1
$\begingroup$

I'm propagating error in the parameters determined by the following growth function...

$$ \hat{y} = ae^\frac{t}{b} + (1- a)e^\frac{t}{c} $$

Say I have another model that uses the parameters {a,b,c} to estimate another rate $z$.

$$ z = a \times b \times c $$

I want to know the 95% confidence interval for z. Can I just assume the true parameter for $z$ falls between the product of the 2.5% limit for {a,b,c} and the product of 97.5% for {a,b,c}?

Improve this question
$\endgroup$
5
  • $\begingroup$ It is hard to determine what your notation means, since "$*$" and "$\pm$" are employed in unusual ways. Could you explain the sense in which "$z=a*b*c$" is a "model" and what your formula for $\epsilon$ means? $\endgroup$
    – whuber ♦
    Commented Aug 20, 2015 at 14:14
  • $\begingroup$ I see what you mean, I edited my question $\endgroup$
    – tomc4yt
    Commented Aug 20, 2015 at 14:36
  • 2
    $\begingroup$ Have you considered reparameterizing your model to include $z$ directly, such as $$y=\frac{z}{bc} e^{t/b} + \left(1 - \frac{z}{bc}\right) e^{t/c}\text{?}$$ $\endgroup$
    – whuber ♦
    Commented Aug 20, 2015 at 18:16
  • $\begingroup$ Really interesting notion, the actual model is a little more complex but I think I will try that out. $\endgroup$
    – tomc4yt
    Commented Aug 20, 2015 at 18:53
  • $\begingroup$ When you fit the model with Maximum Likelihood (assuming that is appropriate), it's a great solution, because you can then obtain a confidence interval for $z$ using your favorite ML-based method. $\endgroup$
    – whuber ♦
    Commented Aug 20, 2015 at 20:10

1 Answer 1

Reset to default
1
$\begingroup$

The product of the confidence intervals for $a,b,c$ will not be a valid confidence interval for $z$, even in the case of $a,b,c$ being independent (which, for the record, is especially unlikely). As a simple example, suppose we only considered $a$ and $b$ and the confidence interval for $a \times b$. If the each of these confidence intervals were (-1,1), then using that formula, we would end up with (1,1) as our confidence interval for $a \times b$. If $a$ and $b$ were independent, this would clearly be false.

If you have the covariance matrix for $a, b, c$, and a fairly large sample size, a straightforward method is to use the delta method (https://en.wikipedia.org/wiki/Delta_method). In general, if $\beta$ is your vector of parameters, if you'd like a confidence interval of $f(\beta)$, it can be created from

$f(\hat \beta) \pm z * ( f'(\hat \beta)^T * \hat \Sigma * f'(\hat \beta))$

where $f$ is your function of interest (in this case, we can think of $f(\beta) = \beta_1 \times \beta_2 \times \beta_3$ where $\beta_1 = a, \beta_2 = b, \beta_3 = c$), $f'$ is the vector of derivatives and $\hat \Sigma$ is the estimated covariance matrix and $z$ is the standard z-score associated with the confidence interval.

Improve this answer
$\endgroup$
3
  • 2
    $\begingroup$ Would you be able to point me to somewhere where I could read up on why what you said in the first paragraph is true? $\endgroup$
    – tomc4yt
    Commented Aug 20, 2015 at 18:27
  • $\begingroup$ Wow, that statement was incorrect! See new edit. I was thinking about the about the additive scale when I wrote that. Thanks for not blindly accepting that statement! $\endgroup$
    – Cliff AB
    Commented Aug 20, 2015 at 19:44
  • $\begingroup$ What do the stars and crosses denote? If simply multiplication, I'd recommend leaving them out, as is standard mathematical notation. $\endgroup$
    – A. Donda
    Commented Aug 21, 2015 at 3:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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