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Consistency is obviously a natural and important property estimators, but are there situations where it may be better to use an inconsistent estimator rather than a consistent one? More specifically, are there examples of an inconsistent estimator which outperforms a reasonable consistent estimator for all finite $n$ (with respect to some suitable loss function)? |
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This answer describes a realistic problem where a natural consistent estimator is dominated (outperformed for all possible parameter values for all sample sizes) by an inconsistent estimator. It is motivated by the idea that consistency is best suited for quadratic losses, so using a loss departing strongly from that (such as an asymmetric loss) should render consistency almost useless in evaluating the performance of estimators. Suppose your client wishes to estimate the mean of a variable (assumed to have a symmetric distribution) from an iid sample $(x_1, \ldots, x_n)$, but they are averse to either (a) underestimating it or (b) grossly overestimating it. To see how this might work out, let us adopt a simple loss function, understanding that in practice the loss might differ from this one quantitatively (but not qualitatively). Choose units of measurement so that $1$ is the largest tolerable overestimate and set the loss of an estimate $t$ when the true mean is $\mu$ to equal $0$ whenever $\mu \le t\le \mu+1$ and equal to $1$ otherwise. The calculations are particularly simple for a Normal family of distributions with mean $\mu$ and variance $\sigma^2 \gt 0$, for then the sample mean $\bar{x}=\frac{1}{n}\sum_i x_i$ has a Normal$(\mu, \sigma^2/n)$ distribution. The sample mean is a consistent estimator of $\mu$, as is well known (and obvious). Writing $\Phi$ for the standard normal CDF, the expected loss of the sample mean equals $1/2 + \Phi(-\sqrt{n}/\sigma)$: $1/2$ comes from the 50% chance that the sample mean will underestimate the true mean and $\Phi(-\sqrt{n}/\sigma)$ comes from the chance of overestimating the true mean by more than $1$.
The expected loss of $\bar{x}$ equals the blue area under this standard normal PDF. The red area gives the expected loss of the alternative estimator, below. They differ by replacing the solid blue area between $-\sqrt{n}/(2\sigma)$ and $0$ by the smaller solid red area between $\sqrt{n}/(2\sigma)$ and $\sqrt{n}/\sigma$. That difference grows as $n$ increases. An alternative estimator given by $\bar{x}+1/2$ has an expected loss of $2\Phi(-\sqrt{n}/(2\sigma))$. The symmetry and unimodality of normal distributions imply its expected loss is always better than that of the sample mean. (This makes the sample mean inadmissible for this loss.) Indeed, the expected loss of the sample mean has a lower limit of $1/2$ whereas that of the alternative converges to $0$ as $n$ grows. However, the alternative clearly is inconsistent: as $n$ grows, it converges in probability to $\mu+1/2 \ne \mu$.
Blue dots show loss for $\bar{x}$ and red dots show loss for $\bar{x}+1/2$ as a function of sample size $n$. |
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