I am calculating the age of lake sediments at the base of a sediment core by dividing the total sediment mass of the core ($\mathrm{mg} \ \mathrm{cm}^{-2}$) by the sediment accumulation rate ($\mathrm{mg} \ \mathrm{cm}^{-2}\ \mathrm{y}^{-1}$).

Both the sediment mass and the accumulation rate have variation associated with them. The sediment mass is a mean of 3 samples and the accumulation rate is reported as $\pm 10\%$.

It is my understanding that I can calculate the error associated with age as:

$\sqrt{\left(\frac{\delta x}{x}\right)^2 + \left(\frac{\delta y}{y}\right)^2}$

where $\delta x$ and $\delta y$ are the relative error of the measurements being divided (i.e., $x$ and $y$).

The error associated with the sedimentation rate is $\pm \ 10%$ but the error associated with the sediment mass in a standard deviation.

Are these forms of error compatible within the above error propagation formula?

Thank you.

  • 1
    $\begingroup$ stats.stackexchange.com/search?q=error%2Bpropagation $\endgroup$ – whuber Jun 9 '11 at 19:48
  • $\begingroup$ @whuber - thanks for suggesting that search. I did try that but unless I am missing it there isn't anything that answers how to propagate the error when one measure is relative error ($\pm$ 10%) and the other is a standard deviation. Perhaps my question isn't clear. $\endgroup$ – DQdlM Jun 9 '11 at 20:51

In some sense this depends on what you mean by $x$ and $\delta x$. Usually people mean that they are modeling $X$ as a random variable with mean $x$ and variance $(\delta x)^2$. Sometimes they mean the stronger condition that $X$ is actually Gaussian, and sometimes they have a broader meaning that $x$ and $\delta x$ can possible be other measures of the center and the spread.

A bit of calculus and handwaving shows that for small variations that are also approximatable as Gaussian, and $X$ and $Y$ independent, $f(X, Y)$ can be approximately described as having mean $f(x, y)$, and $(\delta f)^2 = (\delta x)^2 (\frac{\partial f}{\partial x})^2 + (\delta y)^2 (\frac{\partial f}{\partial y})^2$.

We can do the same thing for $a(m, r) = m /r$, where $a$ is the calculated age, $m$ is the mass, and $r$ is the rate.

$$ \begin{align*} (\delta a)^2 &= (\delta m)^2 / r^2 + (\delta r)^2 m^2 / r^4 \\ a^2 &= m^2 / r^2 \\ (\delta a)^2/a^2 &= (\delta m)^2/m^2 + (\delta r)^2 / r^2 \\ (\delta a)/a &= \sqrt{(\delta m)^2/m^2 + (\delta r)^2 / r^2} \\ \end{align*} $$

This matches the formula you have. You just have to convert between absolute errors and relative errors to be able to use it.

*EDIT*ed to add (incorporating comments): To convert the sedimentation rate to relative error, just use $(\delta r)/r = 10\% = 0.1$. You need to find the $\delta m$ = standard error for the mean. It's not clear whether you have $\delta m_i$ for each individual sediment core measurements. If you do, you want to find $m$ with a weighted mean and calculating the standard error is a bit tricky, but the prescription given above for general $f$ expands fine to three arguments. If it's not, the standard mean can be used and the variance in the sample can be used to calculate the standard error of the mean. The relative error is of course just $(\delta m)/m$.

  • $\begingroup$ thanks for your answer. I guess my real question is what you allude to in your last sentence. How do I convert the error from each term ($\pm 10%) and standard deviation into compatible forms so they can be used in those formulas? Thanks again. $\endgroup$ – DQdlM Jun 9 '11 at 20:53
  • $\begingroup$ @KennyPeanuts: Ah. There are various conventions. The most common by far is that $10\% = (\delta r) / r$, but multiples of this would not be surprising. With three samples, you can use the sample standard deviation to get an estimate for $\delta m$, though it's not quite as simple as the sample standard deviation. en.wikipedia.org/wiki/… $\endgroup$ – wnoise Jun 9 '11 at 22:01
  • $\begingroup$ awesome! thanks. I am assuming you mean standard error for $\delta m$ since thats what you linked to and it seems to make more sense. I edited my question to make it what I was asking a bit more clear and I am accepting your answer. If you are so inclined you may want to add what you have in the comment to the answer itself just to have it all in the the answer. Thanks again! $\endgroup$ – DQdlM Jun 10 '11 at 12:36

See my book: "Propagation of Errors" by Mike Peralta at amazon.com It treats the seldom covered topic of Second Order "Propagation of Errors"

  • 4
    $\begingroup$ Welcome to the site, @Mike. This is against our policies (see our FAQ); please do not advertize your book here. What we welcome you to do is to give a summary of the information there, so that a reader could get the gist of it. You could then reference your book as one place where people could get more information, but people wouldn't need to buy it--it wouldn't be advertizing. $\endgroup$ – gung - Reinstate Monica Feb 27 '13 at 21:37
  • 2
    $\begingroup$ I beg to differ a little, @gung. We always appreciate experts who weigh in and it's natural that when their work is relevant to a question that it should be referenced. But unfortunately a mere reference is inadequate as a stand-alone answer, as I'm sure you will appreciate, Mike. If you could amplify this answer to explain how your treatment answers the present question, I am sure many readers--now and future--would be delighted. $\endgroup$ – whuber Feb 28 '13 at 6:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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