Variance of n discrete time series

I'm currently having problems understanding how to calculate the variance between n given time series, each containing data of exactly one day.

So basically, I'm given one array containing double values (which have at most 4 decimal places) from a specified time interval (granularity of measurements is 30 minutes, which means there are 48 samples a day, defining a time series).

Since the variance needs the probability for each value that may occur (the values reach from 0 to mostly 10, even though all values between 0 to infinity COULD appear, the probability is just very (very) low that the value is greater than 10).

So now I want to ask how to calculate these probabilities for each value, or is there a approximative way (or is it even necessary?).

I'm (obviously) not that advanced in statistics, it is just important in context of a software project.

Thanks for any suggestions or answers in advance.

• I am trying to understand what you are trying to calculate the variance of. For one day, you have a 48-long time series. There you could use the Newey-West estimator that takes into account that the series might be autocorrelated. But I cannot understand what happens beyond one day. Do you need separate estimators for each day? Then do as I said above. Do you need some "overall" estimator of variance? Then how precisely is that defined? – Richard Hardy Jan 17 '18 at 18:32
• Well, a time series is defined as one day which then again is divided in 48 values. My dataset contains n days (400 would be an appropriate value for n). So now I want the variance of all n days. I'm not too familiar with statistical vocabulary, so i hope you now understand what I'm trying to express. – a.j.stu Jan 17 '18 at 18:40
• If you assume the variance is the same on all days, then you could concatenate all your data into one 48*400-long vector and use an estimator that would account for the peculiar autocorrelation structure (where there potentially is intra-day autocorrelation and some inter-day autocorrelation). I do not know of an out-of-the-box solution... – Richard Hardy Jan 17 '18 at 18:46