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

  • 1
    $\begingroup$ 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


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:

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.