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Originally, I was attempting to find a Method of Moments estimator for $\lambda$ given $X_i \sim Geom(e^{-\lambda})$ s.t. $f_\lambda(x) = e^{-\lambda}(1 - e^{-\lambda})^x$.

I found it to be $\hat{\lambda} = \log\left(1 + \frac{\sum_{i=1}^n X_i}{n}\right)$. I am now attempting to verify if it is unbiased ($E\left[\hat{\lambda}\right] \stackrel{?}{=} \lambda$). This leads to a super nasty summation:

$$ E\left[\log\left(1 + \frac{\sum_{i=1}^n X_i}{n}\right)\right] = \sum_{k=0}^\infty \log\left(1 + \frac{k}{n}\right)\cdot \underbrace{\pmatrix{k + n - 1 \\ k} \cdot (1 - e^{-\lambda})^k e^{-n\lambda}}_\text{Negative binomial pmf} $$

Note that $\sum X_i \sim NB(n, e^{-\lambda})$.

My thoughts: I am convinced that the sum is either divergent or not at all equal to $\lambda$.

  • Are there any tools I could use to indirectly state this?
  • Is there a feasible pen/paper way to calculate the exact sum or should this be left for something like Mathematica?
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  • Aren't you missing some factors of $k$ and $n$ in your sum, like $(1-e^{-\lambda})^k e^{-n \lambda}$ instead of plain $(1-e^{-\lambda}) e^{- \lambda}$? These extra factors will make the sum better behaved. – user196574 Sep 06 '23 at 06:36
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    Can you also state if the geometric random variable has support starting at $0$ or $1$? – user196574 Sep 06 '23 at 06:44
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    You can use Jensen's inequality to check unbiasedness. – StubbornAtom Sep 06 '23 at 07:36
  • @user196574 yeah, I was fighting MathJax typing it out. Support starts at 0. – 392781 Sep 06 '23 at 09:10
  • @StubbornAtom I will try that out tomorrow and report back. – 392781 Sep 06 '23 at 09:10
  • Without doing any calculations, I would expect the worst case to be when $n=1$ giving a finite expectation, and (given the non-linear function) for this to be biased – Henry Sep 06 '23 at 09:22

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Thank you to @StubbornAtom for suggesting Jensen's Inequality. We know that $\log$ is concave thus it follows that $E[g(X)] \leq g(E[X])$:

$$ \begin{aligned} E\left[\log\left(1 + \dfrac{\sum_{i=1}^n X_i}{n}\right)\right] &\leq \log\left(E\left[1 + \dfrac{\sum_{i=1}^n X_i}{n}\right]\right) \quad \text{(Jensen's Inequality)} \\ &=\log\left(1 + \dfrac{E\left[\sum_{i=1}^n X_i\right]}{n}\right) \quad \left(\text{Note: } \sum X_i \sim NB(n, e^{-\lambda} \right) \\ &=\log\left(1 + \dfrac{n\left(1 - e^{-\lambda}\right)/e^{-\lambda}}{n}\right) \\ &= \log\left(1 + \dfrac{1}{e^{-\lambda}} - 1\right) \\ &= \log\left(e^{\lambda}\right) = \lambda \end{aligned} $$

This then implies that the original method of moments estimate $\hat{\lambda}$ is greater than, if not equal to, the RHS of the inequality. For the equality to be true, our transformation must be affine which I do not believe is true in this case (someone can correct me on that).

Thus, $\hat{\lambda}$ is a biased estimator of $\lambda$.

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