Dear Jochen and Fabrice,

I do not know if you are still interested in this question, but some clarifications seem motivated.

(1) The error propagation formula for sy contains an error, presumably just a misprint (A-B should be squared).

(2) Inverting the regression probably does not solve any problem, because if m is Gaussian with the right expected value, the inverse regression will yield a biased estimator that is not Gaussian. You must consider what your regression is about.

(3) The error propagation corresponds to approximating the ratio by a linear function of A-B and m. When you talk about degrees of freedom it appears that you think of the t distribution as adequate for a normalized version of this linear function, in particular scaled by the standard error sy as given. However the t distribution only appears exactly if all your three estimated variances, sA^2, sB^2 and sm^2 are proportional to a common chi-2 type variance estimate with given degrees of freedom (which is then the t df). This does not seem to be the case here, but think about it, for this description is often adequate in regression calibration type problems.

(4) If you want to be on the safe side you could use the smallest of the three degrees of freedom of the three standard errors. If you want something less conservative you must derive an approximate formula for the degrees of freedom analogous to the formula in Welch's test, that is the approximate t-test replacing the t-test in the two-sample problem when sample variances are known not to be the same under the null hypothesis.

(5) If you assume you know all three error variances involved, an exact confidence region can be derived by Fieller's method. But I think you know this, already. And it does not help you with the degrees of freedom, unless (again - see (4) above) you have a situation where all three estimated variances are proportional to one and the same estimated variance of chi-2 type. In this case you can introduce the t distribution with the adequate degrees of freedom in the derivation of Fieller's region.

Best wishes

RS

.

if A, B and m can be assumed normal, the distribution of Y is known (and known to be ugly ! see http://en.wikipedia.org/wiki/Ratio_distribution ) which leads to a straigtforward numerical computation of the CI

but this does not help with these degrees of freedom !

.

I can also formulate this question simpler: How do I get the CI of Y?

Would it be possible to used the individual CIs in the error propagation directly instead of the standard errors?

I did a simulation and found that the usual error propagation for the confidence intervals seems to work (the coverage rate of the propagated 95%CIs is around 0.03-0.06). Is someone of the opinion that this was just a conincidence? Or can someone provide a proof or at least a chain of logical arguments that this is (or may be) correct?

Well, Fabrice, another simplification: I can get m from the regression of yi on xi, but i can also get it from the regression of xi on yi. If I take the slope from the "inverse" regression, the values Y is obtained as

Y = -m(A-B)

Now there is no ratio. But this will have a product distribution, what I still find ugly :)

.

i see two issues there :

1) error propagation aims at computing the variance of a function of two or more random variables from the means and covariance of the original random variables ; this works fine with linear combinations but must be used with extreme care in the case of non linear functions, specially in the case of ratios !

(see for instance the example of the reciprocal normal :

http://en.wikipedia.org/wiki/Propagation_of_uncertainty )

2) getting the variance or an approximation of it is not enough to get a CI ... except in some special cases, for instance if the random variable is gaussian ; this is clearly not the case for your Y = (A-B)/m which has a pretty involved distribution

or am i just missing something ?

.

Assume we measure two values A and B, using some apparatus. We know these values are uncertain. By physical reasoning, testing, repeated measurements, or manufacturer’s specifications, we estimate the magnitude of their uncertainties. u{A} is the absolute error in A, and u{B} is the absolute error in B. The relative errors are u{A}/A and u{B}/B.

What are the uncertainties in a calculation that uses A and B? What follows are rules that give the uncertainty expression for basic arithmetic operations using A and B, then what to do for more complicated expressions.

Addition and Subtraction:

The square of the uncertainty in the sum or difference of two numbers is the sum of the squares of individual absolute errors.

ole2.gif(3)

For example, if

• A = 2.5 grams,

• u{A} = 0.4 grams,

• B = 4.1 grams,

• u{B} = 0.3 grams,

then,

ole3.gif

You state the sum of these two masses as, “6.6 ± 0.5 g” (g being the abbreviation for grams). The mass difference is “1.6 ± 0.5 g.”

This procedure gives an error on the sum or difference that is larger than either individual uncertainty, but smaller than u{A} +u{B}. This is appropriate because if our measurement of A strays from the true value in one direction (either greater than or less than), our measurement of B is just as likely to fluctuate in the opposite direction as well as in the same direction. Because B can fluctuate in the same direction, it will make the error on the sum larger than u{A}. But because it can also go in the opposite direction, it will not increase the error as much as the maximum error u{A} + u{B}.

