# Show that $nX_{(1)}$ is not consistent

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.

• Isn't it $X_{(1)}$ rather than $nX_{(1)}$ which is $\exp(n\theta)$? – Jarle Tufto Jun 18 at 13:44
• Yes. Thank you. – Harry Jun 18 at 13:56
• $nX_{(1)}$ is exactly an exponential variable with mean $1/\theta$. If you look at the probability $P\left(\left|nX_{(1)}-\frac{1}{\theta}\right|<\varepsilon\right)$ for some positive $\varepsilon$, it does not even depend on $n$ for you to take the limit (it does not equal $1$ either). – StubbornAtom Jun 18 at 14:06
• Oh I realised my mistake. Thank you. – Harry Jun 18 at 14:18

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