Is there a general interpretation of the ratio of two standard errors, $\frac{se\left(\hat{\theta_1}\right)}{se\left(\hat{\theta_2}\right)}$

We want the standard error to be as small as possible so that that the sample estimate is close to the population estimate. Can someone elaborate on extending it to ratio of standard errors. If $se\left(\hat{\theta_1}\right) = C*{se\left(\hat{\theta_2}\right)}$. Does that mean $se\left(\hat{\theta_1}\right)$ captures $100C\%$ variability in $se\left(\hat{\theta_2}\right)$

Intuitively I want $C$ to be as small as possible so that $\hat{\theta_1}$ is much more closer to the population estimate than $\hat{\theta_2}$. When do I want $C$ to be as large as possible?

  • $\begingroup$ I have never encountered any application where a "high standard error ... is desired." What exactly do you mean by "capture maximum variability" and why is that a good thing? And how does one go about achieving a large standard error--what are you proposing to vary to accomplish this? $\endgroup$ – whuber Dec 19 '14 at 22:15
  • $\begingroup$ It would help the answers if you would provide a specific example where we try to maximize the standard error of an estimator, in an attempt to "capture maximum variability". $\endgroup$ – Alecos Papadopoulos Dec 19 '14 at 22:15
  • $\begingroup$ I edited the question $\endgroup$ – Kumar Dec 19 '14 at 22:58

The ratio you present is very close to a definition of relative efficiency.

The relative efficiency of two unbiased estimators $\hat \theta_1$ and $\hat \theta_2$ is the ratio of their variances $\frac{\mathrm{var}({\hat \theta_1})}{\mathrm{var}({\hat \theta_2})}$. Assuming the estimators are unbiased, the square of the ratio you present will be a sample approximation to the above. A good starting point for understanding the implications of relative efficiency is http://www.cc.gatech.edu/~lebanon/notes/efficiency.pdf and you can follow it up with some Googleing.

I can't think of an application where we use the standard error of one estimator to explain variability in another. In the traditionally setting standard error is just a function of sample size (and asymptotic variance)...as you increase sample size standard error approaches zero. It is also hard to think about maximizing $C$, (I can always increase the std. error of $\hat \theta_2$ by reducing the number of observations or perturbing the model you are using to estimate.)

If your curious about choosing the best estimator, assuming both estimators are unbiased, we always choose the one with smaller variance for obvious reasons.

There are circumstances where we prefer the estimator with higher standard error. But this will be because the estimate of lower std. error is believed to be bias (more generally inconsistent). There is a trade off between efficiency and consistency. It is not so much that we want higher standard error as it is that we expect higher standard error with the consistent\unbiased estimator. An example of this is omitted variable bias. If we have the true regression function

$$ y=\beta_0 + \beta_1x_1 + \beta_2x_2 + \varepsilon $$ where $x_1$ and $x_2$ are correlated and estimate two models

$$ y=\bar \beta_0 + \bar \beta_1x_1 + \bar \varepsilon $$

$$ y=\hat \beta_0 + \hat \beta_1x_1 + \hat \beta_2x_2 + \hat \varepsilon $$

$\hat \beta_1$ would likely have higher standard error than $\bar \beta_1$, but we still prefer $\hat \beta_1$ because $\bar \beta_1$ is a biased estimator of $\beta_1$.


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.