Consider a random sample from exponential distribution with mean $\frac{1}{\theta}$. I have to prove that $nX_{(1)}$ is not consistent for $\frac{1}{\theta}$ . A sufficient condition for consistency is

$$\lim_{n \to \infty}E(\hat\theta)=\frac{1}{\theta}\quad ;\quad \lim_{n \to \infty}\operatorname{Var}(\hat\theta)=0$$

Since $X_{(1)}$ follow $\exp(n\theta)$, the second condition, was seen to be violated as its variance $\frac{1}{\theta^{2}}$ doesn't tend to zero. Is this enough to prove that $nX_{(1)}$ is inconsistent?

Can i say that since $X_{(1)}$ converge in probability to $\frac{1}{n\theta}$, $nX_{(1)}$ converge in probability to $\frac{1}{\theta}$. But this does not prove what i want.

Im confused about the right method to be used.


1 Answer 1


The sufficient condition you use is not strong enough to conclude anything. Proceed from definition.

To restate convergence in probability somewhat roughly, $T_n$ is said to be a consistent estimator of the parametric function $g(\theta)$ if for every positive $\varepsilon$, $P_{\theta}\left[|T_n-g(\theta)|<\varepsilon\right]\to 1$ as $n\to\infty$ for all $\theta$ (or equivalently $P_{\theta}\left[|T_n-g(\theta)|>\varepsilon\right]\to 0$).

Since $nX_{(1)}$ is an exponential variable with mean $1/\theta$, you will find that the probability $P_{\theta}\left[\left|nX_{(1)}-\frac{1}{\theta}\right|<\varepsilon\right]$ does not even depend on $n$. It does not converge to $1$. Hence proved.


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