[R-meta] Estimate variance from time series data

[Arne Janssen]

Dear Wolfgang,

Thanks for your quick reply. The question really is what is the sample 
size. Suppose there are 3 time series of 5 data points through time each 
that I want to combine. Given are the average and s.d. of the 3 series 
per time (so 5 averages and s.d.).

I would like to obtain an overall average  and s.d. of these 3 time 
series. If we consider that N = 15, I can use the standard method to 
combine the 3 series. If we consider N to be 3, because there are only 3 
time series, I would indeed need to know the correlation among the time 
series to estimate the s.d., but this correlation is unknown. Please advise.

Thanks and best wishes,
Arne


On 14-Aug-18 22:50, Viechtbauer, Wolfgang (SP) wrote:
> Hi Arne,
>
> It is not entirely clear to me what you are trying to do. Do you want to know the mean and SD when throwing together the N1 measurements from timepoint 1 and the N1 measurements from timepoint 2 from the same group, such that there are 2*N1 measurements in total now for the group? (or 3*N1 if there were three timepoints and so on).

Reply: This is indeed what I want to do.
>   Then the same equation could be used as if there are independent subgroups.
>
> For example:
>
> ### Suppose we have the mean, SD, and size of several subgroups, but we
> ### need the mean and SD of the total/combined groups. Code below shows
> ### what we need to compute to obtain this.
>
> ### simulate some data
> n.total<- 100
> grp<- sample(1:4, size=n.total, replace=TRUE)
> y<- rnorm(n.total, mean=grp, sd=2)
>
> ### means and SDs of the subgroups
> ni<- c(by(y, grp, length))
> mi<- c(by(y, grp, mean))
> sdi<- c(by(y, grp, sd))
>
> ### want to get mean and SD of the total group
> mean(y)
> sd(y)
>
> ### mean = weighted mean (weights = group sizes)
> m.total<- sum(ni*mi)/sum(ni)
>
> ### SD = sqrt((within-group sum-of-squares plus between-group sum-of-squares) / (n.total - 1))
> sd.total<- sqrt((sum((ni-1) * sdi^2) + sum(ni*(mi - m.total)^2)) / (sum(ni) - 1))
>

Here is my doubt: The sum(ni) is now larger than the number of 
replicates (4 time series, so 4 replicates, n should be 4), am I correct?
> ### check that we get the right values
> m.total
> sd.total
>
> This would be the case for independent subgroups. Now let's simulate data for 50 individuals measured twice:
>
> library(MASS)
>
> Y<- mvrnorm(50, mu=c(0,0), Sigma=matrix(c(1, .8, .8, 1), nrow=2))
> y<- c(t(Y))
> grp<- c(1:50, 1:50)
>
> ### means and SDs of the subgroups
> ni<- c(by(y, grp, length))
> mi<- c(by(y, grp, mean))
> sdi<- c(by(y, grp, sd))
>
> ### want to get mean and SD of the total group
> mean(y)
> sd(y)
>
> ### mean = weighted mean (weights = group sizes)
> m.total<- sum(ni*mi)/sum(ni)
>
> ### SD = sqrt((within-group sum-of-squares plus between-group sum-of-squares) / (n.total - 1))
> sd.total<- sqrt((sum((ni-1) * sdi^2) + sum(ni*(mi - m.total)^2)) / (sum(ni) - 1))
>
> ### check that we get the right values
> m.total
> sd.total
>
> Still works. However, when it comes to computing the sampling variance for m.total (or some function thereof), one cannot treat these two cases as the same. In the first case, we really have sum(ni) independent measurements, so var(y) / sum(ni) would be the correct sampling variance of m.total, but not so for the second case. You would need to know the correlation between the measurements over time to compute an appropriate sampling variance of m.total in the second case.
>
> Best,
> Wolfgang
>
> -----Original Message-----
> From: R-sig-meta-analysis [mailto:r-sig-meta-analysis-bounces using r-project.org] On Behalf Of Arne Janssen
> Sent: Monday, 13 August, 2018 19:23
> To: r-sig-meta-analysis using r-project.org
> Subject: [R-meta] Estimate variance from time series data
>
> Dear list members,
>
> I am doing a meta-analysis with data that are often presented as
> repeated measures of population densities, but authors sometimes also
> give overall averages and s.d. or s.e.. Because I want to combine these
> data into one analysis, I am interested in the overall effect size of
> the repeated measures, so would like to combine all data of the time
> series into one average and s.d. The time series are repeated several
> times, yielding data of the following form:
> Time                Treatment 1                Treatment 2
>                           N    Ave    s.d.                N    Ave    s.d.
> 1                      N1    x1,1    sd1,1          N2    x2,1    sd2,1
> 2                      N1    x1,2    sd1,2           n2    x2,2    sd2,2
> ...
> ...
> ...
>
> What I want to obtain is one average and s.d. per treatment through time.
> The average is straightforward, but I cannot come up with a calculation
> for the s.d.
>
> The formula normally used for calculating the combined variance of two
> series of measurements:
>
> Var = (s1^2(n1 -- 1) + s2^2(n2 -- 1) + n1(X-x1)^22 + n2(X-x2)^22)/( (n1
> + n2 -- 1)
>
> does not seem to apply when combining the measurements through time,
> because this increases the number of replicates, which in my opinion,
> should be the number of time series and not the number of observations.
>
> I hope I made myself clear, and would be very grateful if you could
> advise me on this matter.
>
> Thanks very much in advance.
> Arne Janssen
> .
>
>