Mean squared error of 1/sample mean

Question

Let $n$ be the estimator defined by: $$ n = \frac{1}{\text{sample mean}} $$

What is the mean squared error of the n estimator?

I assume there is no bias since the sample mean is an unbiased estimator.

Is this correct? $$ MSE(n) = \operatorname{Var}(n) = \sum_{i=1}^{n} \Bigl(\frac{1}{x_{i}} - \frac{1}{x_\text{sample mean}}\Bigr)^{2} $$

Before you can define bias and MSE you must first clarify, what $\frac{1}{samplemean}$ is an estimator of. For instance it could be an estimator of $\frac{1}{\mu}$ in which case it is probably biased since usually $\frac{1}{E[X]} \neq E[\frac{1}{X}]$. — Leander Tilsted Kristensen, Jun 26 '20 at 16:24

score 1 · Answer 1 · answered Jun 26 '20 at 16:28

No.

if $\overline{X}_n$ is unbiased, say $\mathbb{E}[\overline{X}_n]=g(\theta)$ it is wellknown that $\frac{1}{\overline{X}_n}$ is biased (this is Jensen's inequality).
To calculate mean, variance, and MSE of $\frac{1}{\overline{X}_n}=\frac{n}{\sum_i X_i}$ you have to know the distribution on $\sum_i X_i$
Sometimes mean and variance $\frac{1}{\overline{X}_n}=\frac{n}{\sum_i X_i}$ can be derived with some brainstomings on the score of the model (for example using Bartlett's identities)

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