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I want to know the asymptotic distribution of $\sqrt{n}(\hat\theta_n-\theta)$ to determine the efficiency of $\hat\theta_n$.

I know there is a theorem with lots of assumptions that immediately concludes that it is efficient and the distribution if $N(0,I(\theta)^{-1})$ but I'm wondering if there is another approach?


I have that $\hat\theta_n = X_{(1)}$ and I have computed $F_{\hat\theta_n}(x)$ and $f_{\hat\theta_n}(x)$ already.

$$f_\theta (x) = e^{\theta-x}1_{x\geq \theta}$$ $$F_{\hat\theta_n} (x) = \left[1-e^{n(\theta-x)} \right] 1_{x\geq \theta}$$ $$f_{\hat\theta_n} (x) = \left[ne^{n(\theta-x)} \right] 1_{x\geq \theta}$$

Is there a way to continue from here that uses the delta method or otherwise?


This is my attempt: We want the distribution of $Y_n = \sqrt{n}(\hat\theta_n-\theta)$. We can rewrite this as $\frac{Y_n}{\sqrt{n}}+\theta = \hat\theta_n$

$$f_{Y_n} (x) = f_{\hat\theta_n} \left(\frac{Y_n}{\sqrt{n}}+\theta\right)$$

$$= ne^{-\sqrt{n}Y_n}1_{Y_n \geq 0}$$

I don't know how to continue from here. I don't see how to arrive from this to a normal approximation so that I can figure out efficiency?! I also thought about using the Delta Method but it seems like it is not applicable in this case?!

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  • $\begingroup$ So, just to clarify- $\hat \theta_n = \min_i X_i$ is the MLE for a parameter $\theta$ in a model with density (proportional to) $e^{\theta-x}$ ($x>0$)? $\endgroup$ – P.Windridge Oct 20 '17 at 13:19
  • $\begingroup$ @P.Windridge Yes, precisely so $\endgroup$ – ChineseStatistician Oct 20 '17 at 13:21
  • $\begingroup$ Apologies, I misread $x > \theta$ in the density as as $x > 0$ on my phone. Anyway, check the conditions for the asymptotic normality of the mle- does it apply in your case? You have $\hat \theta_n > \theta$ so $ $Y_n > 0$ and thus cannot be asymptotically normal $\endgroup$ – P.Windridge Oct 20 '17 at 13:45
  • $\begingroup$ No problem. It should be $x\geq \theta$, hence why $\hat\theta_n = X_{(1)}$ $\endgroup$ – ChineseStatistician Oct 20 '17 at 13:48
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Part of the sufficient conditions for asymptotical normality of the MLE is that all models in the family have the same support. This fails in your example because it is $(\theta,\infty)$.

In particular this means that $\hat \theta_n = X_{(1)} > \theta$ and so $Y_n > 0$. Thus it cannot be asymptotically normal (other than being degenerate, in case you consider that "normal").

We can compute the distribution more explicitly. As you correctly noted we have $$ \mathbb{P}(Y_n < y) = \mathbb{P}(X_{(1)} < \theta + y/\sqrt{n}). $$ (Note you can't just put $\theta + y/\sqrt{n}$ straight into the density for $X_{(1)}$ - you need to differentiate the expression above and you'd find an extra $1/\sqrt{n}$).

This comes out as $$ \mathbb{P}(Y_n < y) = 1 - \exp(-y \sqrt{n}), $$ i.e. Exponential($\sqrt{n}$). This is not asymptotically normal. The variance is $1/n$. Assuming efficiency is defined by the ratio to the Cramer Rao lower bound, we must compute the Fisher Information $I(\theta)$ and examine the ratio $$ n/I(\theta) $$

The Fisher information is $$ I(\theta) = \mathbb{E}[l'(X;\theta)^2] \\ = \mathbb{E}[ \left((d/d\theta)(n\theta - \sum_i X_i)\right)^2] = n^2. $$

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  • $\begingroup$ So the conclusion is that it is not efficient. But according to my exercise, it still approaches some normal distribution and my question was how do you find that normal distribution. $\endgroup$ – ChineseStatistician Oct 20 '17 at 14:12
  • $\begingroup$ Please clarify what you mean by "approaches some normal distribution". $\endgroup$ – P.Windridge Oct 20 '17 at 14:34
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    $\begingroup$ The question literally says: "Determine the asymptotic distribution of $\sqrt{n}(\hat\theta_n-\theta)$". However I think I was confusing "normal" and "asymptotic". The asymptotic distribution is indeed just the exponential as you have show. Then it ask whether $\hat\theta_n$ is efficient. The only definition of "efficient" that I have says that it should go towards a normal distribution with the variance of that distribution being $1/I(\theta)$. So if the distribution is not normal, but the variance is still $1/I(\theta)$, would that still be an efficient estimator? $\endgroup$ – ChineseStatistician Oct 20 '17 at 14:53
  • $\begingroup$ Thanks, your question is clear now. Often I have seen "efficiency" defined in terms of the ratio to $1/I(\theta)$ (the best according to Cramer Rao), which is $1/n^2$ in your example. So the efficiency here is $1/n$, which is quite poor. $\endgroup$ – P.Windridge Oct 20 '17 at 15:01
  • $\begingroup$ Information function is not usually defined for distributions with parameter in their support. As such the Cramer-Rao bound does not apply. Large sample efficiency of $\hat\theta_n$ in general is defined as the ratio $\operatorname{Var}(T)/\operatorname{Var}(\hat\theta_n)$ where $T$ is UMVUE of $\theta$ (assuming it exists). Here UMVUE does exist and it shows that the MLE $\hat\theta_n$ is indeed (asymptotically) efficient, as it is in most cases. $\endgroup$ – StubbornAtom Oct 23 at 14:49
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If in fact $F_{\hat\theta_n} (x) = \left[1-e^{n(\theta-x)} \right] 1_{x\geq \theta}$ then defining $Y_n = n(\hat{\theta}_n - \theta)$ shows that $$ F_{Y_n}(y) = P(\hat{\theta}_n \le \frac{y}{n} + \theta) = 1 - e^{-y}. $$ So $Y_n \sim \text{Exp}(1)$ exactly, not asymptotically, $\hat{\theta}$ converges faster to $\theta$ than root $n$, and you don't need to appeal to any results (just use CDFs). However, do note that I am defining $Y_n$ differently than you are.

Also, the Delta method is typically used when, say a different estimator $\hat{\theta}_n$ is asymptotically Normal. So you assume that $\sqrt{n}(\hat{\theta} - \theta)$ converges already. Then you can use it if you want the asymptotic distribution of some transformed estimator $f(\hat{\theta}_n)$, where $f$ is a function with certain requirements. Typically $\sqrt{n}(f(\hat{\theta}_n) - f(\theta))$ will also be asymptotically Normal with mean $0$ again but with a different variance depending on $f$. Even though it's very popular, I'm not sure it can be of much use here.

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