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Suppose $X \sim \mathcal{N}(\mu_x, \sigma^2_x)$ and $Y \sim \mathcal{N}(\mu_y, \sigma^2_y)$

I am interested in $z = \min(\mu_x, \mu_y)$. Is there an unbiased estimator for $z$?

The simple estimator of $\min(\bar{x}, \bar{y})$ where $\bar{x}$ and $\bar{y}$ are sample means of $X$ and $Y$, for example, is biased (though consistent). It tends to undershoot $z$.

I can't think of an unbiased estimator for $z$. Does one exist?

Thanks for any help.

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This is just a couple of comments not an answer (don't have enough rep. point).

(1). There is an explicit formula for the bias of the simple estimator $min(\bar{x},\bar{y})$ here:

Clark, C. E. 1961, Mar-Apr. The greatest of a finite set of random variables. Operations Research 9 (2): 145–162.

Not sure how this helps though

(2). This is just intuition, but I think such an estimator doesn't exist. If there is such an estimator, it should also be unbiased when $\mu_x=\mu_y=\mu$. Thus any 'downgrading' which makes the estimator less than say the weighted average of the two sample means make the estimator biased for this case.

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    $\begingroup$ conceivably, any correction could end up having mean zero for this case. $\endgroup$ – cardinal Mar 8 '11 at 1:43
  • $\begingroup$ Just to clarify, though, I'm not claiming I believe there is an unbiased estimator. In fact, I agree there likely isn't. $\endgroup$ – cardinal Mar 8 '11 at 5:41
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    $\begingroup$ Yes agree - this is just intuition. The following paper gives conditions for the existence of unbiased estimator for a function of a univariate gaussian mean - maybe can be extended to multivariate: stat.ncsu.edu/library/mimeo.archive/ISMS_1988_1929.pdf $\endgroup$ – Or Zuk Mar 8 '11 at 16:17
  • $\begingroup$ Knowing the bias can help, you can correct for it to obtain an unbiased estimator. I actually went down this route, but computing the exact bias requires that you have $u_x$ and $u_y$ -- which we don't. So naturally I tried to use the sample mean instead to see what happens. It doesn't appear to help. In simulations, the corrected estimator also exhibits bias. I'm leaning towards an unbiased estimator not existing, but I haven't come up with a good proof for it. $\endgroup$ – pazam Mar 9 '11 at 16:52
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You are right that an unbiased estimator doesn't exist. The problem is that the parameter of interest is not a smooth function of the underlying data distribution due to non-differentiability at $\mu_x=\mu_y$.

The proof is as follows. Let $T(X,Y)$ be an unbiased estimator. Then $E_{\mu_x,\mu_y}[T(X,Y)]=\min\{\mu_x,\mu_y\}$. The left-hand side is differentiable everywhere with respect to $\mu_x$ and $\mu_y$ (differentiate under the integral sign). However, the right-hand side is not differentiable at $\mu_x=\mu_y$, which leads to a contradiction.

Hirano and Porter have a general proof in a forthcoming Econometrica paper (see their Proposition 1). Here is the working paper version:

http://www.u.arizona.edu/~hirano/papers/hp4_2011_11_03.pdf

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  • $\begingroup$ Very nice! Thank you for following up on this question. $\endgroup$ – whuber Jan 29 '12 at 23:19
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There is an estimator for the minimum (or the maximum) of a set of numbers given a sample. See Laurens de Haan, "Estimation of the minimum of a function using order statistics," JASM, 76(374), June 1981, 467-469.

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  • $\begingroup$ Unfortunately I don't think the paper you cite addresses this problem. The paper deals with when you have a set of non stochastic variables A, and finding the smallest element in A through sampling. In the context of this problem, each element in A would be a random variable, and therein lies the kicker. You have to find an unbiased estimator of the mean of the smallest random variable in A. $\endgroup$ – pazam Mar 16 '11 at 18:35
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I'd be fairly sure an unbiased estimator does not exist. But unbiased estimators don't exist for most quantities, and unbiasedness is not a particularly desirable property in the first place. Why do you want one here?

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  • $\begingroup$ The samples are expensive to obtain so I can't just up the sample size until the bias goes away. Unbiasedness is desired because I'm using the result of the estimator as the $Y$s in linear regression. Having bias means that $Y$ will contain non-normal disturbance which is equivalent to a specification error and that leads to a mess. I won't be able to accurately estimate the slope, the variance, construct confidence intervals, etc.. $\endgroup$ – pazam Mar 9 '11 at 17:00

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