# Trying to prove consistency, but getting non-sensical limit probabilities

I have

$$X_1 \dots X_n \sim f_\theta(x) = \begin{cases} \exp(\theta-x) & x\geq\theta\\ 0& otherwise \end{cases}$$

And I have the estimator $\hat\theta_n = X_{(1)}=\min\{X_1 \dots X_n\}$

I have found the CDF and PDF of the estimator

$$F_{\hat\theta_n}(x) = 1-e^{n(\theta-x)}$$ $$f_{\hat\theta_n}(x) = ne^{n(\theta-x)}$$

Now I want to test consistency so for $\epsilon >0$

$$\lim_{n\to\infty}\Pr(|\hat\theta_n - \theta| < \epsilon) = 1$$

Then we have $$\Pr(-\epsilon < \hat\theta_n - \theta < \epsilon)$$ $$\Pr(\theta-\epsilon < \hat\theta_n < \theta + \epsilon)$$

$= 1-e^{-n\epsilon}-1+e^{n\epsilon}$

Which goes to $+\infty$ as $n\to\infty$. Did I do everything correctly?

What do I conclude from this? Is the estimator consistent or not? Is this the same as the probability going to $1$?

EDIT: I believe I have found my error. The CDF should be

$$F_{\hat\theta_n}(x) = 1-e^{n(\theta-x)} 1_{x \geq \theta}$$

Then you get the probability: $1-e^{-n\epsilon}$ which goes to $1$ as expected.

• The fact is, the probability that $\hat{\theta}_n$ is less than $\theta$ is $0$. Commented Oct 20, 2017 at 2:40

## 2 Answers

It may be more convenient to work with $P(|\hat{\theta}_n - \theta| \geq \varepsilon)$ (the reason will be clear as the computation flows): \begin{align*} & P(|\hat{\theta}_n - \theta| \geq \varepsilon) \\ = & P(|X_{(1)} - \theta| \geq \varepsilon) = P(X_{(1)} - \theta \geq \varepsilon) \\ = & P(X_{(1)} \geq \theta + \varepsilon) \\ = & P(X_1 \geq \theta + \varepsilon, \ldots, X_n \geq \theta + \varepsilon) \\ = & P(X_1 \geq \theta + \varepsilon)^n \quad \text{by the i.i.d. assumption} \\ = & \left(\int_{\theta + \varepsilon}^\infty e^{\theta - x} dx \right)^n \\ = & e^{-n\varepsilon} \to 0 \end{align*} as $n \to \infty$. This shows that $\hat{\theta}_n$ is a consistent estimate of $\theta$.

I think you have to write $\lim_{n\to\infty}\Pr(|\hat\theta_n - \theta| < \epsilon) =\lim_{n\to\infty}\Pr(\theta < \hat\theta_n < \theta + \epsilon)$ since $\theta$ is the minimum value of $\hat{\theta}$.

Then you can show that \begin{align*} \lim_{n\to\infty}\Pr(\theta < \hat\theta_n < \theta + \epsilon) &= lim_{n\to\infty}\left \{F_{\hat{\theta}}(\theta+\epsilon)-F_{\hat{\theta}}(\theta)\right \} \\&=lim_{n\to\infty}\left \{[1-e^{n[\theta-(\theta+\epsilon)]}]-[1-e^{n(\theta-\theta)}]\right \}\\&=lim_{n\to\infty}\left \{1-e^{-n\epsilon}-1+1\right\}\\&=1 \end{align*}
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