5
$\begingroup$

I have two estimated parameters, $\beta$ and $\alpha$, and the standard error for each of the parameters.

I want to find the standard error of the combined $\frac{\beta}{1-\alpha}$. The standard error can be estimated by using the Taylor series approximation from Mood et al., 1974, p.181, but I don't know the mean values for the fit parameters, or the $\mathrm{Cov}(x,y)$.

$$\mathrm{Var}(X/Y)= (X_{\mathrm{mean}}/Y_\mathrm{{mean}})^2\times(\mathrm{Var}(X)/X_\mathrm{{mean}}^2+\mathrm{Var}(Y)/Y_{\mathrm{mean}}^2-2\times\mathrm{Cov}(X,Y)/X_{\mathrm{mean}}\times Y_{\mathrm{mean}}))$$

$\endgroup$
  • $\begingroup$ Without those quantities you don't know, I'm not sure a lot can be done. Do you have the original data? $\endgroup$ – Glen_b May 12 '15 at 23:23
4
$\begingroup$

Usually when you estimate the model you get the covariance matrix of the parameter estimates. If you assume that the parameter estimates are normally distributed (a standard assumption for large samples and small samples with normal errors), then you have the correlation coefficient between parameter estimates, their means and standard deviations. So, you can use this to plug into Gaussian ratio distribution to get an answer to your question.

Look at this paper: Marsaglia, George. "Ratios of normal variables." Journal of Statistical Software 16.4 (2006): 1-10.

$\endgroup$
2
$\begingroup$

We usually use the exact same symbol to denote the obtained estimate of a parameter (a number) and the estimator we used, which is a random variable (a function). To distinguish, I will use the following notation:
True values of unknown paramters : $\alpha,\beta$
Obtained estimates from a specific sample: $\hat \alpha, \hat \beta$
Estimators used: $a, b$.

We are interested in the variance (and then the standard error), of a function of the estimators, $h[a, b]$. We do indeed say "standard error of the estimate" but this strictly speaking is wrong: estimates are fixed numbers, they do not have a variance or a standard deviation.

We can approximate $h[a, b]$ by a first-order Taylor expansion around the obtained estimates:

$$h[a, b] \approx h[\hat \alpha, \hat \beta]\; + \;\frac {\partial h[a, b]}{\partial a}\Big|_{\{\hat \alpha, \hat \beta\}}\cdot (a - \hat \alpha)\;+\;\frac {\partial h[a, b]}{\partial b}\Big|_{\{\hat \alpha, \hat \beta\}}\cdot (b - \hat \beta)$$

Rearranging,

$$h[a, b] \approx \Big[ h[\hat \alpha, \hat \beta]\; - \;\frac {\partial h[a, b]}{\partial a}\Big|_{\{\hat \alpha, \hat \beta\}}\cdot \hat \alpha\;-\;\frac {\partial h[a, b]}{\partial b}\Big|_{\{\hat \alpha, \hat \beta\}}\cdot \hat \beta\Big]$$

$$+\;\frac {\partial h[a, b]}{\partial a}\Big|_{\{\hat \alpha, \hat \beta\}}\cdot a\;+\;\frac {\partial h[a, b]}{\partial b}\Big|_{\{\hat \alpha, \hat \beta\}}\cdot b$$

Why the re-arrangement? Because, the terms in the big brackets are all fixed numbers. And fixed numbers do not have a variance, and, when they enter additively, they do not affect the variance of the terms that they do. So

$${\rm Var} \left(h[a, b]\right) \approx {\rm Var} \left(\frac {\partial h[a, b]}{\partial a}\Big|_{\{\hat \alpha, \hat \beta\}}\cdot a\;+\;\frac {\partial h[a, b]}{\partial b}\Big|_{\{\hat \alpha, \hat \beta\}}\cdot b\right)$$

In our case

$$h[a,b] = \frac {b}{1-a} \implies \frac {\partial h[a, b]}{\partial a} = \frac {b}{(1-a)^2} \implies \frac {\partial h[a, b]}{\partial a}\Big|_{\{\hat \alpha, \hat \beta\}} = \frac {\hat \beta}{(1-\hat \alpha)^2}$$

and $$\frac {\partial h[a, b]}{\partial b} = \frac {1}{(1-a)} \implies \frac {\partial h[a, b]}{\partial b}\Big|_{\{\hat \alpha, \hat \beta\}} = \frac {1}{(1-\hat \alpha)}$$

Substituting, and using the standard formula for the variance of the sum of two random variables,

$${\rm Var} \left(\frac {b}{1-a}\right) \approx \left(\frac {\hat \beta}{(1-\hat \alpha)^2}\right)^2\cdot {\rm Var}(a)\;+\;\left(\frac {1}{(1-\hat \alpha)}\right)^2\cdot {\rm Var}(b) \\+\; 2\frac {\hat \beta}{(1-\hat \alpha)^2}\frac {1}{(1-\hat \alpha)}{\rm Cov}(a,b)$$

or a bit more compactly

$${\rm Var} \left(\frac {b}{1-a}\right) \approx \frac {\hat \beta^2{\rm Var}(a)}{(1-\hat \alpha)^4}\;+\;\frac {{\rm Var}(b)}{(1-\hat \alpha)^2} \;+\; \frac {2\hat \beta{\rm Cov}(a,b)}{(1-\hat \alpha)^3}$$

The variances and the covariance in the right hand side are unknown. But you have an estimate of them: the "standard errors" -squared-, and the covariance from the estimated covariance matrix obtained from the model. You plug these estimates into the last expression, together with the coefficient estimates themselves, and then you take the square root of the whole to arrive at an estimate for the magnitude you are interested in. Note the more than one source of approximation error here.

$\endgroup$

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.