Estimating the covariance

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  • #1
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Main Question or Discussion Point

Hi everybody! I have to write a protocole for our last experiment about elasticity and torsion (in physics), and as an extra question I am asked to calculate the Poisson ratio and to calculate the correlated error by estimating the covariance. Unfortunately I have never done that before, and I don't really understand what the covariance is. Here is the formula for the Poisson ratio:

##\mu = \frac{E}{2G} - 1##

I imagine that ##Cov(X,Y)## refers to ##Cov(E,G)## in our case. The problem is that we have only one value for E and for G (determined experimentally), which is probably why we were asked to estimate the covariance. I've seen that ##0 < Cov(X,Y) < 1## when the two values grow together, and that's the result we got:

##G = \frac{E}{2(1 + \mu)} = kE = 0.3708 E## (that's the value we got for the Poisson ratio)

Since ##\mu## is supposed to be a constant, a graph of ##G## in function of ##E## should be linear. Can I then estimate ##Cov(E,G) \approx 1##? Or did I completely misunderstand it?


Thanks a lot in advance for your answers.


Julien.
 

Answers and Replies

  • #2
gleem
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I'm not sure I understand the use of covariance in the context of your experiment which is not of a statistical nature since you have only one measurement.. Since G =kE you only have one independent variable E which determines G and for your case G changes by an amount proportional to E, There exists a statistical quantity called the linear correlation coefficient which is one for you case. But again you have only one measurement so it is indeterminate.

The covariance is not constrained to have a value of less than one.

Let me explain a bit.

The uncertainty in the value of a function F is related to the statistical uncertainties in the values of the variables xi by the equation

σF2 = ∑i σI2(∂F/∂xi)2 +2⋅ Σij σij2(∂F/∂xi)(∂F/∂xj)

where σi2 is the variance of xi i.e. the square of the statistical uncertainty in xi

where σij2 is the covariance of xj and xj where σij2 = < (xi- < xi>)⋅( xj- <xj>) >

The brackets < quantity > refer to the average value of quantity,

The covariance is important when the value of one variable affects the value of another. If they are totally independent then in the limit of large number of measurements the covariance approaches zero.

Obviously the the variances and the covariance can only be determined if you have two or more values of the variables.

For you case as a guess since the covariance is the average of the product of the uncertainties of the variables E and G the covariance might be taken as ΔE⋅ΔG your estimated experimental uncertainties in E and G. and the error associated with this is 2∂μ/∂E⋅∂μ/∂GΔEΔG.

Perhaps someone can critique this approach.

If you are continuing in the experimental physical sciences I highly recommend that you find a copy of "Data Reduction and Error Analysis for the Physical Sciences" by Bevington ( original version) or the revised edition by Bevington and Robinson.
 
  • #3
chiro
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Hey JulienB.

You can pick up a good multi-variate statistics book for the actual estimator but if you want to derive it you are going to have use something like MLE.

You can also just use the sample estimates for the different components (equating sample mean to actual mean) and doing a degree of freedom correction.
 
  • #4
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Thanks a lot to both of you for your answers. At the end I used the inequality ## \frac{1}{3} E < G \leq \frac{1}{2} E ## to state that ##0 \leq Cov(E,G) \leq 1##. I calculated the propagation of error for both extremes and found that the correlation had a negligible impact on the uncertainty. I think that's what was asked, as our advisor specifically said that we didn't have to calculate the covariance.

Best regards, thank you for your reading recommendations too.


Julien.
 

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