I've seen more general questions of this nature, but none that discuss specifically the problem setup that I'm interested in, which I believe is substantially simpler.

Suppose that we observe two random variables, $$X \sim N(\mu_{x}, \sigma_{x}^{2})$$ and $$Y \sim N(\mu_{y}, \sigma_{y}^{2}).$$ Now, I'm interested in estimating $$|\mu_{x} - \mu_{y}|.$$

Clearly, $|X - Y|$ is a biased estimate. This bias increases as the true value of $|\mu_{x} - \mu_{y}|$ approaches $0$ and as $\sigma_{y}^{2}$ and $\sigma_{x}^{2}$ increase. Now, since $|X - Y|$ follows a folded normal distribution, one idea I had to decrease the bias of this estimate was to estimate $|\mu_{x} - \mu_{y}|$ as

$$\max \bigg(0, |X - Y| - \sqrt{\sigma_{x}^{2} + \sigma_{y}^{2}} \cdot \sqrt{\frac{2}{\pi}}\bigg),$$

since $\sqrt{\sigma_{x}^{2} + \sigma_{y}^{2}} \cdot \sqrt{\frac{2}{\pi}}$ is the bias of $|X - Y|$ when $\mu_{x} = \mu_{y}$. However, this is still not an unbiased estimator.

So, I'm guessing the answer is no, but is it possible to derive an unbiased estimator to $|\mu_{x} - \mu_{y}|$? What does the MLE look like? Is there any other theory around how one should estimate a quantity like this?

  • $\begingroup$ Why do you need an unbiased estimator, exactly? (rather than, say, one which is generally closer to the thing you're estimating in some typical sense, like minimum mean square error, perhaps) $\endgroup$
    – Glen_b
    Commented Feb 3, 2023 at 0:08
  • $\begingroup$ Sure, I would be interested in a minimum MSE estimator too. $\endgroup$
    – Eric Weine
    Commented Feb 3, 2023 at 6:00
  • $\begingroup$ I was seeking explanation of the reasoning behind the choice (in the hope of gaining some clues that might help lead to a good answer), not seeking to encourage you to make the question broader. $\endgroup$
    – Glen_b
    Commented Feb 3, 2023 at 10:03
  • $\begingroup$ Got it. Well, I’m trying to create a “decision rule” for when one should pool the mean estimates between two groups vs. when one should estimate them separately. Given all parameters are known, this leads to a simple inequality which depends on the true difference in means and the variances. However, in practice one doesn’t know the true difference in means, so I’m interested in estimating that quantity. $\endgroup$
    – Eric Weine
    Commented Feb 3, 2023 at 14:53
  • 1
    $\begingroup$ I encourage you to explain that aim up in your question. $\endgroup$
    – Glen_b
    Commented Feb 3, 2023 at 17:35


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