The Briffa-Osborn Variance Adjustment

16 Comments

  1. Dave Dardinger
    Posted Mar 20, 2007 at 4:09 PM | Permalink

    I assume that in the formula for calculation of effective n there should be a second close paren after “rbar”?

  2. Henry
    Posted Mar 20, 2007 at 5:46 PM | Permalink

    The link to /scripts/proxy/briffa,osborn.adjustment.txt does not work: spot the comma where there should be a dot

  3. Climate Tony
    Posted Mar 20, 2007 at 9:48 PM | Permalink

    Steve,
    This is the best blog on the net. But why is MM03 under links and not articles?

  4. Posted Mar 21, 2007 at 4:52 AM | Permalink

    Thanks Steve! Let’s see, I have a simple question.

    I don’t have the original Osborn paper, but Frank et al seems to explain the method sufficiently:

    In dendrochronology it is common practice to create a mean-value function as the best estimate of the trees’ signal at a site. This averaging process helps eliminate noise particular to individual trees and cores thereby increasing the signal quality.

    IOW, measurement contains wanted part (signal) and unwanted part (noise). Noise term can be reduced by averaging (I wouldn’t use term ‘eliminate’ here). Measurements share a common signal, so averaging does not affect the signal part.

    The variance of the mean-value function, however, depends upon the number of series averaged together and their interseries correlation (Wigley et al. 1984).

    This is obvious. If we use only one tree, variance of the measurement

    y(t)=s(t)+n(t)

    over time is

    \sigma _s ^2+ \sigma _n ^2 ,

    assuming uncorrelated s and n. If there are more trees, we take the average. Average doesn’t affect the signal part, but it reduces the power of the noise. Efficiency of this reduction depends on how correlated the noise term is between the trees. For a given year, the average is

    Y=\frac{1}{N}\sum y = \frac{1}{N}\sum s + \frac{1}{N}\sum n = s + \frac{1}{N}\sum n

    Expectation value is (given s, E(n)=0 )

    E(Y)= s+ E(\frac{1}{N}\sum n)=s

    Looks good to me. If you scale Y, you’ll obtain a biased estimate of s, right? Now, where do I go drunk?

  5. bernie
    Posted Mar 21, 2007 at 5:36 AM | Permalink

    I am not sure I buy the independence of signal(s) and noise(n), where the signal is a pure temperature signal. If y=f(t, p, tp , x), where t is the temperature signal, p the precipitation and tp an interaction term and x is all other factors including random noise then if s = f(t) and n=f(p, tp , x) then s and n can clearly be correlated. Can you really separate strong interaction effects between t and any other factors, with p being the obvious one, by this approach? Does this make sense? For example, and perhaps simple mindedly, if a given ring width is produced by average temperatures and average precipitation or by above average temperatures and above average precipitation, how do you separate temperature and precipitation? But I assume that this is so obvious a point dendrochronologists must have addressed it, would they not? It sounds like, for example, they choose sites based on some assumptions that attempt to control for other factors like precipitation – but frankly the logic of assuming constant that which inherently fluctuates is very puzzling. This is, I assume, part of the argument for up-to-date records so that this assumption of independence can be tested.

  6. Martin à…
    Posted Mar 21, 2007 at 2:35 PM | Permalink

    Steve,

    Am I readin this correct: “…this effect was corrected for by scaling by the square root of the effective number of independent samples available in each year”?

    If one assumes that each sample contains a signal plus noise, doesn’t different scaling for different years distort the signal? After averaging chronologies of different length, the result must be the signal plus different amount of noise for different time periods, depending on the number of effective samples in each time period. Isn’t the effect of the correction that you lower the signal amplitude for periods where you have less data!? Instead of increasing the error bars!!

  7. Nicholas
    Posted Mar 21, 2007 at 10:09 PM | Permalink

    I think Martin is right, the difference in variance is due to different amount of noise cancellation, correct? If so, you just have to live with the higher noise when you have less samples, and represent this as larger error bars. Scaling will affect both the signal and the noise, thus masking the potential signal in the periods where there are less samples. I don’t see how this can possibly be justified. Why is this procedure being used repeatedly if it hasn’t been shown to be a valid statistical technique?