Another example of the application of the rule for the error on a difference is when you are asked if two measurements are consistent. Are they the same numbers? You perform the SET. Suppose your first measurement of the oscillation period of a pendulum is t1 = 4.0 ± 0.1 seconds and the second is t2 = 3.85 ± .05 seconds. To find if the two measurements are consistent, you calculate the positive difference (t1 - t2 ) = 0.15 seconds and the error on that difference:

ole4.gif

Since the difference is less than twice the error (0.15 ≤ 2×0.11 = 0.22), the SET is true, and you say that the two measurements are consistent.

.

Jochen, having Y as a product solves point (1) above as the exact formula for the variance is known

now, point (2) is still open ...

.

The distributions are gaussian. And simulations show that the so propagated 95%CIs do in fact have a coverage of ~95%.

.

the product of two gaussian random variables X and Y is distributed as a linear combination of two (correlated) chi-square random variables:

2*XY = (X+Y)^2 − (X−Y)^2

X+Y and X−Y being gaussian random variables, therefore (X+Y)^2 and (X−Y)^2 are chi-square distributed with 1 degree of freedom

this linear combination is not gaussian in general (take X = Y !) so point (2) remains open ; what your simulations show, however, is that, in your case, it may be not so far away from gaussian (95% is not an extreme value ...)

.

Thanks, Fabrice. We are talking about real data analysis, and here any distributional model is only an approximation. The key is that the approximation is reasonable and "good enough". Since the real data are not exactly normal distributed anyway, any derivation is again approximate. So, also for the derivations the question remains: are they "good enough". With sufficient math skills one may be able to answer this question theoretically (at least get some boundaries for the problem and then let an expert decide if this is tolerable), but if not, a simulation may show its adequacy (or its impracticality). Do you agree?

.

given two _independent_ normal random variables X~N(mX, sX^2) and Y~N(mY, sY^2), it is easy to show from its moment generating function that if mX/sX and mY/sY are _both_ large, the distribution of the product XY is close to a normal N(mX*mY, mX^2*sY^2+mY^2*sX^2)

fig 16 of the reference below shows an example of the "true" (obtained from monte carlo simulation or numerical integration) distribution and the normal approximation when mX/sX = 5 and mY/sY=2 ; although 5 and 2 cannot be considered so large, the normal approximation does not look so bad ...

looking at the skewness as a function of the means and variances of X and Y (formula 36 of the reference) may be a way to check if the normal approximation is acceptable

http://www.math.canterbury.ac.nz/research/ucdms2003n15.pdf

.

Dear Jochen and Fabrice,

I do not know if you are still interested in this question, but some clarifications seem motivated.

(1) The error propagation formula for sy contains an error, presumably just a misprint (A-B should be squared).

(2) Inverting the regression probably does not solve any problem, because if m is Gaussian with the right expected value, the inverse regression will yield a biased estimator that is not Gaussian. You must consider what your regression is about.

(3) The error propagation corresponds to approximating the ratio by a linear function of A-B and m. When you talk about degrees of freedom it appears that you think of the t distribution as adequate for a normalized version of this linear function, in particular scaled by the standard error sy as given. However the t distribution only appears exactly if all your three estimated variances, sA^2, sB^2 and sm^2 are proportional to a common chi-2 type variance estimate with given degrees of freedom (which is then the t df). This does not seem to be the case here, but think about it, for this description is often adequate in regression calibration type problems.

(4) If you want to be on the safe side you could use the smallest of the three degrees of freedom of the three standard errors. If you want something less conservative you must derive an approximate formula for the degrees of freedom analogous to the formula in Welch's test, that is the approximate t-test replacing the t-test in the two-sample problem when sample variances are known not to be the same under the null hypothesis.

(5) If you assume you know all three error variances involved, an exact confidence region can be derived by Fieller's method. But I think you know this, already. And it does not help you with the degrees of freedom, unless (again - see (4) above) you have a situation where all three estimated variances are proportional to one and the same estimated variance of chi-2 type. In this case you can introduce the t distribution with the adequate degrees of freedom in the derivation of Fieller's region.

Best wishes

RS