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So we've got a sample data coming from exponential distribution with parameter $\lambda$, and we take an estimator $\lambda_n = \frac{n}{X_1+X_2+\cdots+X_n}$.

I need to show that this is a biased and consistent estimator. Normally what you'd do is say that the sample mean follows gamma distribution and find the expectation and variance, but we've not covered that topic yet, so using gamma distribution is not allowed. How else could we compute the expectation of the estimator?

$\frac{n}{E(X_1+X_2+\cdots+X_n)}$ gets me stuck as it's in the denominator

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  • $\begingroup$ Just FYI, the expectation is $\mathbb{E} (n /( X_1 + X_2 + \cdots + X_n))$, not $n / \mathbb{E}( X_1 + X_2 + \cdots + X_n)$. $\endgroup$
    – jbowman
    Commented Oct 27, 2021 at 0:26

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A special case of Jensen's inequality is that for a positive-valued random variable $\overline X,$ you have $$ \frac 1 {\operatorname E\left(\overline X\right)} \le \operatorname E\left( \frac {\,\,1\,\,} {\overline X}\right). $$ The inequality is strict if $\operatorname{var}\left( \overline X \right)>0.$

If $\overline X = (X_1+\cdots+X_n)/n,$ you therefore have $\operatorname E\left( 1\left/\overline X\right.\right) > 1 / \operatorname E\left( \overline X\right) = \lambda.$

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  • $\begingroup$ ok I see. how would I show the consistency then? By plugging X squared to the inequality and calculating the variance? $\endgroup$
    – narek
    Commented Oct 27, 2021 at 4:51
  • $\begingroup$ @narek : I think first I'd try finding an upper bound on the variance (as opposed to computing the variance exactly) and then applying Chebyshev's inequality. $\endgroup$
    – Michael Hardy
    Commented Oct 27, 2021 at 5:24

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