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What relation are between minimum contrast estimate and minimum distance estimate?

If I understand correctly, these two are different methods? or are they equivalent?

Thanks and regards!


Minimum distance estimate from Wikipedia

Let $\displaystyle X_1,\ldots,X_n$ be an independent and identically distributed (iid) random sample from a population with distribution $F(x;\theta)\colon \theta\in\Theta$ and $\Theta\subseteq\mathbb{R}^k (k\geq 1)$.

Let $\displaystyle F_n(x)$ be the empirical distribution function based on the sample.

Let $\hat{\theta}$ be an estimator for $\displaystyle \theta$. Then $F(x;\hat{\theta})$ is an estimator for $\displaystyle F(x;\theta)$.

Let $d[\cdot,\cdot]$ be a functional returning some measure of "distance" between the two arguments. The functional $\displaystyle d$ is also called the criterion function.

If there exists a $\hat{\theta}\in\Theta$ such that $d[F(x;\hat{\theta}),F_n(x)]=\inf\{d[F(x;\theta),F_n(x)]; \theta\in\Theta\},$ then $\hat{\theta}$ is called the minimum distance estimate of $\displaystyle \theta$.


Minimum contrast estimate from Bickel and Doksum's Mathematical Statistics Vol1

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The minimum distance estimator is equivalent to the minimum contrast estimator, provided you pick $d$ and and $\rho$ such that $$ d[F(x; \theta), \, F_n(x)] = \rho(X, \theta). $$

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  • $\begingroup$ Thanks! So do you mean MDE belongs to MCE? Does MCE belong to MDE, so that they are equivalent to each other? $\endgroup$ – Tim Apr 26 '13 at 17:52
  • $\begingroup$ I guess, technically, it depends on what kinds of functions d and $\rho$ are allowed $\endgroup$ – Stefan Wager Apr 26 '13 at 23:13
  • $\begingroup$ Could you be more specific? $\endgroup$ – Tim Apr 26 '13 at 23:23
  • $\begingroup$ So, in practice, MCE and MDE are the same thing. From the definitions you gave, it is possible to construct pathological examples where this is not true (e.g. what if two parameters theta lead to the same distribution). But I don't think there's much point in worrying about those pathologies. $\endgroup$ – Stefan Wager Apr 27 '13 at 1:25

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