Tell me more ×
Cross Validated is a question and answer site for statisticians, data analysts, data miners and data visualization experts. It's 100% free, no registration required.
up vote 3 down vote favorite
1

I collected 3D velocity data in a river. the recorded timeseries of velocity components $(u,v,w)$ contains turbulent and mean part. Reynolds Shear Stress can be estimated from

$$u=u'+U$$

$$\langle u'w' \rangle=\text{cov}( u'_{t}, w'_{t} )$$

and to extract velocity fluctuation

$$\langle u' \rangle=\sqrt{\text{var}( u'_{t})}$$

$$\langle w' \rangle=\sqrt{\text{var}( w'_{t})}$$

where $\langle\rangle$ shows the averaging over the time span of $t$.

My question is why $\langle u' \rangle\langle w' \rangle$ is not equal to $\langle u' w' \rangle$

Example:

$\langle u'w' \rangle=$ {55.0336, 24.3896, 22.4693, 21.9123, 31.9418, 34.791, 30.9995, -1.12979, -2.76629, -5.60678, -7.23715, -11.1097, -24.4944, -31.5994, -49.414, -92.9571, -97.5096, -91.7745, -141.235, -110.685}

$\langle u' \rangle^2=$ {691.168, 438.195, 402.749, 332.111, 470.982, 565.431, 454.04, 192.986, 94.7298, 110.964, 81.9545, 128.034, 198.726, 297.308, 508.916, 910.229, 922.968, 983.665, 1311.84, 1011.1}

$\langle w' \rangle^2=${46.2054, 28.1408, 43.3248, 32.9785, 32.092, 40.1115, 28.038, 13.3292, 6.74369, 7.3284, 6.98387, 11.5178, 18.7025, 29.0199, 39.9626, 60.4201, 74.1318, 80.8566, 89.0198, 78.8576}

share|improve this question

1 Answer

up vote 5 down vote

The notation is a bit convoluted, but if I read it correctly, the answer is: because $u'$ and $v'$ are not independent. With your notations we have

$$\langle u' \rangle\langle w' \rangle=\sqrt{\text{var}( u'_{t})\text{var}(w'_t)},$$

but

$$\langle u' w' \rangle=\sqrt{\text{var}(u_t'w_t')}$$

The relation $var(XY)=var(X)var(Y)$ holds for zero mean uncorrelated random variables $X$, $Y$. If variables $u'$ and $v'$ do not satisfy that, the relation needs not to hold.

Update I did not notice that you define $\langle u'w'\rangle=cov(u_t',w_t')$. Then it is not clear what $\langle \cdot\rangle$ really means. But if we use this definition, then the claim is that

$$cov(u_t',w_t')=\sqrt{var(u_t')var(w_t')}$$

which means that the correlation between $u_t'$ and $v_t'$ should be 1. Which in turn means that they are perfectly linear related.

share|improve this answer
$\langle \cdot\rangle$ is standard terminology in physics. It can represent both expectation and inner product. The OP abuses this notation with expressions like $\langle u' \rangle=\sqrt{\text{var}( u'_{t})}$; this would usually be written as a norm, $||u'||$ = $\sqrt{\langle u', u' \rangle}$ = $\sqrt{\text{cov}(u', u')}$ = $\sqrt{\text{var}(u')}$. This suggestive (and correct) terminology shows the question is really asking why the cosine of an angle does not always equal $1$. – whuber♦ Aug 5 '11 at 15:56
@mpiktas I dont think ${\rm var}(XY) = {\rm var}(X) {\rm var}(Y)$ for independent variables $X,Y$, in general. This would imply that $E(X^2){\rm var}(Y) - E(X)^{2} {\rm var}(Y) = E(X^2)E(Y^2)- E(X)^{2} E(Y)^{2}$ has to be true any independent variables $X,Y$ - it clearly is not - $X,Y$ being independent uniform(0,1) variables is a simple counterexample, since $E(X^2)=E(Y^2)=1/3$, $E(X)=E(Y)=1/2$, and ${\rm var}(X) = {\rm var}(Y) = 1/12$ and $(1/3)(1/12) - (1/4)(1/12)$ clearly doesn't equal $(1/9) - (1/16)$ – Macro Aug 5 '11 at 17:05
@Macro, yes the relation will hold if means are zero. For some reason I thought that it will hold in general case. I did not check carefully because my main point was that the relation holds only in some very particular case, so there is no surprise that it did not hold. That is however not an excuse for leaving such an obvious mistake. Thanks for catching it. – mpiktas Aug 5 '11 at 20:13
@whuber, I am somewhat familiar with such notation. I suspected that there is some abuse of notation, but on the other hand my interpretation was straightforward. I will probably remove my answer when the poster clarifies the question, since the answer is more of a comment. – mpiktas Aug 5 '11 at 20:18
@mp I think your answer's fine. – whuber♦ Aug 5 '11 at 20:32
show 1 more comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.