  8. Posted Mar 22, 2007 at 12:58 AM | Permalink

    #6

    If one assumes that each sample contains a signal plus noise, doesn’t different scaling for different years distort the signal?

    Yes. This adjustment leads to a biased estimate.

    Isn’t the effect of the correction that you lower the signal amplitude for periods where you have less data!? Instead of increasing the error bars!!

    Yes. In the past, we have sparser data. Past variations will be scaled towards zero. Increasing the error bars is not a legal move in climate science. Those bars might reach the current temperature levels, that wont do.

    #7

    Why is this procedure being used repeatedly if it hasn’t been shown to be a valid statistical technique?

    Because it makes the results look nice.

  9. Martin à…
    Posted Mar 22, 2007 at 5:08 PM | Permalink

    Ok , so

    Y(t) = X(t) \frac{1}{\sqrt {n('t)}}

    that means that they are doing the opposite to what I said earlier, in fact amplifying the mean where you have fever effective samples.

    What I don’t understand is how this can cause “…the variance Var(Y) to be independent of sample size” if

    Var(\overline{X})=\frac{1}{n'}

  10. bernie
    Posted Mar 22, 2007 at 7:16 PM | Permalink

    Whichever way you cut it, the error term is correlated with one or more of the indpendent variables and you have a big problem.

  11. Steve McIntyre
    Posted Mar 27, 2007 at 3:41 PM | Permalink

    UC, I posted on this topic a long time ago here http://www.climateaudit.org/?p=418

  12. jae
    Posted Mar 27, 2007 at 4:24 PM | Permalink

    #5:

    For example, and perhaps simple mindedly, if a given ring width is produced by average temperatures and average precipitation or by above average temperatures and above average precipitation, how do you separate temperature and precipitation? But I assume that this is so obvious a point dendrochronologists must have addressed it, would they not? It sounds like, for example, they choose sites based on some assumptions that attempt to control for other factors like precipitation – but frankly the logic of assuming constant that which inherently fluctuates is very puzzling. This is, I assume, part of the argument for up-to-date records so that this assumption of independence can be tested.

    Amazingly, I don’t think the dendrochronologists HAVE addressed this fundamental and extremely important issue. They seem to avoid the question like the plague. Tree rings are often very good proxies for moisture. I don’t think they are generally valuable as “thermometers” for many reasons.

  13. Posted Mar 28, 2007 at 6:11 AM | Permalink

    Steve, Martin

    Frank et al:

    Multiplication of the mean timeseries with the square root of Neff at every time t theoretically results in variance that is independent of sample size.

    Seems that Eq (6) is not correct, X and Y mixed (??)

    Steve wrote:

    Because there is little reason to believe that the annual variance in the early period was substantially greater than at present, Briffa and Osborn [1999] proposed a variance adjustment methodology (applied here) as follows.

    Because Briffa and Osborn have never heard of filtering theory (specifically, the problem of estimating the state of a stochastic dynamical system from noisy observations), they decided to go the easy way and just scale the observation so that the result looks good.

  14. Steve Sadlov
    Posted Mar 28, 2007 at 9:57 AM | Permalink

    RE: #12 – I would however concede that some species in Marine West Coast, and in the wetter coastal portions at the northern margins of Mediterranean climates may be local temperature proxies. But how many such places are there on earth and how few of the overall claimed set of global tree ring proxies are actually found in such places?

  15. Martin à…
    Posted Mar 30, 2007 at 2:35 AM | Permalink

    This would have made sense:

    A site mean with fewer effective samples might be expected to have a lower (temperature)signal to noise ratio. Then the signal part has lower amplitude after normalization than in a mean from a site with more effective samples. If all site means were adjusted to compensate for this before the total mean is calculated, the signal part in the total mean would be independent of which sites that are included at a specific time.

    How this compensation should look I don’t know, but multiplying with n’ is probably in the right direction.

    But, this compensation should be done uniformally over time but different on different site means before calculating the total mean. I can’t see that this is what they do.

  16. Martin à…
    Posted Mar 30, 2007 at 2:40 AM | Permalink

    A correction to my previous post.

    I wrote multiplying with n’ is probably in the right direction. I meant dividing with square root of n’ is probably be in the right direction.

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