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A random sample of size $n$ is being drawn from a population with pdf as:

$$f(x) = \begin{cases} (a + 1)x^a & \text{for }0<x<1, \\ 0 & \text{otherwise.} \end{cases}$$

Can we express the method of moments estimator of $1/a$ for this distribution in the following form?

$(1-H)/H$ ; where $H$ is the harmonic mean.

My attempt:

By equating the population mean to sample mean, I got estimator for $1/a$ as

$$\frac 1 a = \frac{1 - \overline X}{2\overline X - 1}$$

I am not able to express this in the form $(1-H)/H$; where $H$ is the harmonic mean.

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    $\begingroup$ How did you arrive at $(1-H)/H$ as your method of moment estimator? Also read stats.stackexchange.com/tags/self-study/info $\endgroup$ Oct 31, 2021 at 10:51
  • $\begingroup$ I have edited the question. I've added the answer I got for method of moments estimator. I just want to know, can we somehow express it in the above mentioned form of (1-H)/H ? $\endgroup$
    – Kcd
    Oct 31, 2021 at 14:21

2 Answers 2

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A simple counterexample, e.g. $x_1=0.1, x_2=0.7$ shows that in general $$\frac{1-H}{H} \neq \frac{1-\overline{X}}{2\overline{X}-1}$$ If this is a homework question, the teacher presumably meant that you shall equate the empirical estimator of the harmonic mean with the theoretical "true" value of the harmonic mean, i.e. $1/E(1/X)$.

Edit: As this answer is downvoted, there must be something wrong. I would be helpful to discuss this issue in the comments so that this can be corrected.

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  • $\begingroup$ Thank you. I understood that equating emperical harmonic mean to theoretical harmonic mean will give the required result. But, I just had to ask, whether it is correct to equate empirical harmonic mean to theoretical one? Because, method of moments estimation is about equating sample and population moments. $\endgroup$
    – Kcd
    Oct 31, 2021 at 15:31
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    $\begingroup$ The method of moments is open to deliberate choices of the moments to be equated with their estimators. In many cases, the choice is based upon which observable is easily to be measured. I am not sure, however, whether the harmonic mean is a good candidate. Is its empirical estimator even unbiased? $\endgroup$
    – cdalitz
    Oct 31, 2021 at 15:44
  • $\begingroup$ I am not sure about the unbiased part. But, the doubt I had is clarified, thank you. $\endgroup$
    – Kcd
    Oct 31, 2021 at 16:12
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Use the expression for the harmonic mean instead of the expression for the arithmetic mean. https://en.m.wikipedia.org/wiki/Beta_distribution#Harmonic_mean

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