncated data the estimate would be
$$\frac{\beta+T}{\alpha+n-1}$$
while for a full buffer the estimate would be
$$\frac{\beta+\sum_{i=1}^n x_i}{\alpha+n}.$$
These can be described from a Bayesian point of view as the inverse of the posterior mean rate using a textbook Gamma conjugate prior as described on Wikipedia here:  https://en.wikipedia.org/wiki/Exponential_distribution#Bayesian_inference
(It's a bit dodgy to use the inverse of a mean, but I didn't want to get an extra $-1$ in the denominator, and the result might not be being used in a linear way anyway.)
