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Consider the family of probability density functions indexed by parameter $\theta \geq 1$ and given by:

$$f(x, \theta) = \frac{x}{\theta} e^{\theta - 1 - x}, x \geq \theta -1$$

For a random sample of size n, and justifying all steps:

Derive the method of moments estimator for $\theta$

$$E[X] = \int_{\theta -1}^\infty \frac{x^2}{\theta}e^{\theta - 1 - x}dx = ... = \frac{\theta^2 + 1}{\theta} \text{ (using integration by parts twice or WolframAlpha) and set this } = \overline{X}$$

$$\hat \theta_{MM} + \frac{1}{\hat \theta_{MM}} = \overline{X}$$

Is there a way to just isolate the estimator or is it okay to leave it like this?

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    did you try using any higher moment? – 2017-01-09
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    Using only the first moment, one cannot distinguish between the parameters $\theta$ and $1/\theta$, except if one of them violates the condition that $\theta$i$. Thus, in general, $$\hat\theta=\tfrac12\left(\bar x\pm\sqrt{\bar x^2-4}\right)$$ and, if $$x_{\mathrm{min}}+1<\tfrac12\left(\bar x+\sqrt{\bar x^2-4}\right)$$ then $$\hat\theta=\tfrac12\left(\bar x-\sqrt{\bar x^2-4}\right)$$ – 2017-01-09
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    @Did I don't understand how you got this? – 2017-01-09
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    Where did the whole $+- \sqrt {\overline{x}^2 - 4}$ thing come from? – 2017-01-09
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    From the solutions of the quadratic equation $\hat\theta^2-\bar x\hat\theta+1=0$. – 2017-01-09
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    @Did Oh it makes much more sense now thank you so much – 2017-01-09

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