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Suppose we have sample $y_1,y_2,...y_n$ from a borel distribution

$$P(Y=y;\alpha)= \dfrac{1}{y!}(\alpha y)^{y-1}e^{-\alpha y} , y=1,2..$$

The MLE of $\alpha$ is $\hat{\alpha} = 1-\dfrac{1}{\bar{y}} $

By using that $\sqrt{n}\Big(\bar{y} - \dfrac{1}{1-\alpha} \Big) \rightarrow N \Big(0,\dfrac{\alpha}{(1-\alpha)^3} \Big)$

How can I find the asymptotic distribution for $\hat{\alpha}$ (MLE)?

added extra text

My thought is that I can make use of the following: Set $\bar{y}(\alpha)= \dfrac{1}{1-\alpha}$. Then we have

$$\sqrt{n}\Big( \bar{y}({\hat{\alpha}}) - \bar{y}(\alpha)\Big) \rightarrow N(0,I(\bar{y})^{-1}). $$

$I(\alpha) = (\dfrac{d \bar{y} }{d\alpha})^2 \cdot I(\bar{y}(\alpha)) = \dfrac{1}{(1-\alpha)^4} \cdot \dfrac{(1-\alpha)^3}{\alpha} = \dfrac{1}{(1-\alpha)\alpha} $ (by reparametrization lemma).

Finally we obtain:

$$\sqrt{n}\Big( \hat{\alpha} - \alpha\Big) \rightarrow N(0,I(\alpha)^{-1}) = N(0,(1-\alpha)\alpha) $$

but that is exactly the distribution you got :) @Ben

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    $\begingroup$ I think you mean: $$\sqrt{n}\left(\overline Y - \dfrac{1}{1-\alpha}\right) \stackrel{d}{\to} N\left(0, \dfrac{\alpha}{(1-\alpha)^3}\right).$$ Notice the usage of capital letter $\overline Y$, since this is a random variable. (Otherwise the 'asymptotic' has no meaning). Also the factor $\sqrt{n}$ is needed. To find the asympotic distribution you must then use the delta method $\endgroup$
    – dietervdf
    Aug 8, 2018 at 1:06

1 Answer 1

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From the form of your estimator, its distribution function can be written as:

$$F_\hat{\alpha}(a) = \mathbb{P}(\hat{\alpha} \leqslant a) = \mathbb{P} \Big( 1 - \frac{1}{\bar{Y}} \leqslant a \Big) = \mathbb{P} \Big( \bar{Y} \leqslant \frac{1}{1 - a} \Big) = F_\bar{Y}\Big( \frac{1}{1 - a} \Big).$$

Let $\Phi$ denote the standard normal distribution function and $\phi$ denote the standard normal density function. Using the asserted limiting result (corrected in the comment by dietervdf) you then have the asymptotic result:

$$\begin{equation} \begin{aligned} F_\hat{\alpha}(a) &= \mathbb{P} \Big( \bar{Y} \leqslant \frac{1}{1 - a} \Big) \\[6pt] &= \mathbb{P} \Big( \bar{Y} - \frac{1}{1 - \alpha} \leqslant \frac{1}{1 - a} - \frac{1}{1 - \alpha} \Big) \\[6pt] &\approx \Phi \Bigg( \sqrt{n} \cdot \frac{\tfrac{1}{1 - a} - \tfrac{1}{1 - \alpha}}{\sqrt{\tfrac{\alpha}{(1-\alpha)^3}}} \Bigg) \\[6pt] &= \Phi \Bigg( \sqrt{n \cdot \frac{1-\alpha}{\alpha}} \cdot \Big( \frac{1-\alpha}{1 - a} - 1 \Big) \Bigg). \\[6pt] \end{aligned} \end{equation}$$

Differentiating and applying the chain rule yields the asymptotic density function:

$$\begin{equation} \begin{aligned} f_\hat{\alpha}(a) &= \frac{dF_\hat{\alpha}}{da} (a)\\[6pt] &\approx \sqrt{n \cdot \frac{1-\alpha}{\alpha}} \cdot \frac{1-\alpha}{(1 - a)^2} \cdot \phi \Bigg( \sqrt{n \cdot \frac{1-\alpha}{\alpha}} \cdot \Big( \frac{1-\alpha}{1 - a} - 1 \Big) \Bigg) \\[6pt] &= \sqrt{\frac{n}{2 \pi} \cdot \frac{1-\alpha}{\alpha}} \cdot \frac{1-\alpha}{(1 - a)^2} \cdot \exp \Bigg( - \frac{n}{2} \cdot \frac{1-\alpha}{\alpha} \cdot \Big( \frac{1-\alpha}{1 - a} - 1 \Big)^2 \Bigg). \\[6pt] \end{aligned} \end{equation}$$

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