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Let $X_{i}, i=1,\ldots, n$ denote a random simple of size $n$ from a population with mean $3$. Assume that $\hat{\theta}_{2}$ is an unbiased estimator of $\mathbb{E}[X^{2}]$ and that $\hat{\theta}_{3}$ is an unbiased estimator for $\mathbb{E}[X^3]$. Give an unbiased estimator for the third central moment of the underlying distribution.

My approach:

I know that $$\mu_{3}=\mathbb{E}[(X-3)^{3}]=\mathbb{E}[X^{3}]-9\mathbb{E}[X^{2}]+54$$ so, we can see that $\hat{\mu_{3}}=\hat{\theta}_{3}-9\hat{\theta}_{2}+54$ is an unbiased estimator for $\mu_{3}$ of the underlying distribution.

Is it correct?

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Yes it is unbiased and so correct. To check:

$$\mathbb E[\hat{\mu_{3}}]\\=\mathbb E[\hat{\theta}_{3}]-9\mathbb E[\hat{\theta}_{2}]+54 \\ = \mathbb E[X^3]-3\cdot3\cdot\mathbb E[X^2]+3 \cdot3^2\cdot\mathbb E[X] - 3^3 \\ = \mathbb E[(X-3)^3]$$

It is not the only unbiased estimator.

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  • $\begingroup$ Thank you so much for your verification. $\endgroup$
    – mathproof
    Nov 19 '20 at 11:17

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