Let $X_1, X_2, \ldots, X_n$ be a random sample from a normal distribution with mean $\mu$ and variance $\sigma^2$. Consider the problem of estimating $P(X > 100)$.

One way to accomplish this is to calculate $n^{-1}\sum_{i=1}^n \mathbb{1}(X_i > 100) $. This "plug-in" estimator is consistent, and its bias and MSE are easy to compute.

A smaller group of my students came up with another way to approach the problem: calculate $$ 1 - \Phi\left(\frac{100 - \bar{x}}{s} \right). $$ This can be motivated by the fact that $P(X > 100) = 1 - \Phi[(100 - \mu)/\sigma].$ This estimator is also consistent, but its bias and MSE are more difficult to compute.

My question is this: does this kind of strategy have a name? I ask because we are still plugging things in, but this is not a so-called plug-in estimator.

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    $\begingroup$ Isn't this just the Gaussian MLE? $\endgroup$ – shadowtalker Oct 3 '18 at 3:40
  • $\begingroup$ @shadowtalker yeah it’d be the invariance principle if you were using the MLE for the variance, but we are not as long I define this sample variance as the one where you divide by $n-1$. $\endgroup$ – Taylor Oct 3 '18 at 3:45
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    $\begingroup$ I would call the second one the "plug-in" estimator, while the first one is the moment estimator. $\endgroup$ – Xi'an Oct 4 '18 at 20:35
  • $\begingroup$ It's Rao-Blackwellized...but that isn't very specific $\endgroup$ – Taylor Oct 11 '18 at 13:08
  • $\begingroup$ The second method is much more powerful because it uses the distributional assumption of normality. However $\frac{100-\bar x}{s}$ is not normally distributed due to the uncertainty in $s$. Look up the $t$ distribution with $n$ degrees of freedom. $\endgroup$ – dave fournier Jan 27 '19 at 21:20

Your second estimator is the "plug-in" estimator, based on the invariance property of MLE's it is the maximum-likelihood estimator (under the normal assumptions). The first estimator could be called a moments estimator, but could also be seen as non-parametric, as it is unbiased without need for normality assumption.

So you could try to find a better unbiased estimator using Rao-Blackwell theorem.

  • $\begingroup$ thanks (+1) I'm not sure if you saw the discussion above, but I believe $s^2 = \sum_i(x_i - \bar{x})^2/(n-1)$ is not the MLE for iid normal data. $\endgroup$ – Taylor Jan 27 '19 at 20:29
  • $\begingroup$ Yes, so replace with $n$ as divisor. Rao-Blackwellization at least can be done approximately by simulation, exact calculation I am not so sure ... $\endgroup$ – kjetil b halvorsen Jan 28 '19 at 1:54

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