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Given a sample $X_1, \ldots, X_n$ that is geomterically distributed, i.e. $X_i$ ~ $G_{\theta}$, how does one have to choose the constants $a$ and $b$, such that the following estimating function $T$ is unbiased for $\tau = \frac{1}{\theta}$:

$$ T = b \sum_{i=1}^{n-1} X_i(aX_{i+1}-X_i) $$

Furthermore, is $T$ consistent for $\theta$?

To Show that $T$ is unbiased, I'd have to Show that $E[T] = \frac{1}{\theta}$ right?

Therefore I'd get:

$$ E[T] = b \sum_{i=1}^{n-1} E[X_i(aX_{i+1}-X_i)]$$

Is it allowed to assume Independence here or is there another way to determine $a$ and $b$?

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    Samples are typically assumed to be independent (except when they are without replacement from a finite population) – 2017-01-12
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    Therefore I'd get $E[T] = b \sum \frac{1}{\theta}(\frac{a}{\theta}-\frac{1}{\theta})$. If I take for example $a=\theta +1$ and $b=1/n$, my expectation would be equal to $\frac{1}{\theta}$. But is this correct? Furthermore, what about the consistency? – 2017-01-15
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    $X_i$ is probably intended to be independent of $X_{i+1}$ but not of $X_i$. And you are taking the sum over $n-1$ terms not $n$ – 2017-01-15
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    So I have $E[T] = b \sum (a E[X_i]E[X_{i+1}] - E[X_i^2]) = b \sum(\frac{a}{\theta^2}-\frac{2-\theta}{\theta^2}) = b(n-1)\frac{a-2+\theta}{\theta^2}$. How do I proceed then...? – 2017-01-16
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    That rather depends on whether your geometric distribution can take the value $0$ or not. You seem to assume not. In which case the obvious values for unbiasedness seem to be $a=2$ and $b=\frac1{n-1}$ to give $\frac1{\theta}$ – 2017-01-16

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